Schematic for 18650 battery charging circuit

Terminal
An electrical connector acting as reusable interface to a conductor and creating a point where external circuits can be connected.
natarius
Ground
A common return path for electric current. Commonly known as ground.
jharwinbarrozo
20.5M
Net Portal
A net identifier used to create connectivity to other Net Portal with the same designator. #portal
jharwinbarrozo
43.0M
Power Net Portal
Power net portal is a special schematic object used to define a power or ground net. Power ports allow you to conveniently indicate a power net at any location in the design. Power nets of the same designator automatically connect throughout your design #portal #power
jharwinbarrozo
11.4M
Generic Resistor
A generic fixed resistor for rapid developing circuit topology. Save precious design time by seamlessly add more information to this part (value, footprint, etc.) as it becomes available. Standard resistor values: 1.0Ω 10Ω 100Ω 1.0kΩ 10kΩ 100kΩ 1.0MΩ 1.1Ω 11Ω 110Ω 1.1kΩ 11kΩ 110kΩ 1.1MΩ 1.2Ω 12Ω 120Ω 1.2kΩ 12kΩ 120kΩ 1.2MΩ 1.3Ω 13Ω 130Ω 1.3kΩ 13kΩ 130kΩ 1.3MΩ 1.5Ω 15Ω 150Ω 1.5kΩ 15kΩ 150kΩ 1.5MΩ 1.6Ω 16Ω 160Ω 1.6kΩ 16kΩ 160kΩ 1.6MΩ 1.8Ω 18Ω 180Ω 1.8KΩ 18kΩ 180kΩ 1.8MΩ 2.0Ω 20Ω 200Ω 2.0kΩ 20kΩ 200kΩ 2.0MΩ 2.2Ω 22Ω 220Ω 2.2kΩ 22kΩ 220kΩ 2.2MΩ 2.4Ω 24Ω 240Ω 2.4kΩ 24kΩ 240kΩ 2.4MΩ 2.7Ω 27Ω 270Ω 2.7kΩ 27kΩ 270kΩ 2.7MΩ 3.0Ω 30Ω 300Ω 3.0KΩ 30KΩ 300KΩ 3.0MΩ 3.3Ω 33Ω 330Ω 3.3kΩ 33kΩ 330kΩ 3.3MΩ 3.6Ω 36Ω 360Ω 3.6kΩ 36kΩ 360kΩ 3.6MΩ 3.9Ω 39Ω 390Ω 3.9kΩ 39kΩ 390kΩ 3.9MΩ 4.3Ω 43Ω 430Ω 4.3kΩ 43KΩ 430KΩ 4.3MΩ 4.7Ω 47Ω 470Ω 4.7kΩ 47kΩ 470kΩ 4.7MΩ 5.1Ω 51Ω 510Ω 5.1kΩ 51kΩ 510kΩ 5.1MΩ 5.6Ω 56Ω 560Ω 5.6kΩ 56kΩ 560kΩ 5.6MΩ 6.2Ω 62Ω 620Ω 6.2kΩ 62KΩ 620KΩ 6.2MΩ 6.8Ω 68Ω 680Ω 6.8kΩ 68kΩ 680kΩ 6.8MΩ 7.5Ω 75Ω 750Ω 7.5kΩ 75kΩ 750kΩ 7.5MΩ 8.2Ω 82Ω 820Ω 8.2kΩ 82kΩ 820kΩ 8.2MΩ 9.1Ω 91Ω 910Ω 9.1kΩ 91kΩ 910kΩ 9.1MΩ #generics #CommonPartsLibrary
jharwinbarrozo
1.5M
Generic Capacitor
A generic fixed capacitor for rapid developing circuit topology. Save precious design time by seamlessly add more information to this part (value, footprint, etc.) as it becomes available. Standard capacitor values: 1.0pF 10pF 100pF 1000pF 0.01uF 0.1uF 1.0uF 10uF 100uF 1000uF 10,000uF 1.1pF 11pF 110pF 1100pF 1.2pF 12pF 120pF 1200pF 1.3pF 13pF 130pF 1300pF 1.5pF 15pF 150pF 1500pF 0.015uF 0.15uF 1.5uF 15uF 150uF 1500uF 1.6pF 16pF 160pF 1600pF 1.8pF 18pF 180pF 1800pF 2.0pF 20pF 200pF 2000pF 2.2pF 22pF 20pF 2200pF 0.022uF 0.22uF 2.2uF 22uF 220uF 2200uF 2.4pF 24pF 240pF 2400pF 2.7pF 27pF 270pF 2700pF 3.0pF 30pF 300pF 3000pF 3.3pF 33pF 330pF 3300pF 0.033uF 0.33uF 3.3uF 33uF 330uF 3300uF 3.6pF 36pF 360pF 3600pF 3.9pF 39pF 390pF 3900pF 4.3pF 43pF 430pF 4300pF 4.7pF 47pF 470pF 4700pF 0.047uF 0.47uF 4.7uF 47uF 470uF 4700uF 5.1pF 51pF 510pF 5100pF 5.6pF 56pF 560pF 5600pF 6.2pF 62pF 620pF 6200pF 6.8pF 68pF 680pF 6800pF 0.068uF 0.68uF 6.8uF 68uF 680uF 6800uF 7.5pF 75pF 750pF 7500pF 8.2pF 82pF 820pF 8200pF 9.1pF 91pF 910pF 9100pF #generics #CommonPartsLibrary
jharwinbarrozo
1.5M
Generic Inductor
A generic fixed inductor for rapid developing circuit topology. *You can now change the footprint and 3D model at the top level anytime you want. This is the power of #generics
jharwinbarrozo
14.4k
RMCF0805JT47K0
47 kOhms ±5% 0.125W, 1/8W Chip Resistor 0805 (2012 Metric) Automotive AEC-Q200 Thick Film #forLedBlink
jharwinbarrozo
1.2M
875105359001
10uF Capacitor Aluminum Polymer 20% 16V SMD 5x5.3mm #forLedBlink #commonpartslibrary #capacitor #aluminumpolymer #radialcan
jharwinbarrozo
1.2M
CTL1206FYW1T
Yellow 595nm LED Indication - Discrete 1.7V 1206 (3216 Metric) #forLedBlink
jharwinbarrozo
1.1M

What is the effect of resistor R7?

@copilot What is the effect of this resistor R7?

11mo

Resistor R7 in your circuit is part of the net labeled "Net: GND", and it's connected to the anode of the LED (D12-LED1) on one side and pin P1 on the other. Here's the relevant connection:

R7: P1 is connected to the cathode of D12-LED1 (K) via Net 10
R7: P2 is connected to GND via Net GND

Function of R7

This resistor appears to be a current-limiting resistor for the LED, D12-LED1. The purpose of this resistor is to limit the amount of current flowing through the LED, thus protecting it from damage due to excess current.

Calculation

To determine the effect precisely, you can calculate the current through the LED using Ohm's Law:

I = \frac{V_{supply} - V_f}{R}

Where:

$I$ is the current through the LED.
$V_{supply}$ is the supply voltage.
$V_f$ is the forward voltage of the LED.
$R$ is the resistance value of R7.

Given:

Forward Voltage of the LED ( $V_f$ ): 1.85V
Resistor Value $R$ : 3.3k $\Omega$ (R7)
Assume the supply voltage ( $V_{supply}$ ) depends on your specific setup; for simplicity, let's consider a typical value:

Assume a supply voltage of 12V:

I = \frac{12V - 1.85V}{3.3k\Omega} = 3.1mA

If the supply voltage is different, the current will change proportionally.

Conclusion

R7 ensures that the LED (D12-LED1) is not over-driven by limiting the current, thus prolonging the LED's lifespan and ensuring safe operation. If you have a specific supply voltage value in mind, substitute it into the formula to get the correct current.

11mo

yudyud

What is the effect of resistor R7?

11mo

Someone is typing...

what resistance value should this part have?

yudyud

@copilot what resistance value should this part have?

11mo

copilot

Sure, I can help with that. However, I'll need some additional information to proceed:

What is the purpose of the resistor in your circuit? e.g., pull-up, pull-down, current limiting
What is the voltage and current rating for the part or the circuit it's being used in?
Are there any constraints or specific requirements for tolerance, power rating, or temperature coefficient?

If you could provide these details, I can calculate or recommend an appropriate resistance value.

11mo

yudyud

Vsupply = 15V I = 1A

11mo

copilot

Given a supply voltage (

V_{\text{supply}}

) of 15V and a current of 1A, we can use Ohm's Law to determine the required resistance.

Ohm's Law is given by:

V = IR

where:

$V$ is the voltage (15V)
$I$ is the current (1A)
$R$ is the resistance

Rearranging for

R

R = \frac{V}{I}

Substituting in the given values:

R = \frac{15\text{V}}{1\text{A}} = 15\text{ ohms}

So, the required resistance value is 15 ohms.

Let me know if you need more detailed information or have any specific constraints regarding this resistor, such as power dissipation, tolerance, or other parameters.

11mo

yudyud

The source voltage is 15V with a 1A charger. Find R5 and R4 to bias transistors Q1 and Q6 so that the voltage drop of each is 4.1V.

11mo

copilot

To bias the transistors $Q1$ and $Q6$ such that each has a voltage drop of 4.1V, we need to select appropriate values for

R4

and

R5

considering that the total voltage drop across both these components will equal the source voltage (15V).

Steps to Determine

R4

and

R5

Voltage across Transistors:
- Drops across $Q1$ and $Q6$: 4.1V each
- Total voltage drop across both transistors: $4.1\text{V} + 4.1\text{V} = 8.2\text{V}$
Voltage across Resistors:
- Voltage drop across $R4$ and $R5$ : $15\text{V} - 8.2\text{V} = 6.8\text{V}$
Determine Current through Resistors:
- Since these are in series and the current is 1A, the same current will flow through both $R4$ and $R5$ .
Calculate Resistor Values using Ohm's Law:
- Resistor value $R = \frac{V}{I}$
For $R4$ and $R5$ :
- Voltage drop across $R4$ : 3.4V
- Voltage drop across $R5$ : 3.4V
(Here we are distributing the voltage drop equally for simplicity, unless dictated otherwise by specific circuit conditions.)

Calculations:

R4 = R5 = \frac{3.4\text{V}}{1\text{A}} = 3.4\text{ ohms}

So:

R4 = 3.4\text{ ohms}, \quad R5 = 3.4\text{ ohms}

Assumptions:

The transistors $Q1$ and $Q6$ are identical and require the same base current.
The source provides a consistent current of 1A.

These values ensure that $Q1$ and $Q6$ each have a voltage drop of 4.1V, and the resistors

R4

and

R5

have appropriate drops to meet the source voltage.

Would you like more detailed information or specific considerations for the transistor types and circuit configuration?

11mo

Someone is typing...

What is the effect of this resistor R25?

yudyud

@copilot What is the effect of this resistor R25?

11mo

copilot

To analyze the effect of resistor R25, let's first look at its connections:

R25 (3.3k) is part of the net Net 9, which connects to:
- D4-LED2 (A),
- R3 (P1).
R25 (P1) is part of net Net 5, which connects to:
- R1 (P1),
- R4 (P1),
- Q1 (E),
- L1 (1),
- L5 (Pin_1),
- R8 (P1),
- R13 (P1),
- R6 (P2).

Analysis:

Placing in the Circuit:
- Net 9 suggests R25 is connected in series with D4-LED2 (anode) and R3.
- Net 5 suggests R25 is connected in parallel with components on Net 5.
Behavior:
- Current Flow: R25 is in the current path of LED D4-LED2. It will limit the current flowing through the LED, protecting it from overcurrent which could damage the LED.
- Voltage Drop: It causes a voltage drop which ensures the LED operates at its required forward voltage.
Component Interaction:
- Since resistors R3 and R25 are involved, they form a voltage divider with other resistors in the circuit leading to R25.

Effect Summary:

Current Limiting: R25 limits the current through D4-LED2, preventing it from exceeding its maximum rated current.
Voltage Regulation: R25 helps maintain a proper operating voltage across the LED.

Assumptions:

Assuming standard LED forward current $10\,mA$ to $20\,mA$ and forward voltage ~$2,V$.

In a real design, these values might be adjusted depending on the specific LED characteristics and desired brightness. Ensure that these assumptions reflect your actual circuit requirements for accurate functioning.

11mo

Someone is typing...

https://cdn2.flux.ai/flux-comment-files/files/commentFileId-dfd0f2e5e8de7ee8e325...

yudyud

@copilot

Image

https://cdn2.flux.ai/flux-comment-files/files/commentFileId-dfd0f2e5e8de7ee8e325e1a0d11dc22c5a8411cd5fc5dd9078568489648743e2.png?GoogleAccessId=graviton-mvp%40appspot.gserviceaccount.com&Expires=4872761097&Signature=KcVRRt04pQXezihTzapQbo9P8d345GyEEVt%2FI1vC2V0C5t4qF6TfJ90Ef%2FpnksTK3txVxHgAwF0kvaeScDCuVySCwid382UFRTuwYh7OZh44skxdJGD5GDReXcyBrJbo0qiUPppI4ffsScga11UlS%2FfPOY9SPnAuR0mnEJKaX9dCuWTthrGxf0hGf8uxDvv5mqNinMxHls6c%2BQ1JG3GQD%2Bl10jPJEqSOy%2BOhL%2FljdKguk2qNsv7pj1Ecpn%2F5DNJWDXBtIXKrat0mHppnns90FM7y89ATTbW9ZURQ1xBh8Ug1AGMmM3sisOsBfqAeezYFfnqbXprqooJKN6jIKHZIew%3D%3D

11mo

copilot

Thank you for providing the schematic image. Let's detail the connections based on the components and nets described in your project.

Here's a breakdown of the key connections:

Power Input and Protection:
- Connect a DC power supply to L5:Pin_1 and L5:Pin_2 for input.
- Place a fuse (F2) between the DC input and the rest of the circuit for protection.
- Add the diode (D13) for reverse polarity protection.
LED Indicators:
- Connect the LEDs and their respective current-limiting resistors as follows:
  - First LED Circuit:
    - Connect D12-LED1:K to R7:P1.
    - Connect R7:P2 to GND.
    - Connect D12-LED1:A to Q8:C.
  - Second LED Circuit:
    - Connect D4-LED2:A to R25:P2.
    - Connect D4-LED2:K to R3:P2.
    - Connect R25:P1 to Net 8.
Transistor Circuits:
- Connect the bases, collectors, and emitters of the transistors:
  - For Q4 (SS8550BBU):
    - Base: Q4:B to R6:P1.
    - Collector: Q4:C to D12-LED1:A.
    - Emitter: Q4:E to F2:P2.
  - For Q6 (SS8550BBU):
    - Base: Q6:B to R5:P2.
    - Collector: Q6:C to Q1:C.
    - Emitter: Q6:E to Q1:B.
Voltage Regulation and Control (using TL431):
- Connect the TL431 adjustable shunt regulators:
  - D1-TL431:
    - REF: D1-TL431:REF to R1:P2.
    - Cathode: D1-TL431:K to Net 15.
    - Anode: D1-TL431:A to Net 17.
  - D2-TL431:
    - REF: D2-TL431:REF to C2:P1.
    - Cathode: D2-TL431:K to Net 18.
    - Anode: D2-TL431:A to Net 19.
Capacitors for Filtering:
- Connect the capacitors for noise filtering and stability:
  - C1: Connect C1:P1 to Net 15 and C1:P2 to Net 16.
  - C4: Connect C4:P1 to Net 6 and C4:P2 to Net 7.
Resistors Network:
- Wire up the resistors in their respective positions to form filter networks and voltage dividers as outlined in the initially provided net lists.

This level of detail should assist in accurately placing all your components and connecting them in the correct manner. Should you need further clarification on any specific part or connection, feel free to ask!

11mo

yudyud

How to calculate each resistor value so that the charging source for each battery is 1A and 4.2V

11mo

copilot

To ensure a charging current of 1A at a charging voltage of 4.2V for each 18650 lithium-ion battery, there are a few critical components and calculations needed. The key parameters to design for are:

Voltage Regulation: Ensuring that the output voltage is precisely 4.2V.
Current Limiting: Ensuring the current does not exceed 1A during the charging process.

For a lithium-ion battery charging circuit, typically an adjustable shunt regulator like the TL431 can be used for precise voltage regulation and a series resistor or a current limiting circuit for controlling the current.

Voltage Regulation with TL431

The TL431 acts as an adjustable Zener diode with a reference voltage of 2.495V. In the typical application, a voltage divider sets the output voltage.

Voltage Divider Calculation:

V_{OUT} = V_{REF} \left(1 + \frac{R1}{R2}\right) + \frac{I_{REF} \cdot R1}{V_{OUT}}

Since

I_{REF}

is tiny, it can often be neglected.

Given:

$V_{OUT} = 4.2V$
$V_{REF} = 2.495V$

We need to solve for

R1

and

R2

4.2 = 2.495 \left(1 + \frac{R1}{R2}\right)

Solving gives:

1.684 \approx \frac{R1}{R2}

We can choose standard resistor values. Suppose

R2 = 1k\Omega

R1 \approx 1.684 \times 1k\Omega \approx 1.68k\Omega

Current Limiting with Resistor

For current limiting, we aim to design in such a way that the current does not exceed the desired value. A simple way to do this is by using a resistor in series with the load.

Ohm’s Law:

R_{LIMIT} = \frac{V_{LIMIT}}{I_{LIMIT}}

In your case:

$I_{LIMIT} = 1A$
If we assume a small headroom (say 0.2V) beyond the regulated 4.2V, then $V_{LIMIT} = 0.2V$

Hence:

R_{LIMIT} = \frac{0.2V}{1A} \approx 0.2\Omega

Circuit Implementation

Voltage Regulation Circuit:
- Use TL431:
- Connect resistors $R1 = 1.68k\Omega$ and $R2 = 1k\Omega$ between the REF pin of TL431 and GND/Feedback point respectively.
Current Limiting:
- Place a 0.2Ω resistor in series with the output to limit the current to 1A.

Example Component Connections:

Voltage Regulator:

Connect $R1:(P1)$ to $TL431:REF$ and $R1:(P2)$ to $V_{OUT}$ .
Connect $R2:(P2)$ to $TL431:A$ and $R2:(P1)$ to $TL431:K$ .

Current Limiting Resistor:

Connect the positive output of the regulated voltage to one end of $R_{LIMIT}$ .
Connect the other end of $R_{LIMIT}$ to the positive terminal of your battery.

This will ensure that the voltage across the battery terminals is 4.2V with a current limit of 1A, adhering to the lithium-ion battery charging requirements.

Assumptions:

Ideal components without tolerance variation.
The power supply can deliver beyond 4.2V for headroom and current limiting.
A more sophisticated current-limiting circuit might be preferred in practical applications to dynamically adjust the current.

Feel free to correct any assumptions as required or provide additional constraints!

11mo

yudyud

Operating principle of the circuit

11mo

copilot

The circuit you have is a lithium-ion battery charging circuit. Let's break down the operating principle based on key stages and components in the circuit:

Key Components and Their Roles:

Power Supply and Protection:
- L6 and F2: The power supply (input through L6) and a fuse (F2) for protection against overcurrent.
- D13: A diode for reverse polarity protection, ensuring the correct direction of current flow into the circuit.
LED Indicators:
- D12-LED1 and D4-LED2: LEDs with their respective current-limiting resistors for visual indication of the circuit's status.
Transistors:
- Q1, Q2, Q3, Q4, Q6, Q7, Q8: Bipolar Junction Transistors (BJTs) used for switching and regulation.
- Q1 & Q2 (TIP42A, PNP): Act as active switches or pass elements in regulating current and voltage.
- Q4-Q8 (SS8550BBU, PNP): Function in driving the LEDs and providing necessary amplification stages for current control.
Shunt Regulators (TL431):
- Each TL431 is an adjustable shunt regulator, used to maintain a precise reference voltage, which is crucial for achieving the desired charging voltage of 4.2V.
- Multiple TL431 regulators are employed in the circuit to regulate and stabilize the voltage across the battery.

Operating Principle:

Power Supply and Initial Protection:
- The input voltage is supplied through the header (L6). The fuse (F2) ensures protection against overcurrent. If the current exceeds a certain threshold, the fuse will blow and protect the rest of the circuit.
Voltage Regulation with TL431 and Transistors:
- TL431 Shunt Regulators:
  - These are configured to establish a precise regulation point (4.2V) using a voltage divider formed by resistors. The output voltage is controlled to be stable and accurate, crucial for proper battery charging.
- Transistors Q1 and Q2 (as Pass Transistors):
  - These transistors (TIP42A) work in conjunction with the shunt regulators to provide the necessary current to the battery while maintaining the voltage at 4.2V. They pass the required current according to the regulation set by the TL431.
Current Limiting Mechanism:
- A low-value resistor (e.g., 0.2Ω) is placed in series with the charging path to limit the current to 1A by Ohm's Law. If the current exceeds 1A, the voltage drop across this resistor increases, thereby reducing the effective charging voltage and preventing overcurrent.
Indicator LEDs and Secondary Transistors:
- D12-LED1 and D4-LED2: Indicate the operational status of charging.
- Q4-Q8: Drive the LEDs and amplify control signals for regulating the charging current and voltage.

Summary of the Charging Process:

Initialization:
- The power supply provides a voltage higher than 4.2V for regulation.
- The fuse and diode ensure protection against overcurrent and reverse polarity.
Voltage Regulation:
- TL431 regulators and their associated resistor networks maintain the output voltage at a precise 4.2V.
- The pass transistors (Q1 and Q2) modulate the current to the battery based on the feedback from TL431 regulators.
Current Limiting:
- The series resistor ensures that the charging current does not exceed 1A, protecting the battery from overcurrent.
Indication:
- LEDs indicate the status, and transistors drive these LEDs based on the charging condition.

This configuration ensures that the battery is charged safely with a controlled voltage of 4.2V and a current limit of 1A, protecting the battery and extending its lifespan. If you have any further queries or need additional details, feel free to ask!

11mo

yudyud

How to calculate the resistor values to bias the transistor and tl431 knowing the charging current is 1A and the voltage is 4.2V 2 battery terminals set R5 R12 R21 to 1k

11mo

copilot

To bias the transistors and TL431 to achieve a charging current of 1A and a voltage of 4.2V, you need to calculate the resistor values in such a way that the desired output characteristics are achieved.

Key Points to Address:

Setting the Voltage to 4.2V:
- Configure the TL431's feedback network to set the output voltage to 4.2V.
Ensuring a 1A Charging Current:
- Use a current sense resistor to limit the charging current.

Voltage Regulation using TL431:

We need to use a voltage divider to set the reference voltage for the TL431 to ensure the output voltage is stabilized to 4.2V.

Given:

$V_{\text{OUT}} = 4.2V$
$V_{\text{REF}} = 2.495V$

The relationship is:

V_{\text{OUT}} = V_{\text{REF}} \left(1 + \frac{R1}{R2}\right)

where:

$R1$ is the resistor from REF to the output voltage (and anode).
$R2$ is the resistor from REF to ground.

Given that

V_{\text{OUT}} = 4.2V

4.2 = 2.495 \left(1 + \frac{R1}{R2}\right)

\frac{R1}{R2} = \frac{4.2}{2.495} - 1 = 0.683

Choosing Standard Resistor Values:

To simplify, assume standard resistor values close to the calculated ratio. If

R2 = 1k\Omega

R1 = 0.683 \times 1k\Omega \approx 683\Omega

Since 683Ω is not a standard value, you can use the closest standard value, which is 680Ω.

Current Limiting:

To limit the current to 1A, use a sense resistor in series with the battery. Assume the sense resistor is connected to an emitter resistor that sets the current limit.

The voltage drop for 1A current should be a small value, e.g., 0.2V:

R_{\text{LIMIT}} = \frac{V}{I} = \frac{0.2V}{1A} = 0.2\Omega

Biasing Transistors:

The transistors are generally biased to ensure they are in the active region for switching and amplification. Assuming basic BJT biasing:

The base resistor ( $R_B$ ) needs to ensure sufficient base current ( $I_B$ ) to drive the desired collector current ( $I_C$ ).

Given:

I_C = 1A

Typical current gain (hFE) for TIP42A:

hFE \approx 50

I_B = \frac{I_C}{hFE} = \frac{1A}{50} = 20mA

Given

V_{BE} \approx 0.7V

To calculate

R_B

If base voltage is provided by the output of TL431 and suppose output voltage of TL431 is 2.495V:

$R_B = \frac{V_{OUT \,(TL431)} - V_{BE}}{I_B} = \frac{2.495V - 0.7V}{20mA} \approx 90\Omega$

Values for Your Specific Resistors (R5, R12, R21):

Given

R5 = R12 = R21 = 1k\Omega

These are used in the feedback network or biasing the TL431:

They would directly interrelate with the other feedback resistors (e.g., $R4$ , $R2$ , etc.) affecting the reference voltage settings.
Assume they form a part of the voltage setting feedback directly affecting REF pins of TL431.

Example Adjustments:

Set $R1 \approx 680 \Omega$ and $R2 = 1k\Omega$ (Standard values around calculated)
$R_B \approx 90\Omega$ for each base bias of transistors, ensuring sufficient drive.
$R_{\text{LIMIT}} = 0.2 \Omega$ for current limiting 1A.

Summary:

Feedback Resistors: $R1$ = 680Ω, $R2 = 1kΩ$
Current Sense Resistor: $R_{\text{LIMIT}} = 0.2 \Omega$
Bias Resistor for Base: $R_{B} = 90\Omega$

Integrating properly configured current and voltage limiting ensures your charging circuit safely charges the batteries at 4.2V and 1A. Adjust any precise values based on final testing and expected variations in actual implementation.

Would you like more specific insights into individual connections or additional considerations in the circuit?

11mo

yudyud

Calculate the value of R4 R5 knowing that at Q1 and Q6 the voltage drop is 4.1V and the current is 1A

11mo

copilot

To calculate the values of resistors ( R4 ) and ( R5 ), we need to ensure that they provide the proper base current to the transistors ( Q1 ) and ( Q6 ) to maintain a voltage drop of 4.1V across them while passing a current of 1A.

Here's a step-by-step process to determine the resistor values:

Understanding the Requirements

Transistor Operation:
- ( Q1 ) and ( Q6 ) are PNP transistors, with a voltage drop across their collector-emitter junctions of 4.1V and a current of 1A.
- For proper operation in the active region, the base-emitter voltage ( V_{BE} ) is approximately 0.7V.
Current Gain (hFE):
- Assume the current gain (hFE) of the transistors ( Q1 ) and ( Q6 ) is 50 (typical for TIP42A).
Base Current Calculation:
- The base current ( I_B ) is needed to support a collector current ( I_C ) of 1A: $I_B = \frac{I_C}{hFE} = \frac{1A}{50} = 20mA$
Base Resistor Calculation:
- The base resistors ( R4 ) and ( R5 ) are connected to the base of ( Q1 ) and ( Q6 ) respectively.
- The voltage at the base of ( Q1 ) and ( Q6 ) needs to be 4.1V + 0.7V = 4.8V.

Calculation for ( R4 ) (Base Resistor for ( Q1 )) and ( R5 ) (Base Resistor for ( Q6 ))

Assume the voltage at the base is 4.8V and the base-emitter voltage ( V_{BE} ) is 0.7V.
The power supply voltage is higher than the emitter voltage by 4.1V:
The voltage difference across ( R4 ) and ( R5 ) is the supply voltage minus the base voltage:

Typically, a resistor connected from a higher base supply voltage:

R = \frac{(V_{supply} - V_{BE})}{I_B}

If the supply voltage is 4.8V (to provide sufficient)

Example Calculation:

Assume Calculation Step:

For ( V_{OUT} ) (Supply higher than the emitter voltage)

R4 = \frac{V_{supplied} - V_{BE}}{I_B} = \frac{(4.8V - 0.7V)}{0.02A }= \approx 205 \Omega

For ( R5 ): 4.8V same method used to example calculated ( R4 ) = (similar calculation approach used)

Summary check:

Calculate and consideration approach:

Select example standard resistor values and calculated components for application match.

Estimate more precision test along Environment can ensure a value compatibility.

Conclusion and Result

This approach ensures selected ( R4 ) and ( R5 ) resistors value match your criteria and ensure correct values. Residual any concern with your provision be adaptable.

Conduction checking constraints may conserve values are correct.

To calculate the exact values of ( R4 ) and ( R5 ) given the conditions, let's go through the steps in a clear and detailed manner.

Given Parameters:

Voltage drop across ( Q1 ) and ( Q6 ) (( V_{CE} )) is 4.1V.
Current passing through ( Q1 ) and ( Q6 ) (( I_C )) is 1A.
Transistor current gain (hFE) is assumed to be 50 (a common value for TIP42A).

Steps to Determine ( R4 ) and ( R5 ):

Determine the Required Base Current (( I_B )):
- Using the current gain (hFE): $I_B = \frac{I_C}{hFE} = \frac{1A}{50} = 0.02A = 20mA$
Voltage at the Base (assuming typical ( V_{BE} \approx 0.7V )):
- The base voltage (( V_B )) for ( Q1 ) and ( Q6 ): $V_B = V_E + V_{BE} \approx 4.1V + 0.7V = 4.8V$
Calculate the Base Resistor (( R_B )):
- Assuming the base voltage ( V_B = 4.8V ) and the supply voltage ( V_{SUPPLY} ) must be known or assumed. Here we assume a higher voltage supply, let’s assume it to be around 12V for a plausible design.
- The voltage drop across the base resistors ( R4 ) and ( R5 ): $V_{R_B} = V_{SUPPLY} - V_B = 12V - 4.8V = 7.2V$
- Using Ohm's Law to determine the resistor value: $R_B = \frac{V_{R_B}}{I_B} = \frac{7.2V}{0.02A} = 360\Omega$

Resulting Values:

Resistor ( R4 ):
- Calculated value: 360Ω.
Resistor ( R5 ):
- As the conditions for ( R4 ) and ( R5 ) are identical, ( R5 ) would also be 360Ω.

Thus, both ( R4 ) and ( R5 ) should be 360Ω to provide the necessary base current to ( Q1 ) and ( Q6 ) for maintaining a voltage drop of 4.1V across the transistors while allowing 1A to flow through the collector.

Summary:

( R4 = 360\Omega )
( R5 = 360\Omega )

These resistors ensure that the ( Q1 ) and ( Q6 ) transistors are correctly biased to maintain the desired voltage drop and current flow.

If your power supply has a different voltage or specific design constraints, adjust accordingly while maintaining the base current requirements and recalculating as needed. If you have any different assumptions or elements in the setup, feel free to ask!

11mo

Someone is typing...

how to calculate R6 R7 R25 R3

yudyud

@copilot how to calculate R6 R7 R25 R3

11mo

Someone is typing...

What effect does resistor R3 have here?

yudyud

@copilot What effect does resistor R3 have here?

11mo

copilot

Resistor R3, as a part of a circuit, can have varying effects depending on its role and placement. Since there's no specific mention of your full circuit, I will explain in general terms.

Here are some common roles R3 might be playing in your circuit:

Current Limiting:
- If R3 is in series with a component like an LED or another active component, it would limit the current flowing through that component, protecting it from excessive current.
Voltage Division:
- If R3 is part of a voltage divider network, it would help in scaling down the voltage to a desired level.
Pull-up or Pull-down:
- If R3 is connected between a pin (say a GPIO) and a supply voltage (VCC) or ground (GND), it could act as a pull-up or pull-down resistor, ensuring that the pin is at a defined logic level when not actively driven.
Biasing:
- In transistor circuits, R3 could be used for biasing the transistor base or emitter to set the correct operating point.

Assuming a typical scenario such as part of a voltage regulation circuit,

Series Resistor in Zener Diode Circuit:
- In a Zener diode voltage regulation circuit, R3 can act as the series resistor crucial for defining the current through the Zener diode, hence influencing the voltage regulation effectiveness $1d5864d1-c9df-59d3-9195-30ad91393842$ .

To give a precise answer, I'd need more details about your circuit especially how R3 is connected and which additional components it's interacting with. Can you provide further context or a snippet of your schematic?

11mo

yudyud

R3 is connected to LED D4

11mo

copilot

Great, thanks for the additional information. If R3 is connected to LED D4, it most likely serves as a current-limiting resistor. Its main function would be to limit the amount of current flowing through the LED to prevent it from burning out.

Here's how it works:

Purpose: LEDs require current limiting to avoid excessive current flowing through them, which can cause overheating and eventual failure.
Calculation: The value of R3 is chosen based on the supply voltage ( $V_{supply}$ ), the forward voltage of the LED ( $V_{forward}$ ), and the desired current through the LED ( $I_{LED}$ ).

The formula to calculate the resistor value is:
$R3 = \frac{V_{supply} - V_{forward}}{I_{LED}}$

Assuming typical values:

Supply Voltage $V_{supply}$ : 5V
Forward Voltage $V_{forward}$ of the LED: 2V
Desired LED Current $I_{LED}$ : 20mA (0.02A)

The resistor value would be calculated as:

R3 = \frac{5V - 2V}{0.02A} = 150 \Omega

Connecting LED D4 and Resistor R3 to Arduino

Here are the connection steps for a typical scenario where the LED is connected to a digital pin on an Arduino:

Anode (longer leg) of LED: Connect D4:Anode to Arduino:Digital Pin (e.g., Pin 13).
Cathode (shorter leg) of LED: Connect D4:Cathode to one end of R3.
Other end of R3: Connect R3:Other end to Arduino:GND.

Specific Connection Format

Connect LED D4:Anode to Arduino:Digital Pin 13.
Connect D4:Cathode to R3:1.
Connect R3:2 to Arduino:GND.

Example using a breadboard (cite if applicable):

Connect the 220-ohm resistor (R3) between the cathode (shorter leg) of the LED and ground (GND) on the Arduino board.

This setup ensures that the LED receives current within safe limits when controlled by the Arduino. If you have other specific parameters in your circuit, make sure to adjust the resistor value accordingly.

11mo

yudyud

R3 is connected in parallel with LED D4

11mo

copilot

If R3 is connected in parallel with LED D4, the resistor will affect the current distribution and voltage across the LED. Here are the key effects:

Current Sharing:
- Parallel Configuration: When components are connected in parallel, the voltage across each component is the same. The current through each component, however, can differ.
- Bypassing the LED: Some of the current that might normally flow through the LED will be diverted through the resistor, depending on the resistor's value. The smaller the resistance, the more current will be diverted away from the LED.
LED Brightness:
- Reduced Current: Since current is shared between the resistor and the LED, the current flowing through the LED may decrease, potentially reducing its brightness.
Protection:
- Voltage Dropping: If any transient spikes occur, R3 can help to protect the LED by providing an additional path for the current, thereby limiting the potential damage to the LED.

Example Calculation

For better insight, let’s assume some values and analyze the situation:

Voltage across LED (V_LED): 2V (typical for a red LED)
Supply Voltage (V_SUPPLY): 5V
R3 Value: 220 ohms

Since the resistor is parallel with the LED, the following can be considered:

Voltage Across and Current Through LED and Resistor LED

If the LED forward voltage is 2V, it will draw a current based on its internal characteristics (e.g., 20mA is typical for many small LEDs).

Resistor R3

The voltage across R3 is the same as the LED, so it would be 2V.
The current through R3: $I_R3 = \frac{V_{LED}}{R3} = \frac{2}{220} = 9.09 \, mA$

Total Current

The total current from the power source will be the sum of the current through the LED and R3:

I_{TOTAL} = I_{LED} + I_{R3}

Assuming the LED current ($I_{LED}$) is 20mA (typical),

$I_{TOTAL} = 20 \, mA + 9.09 \, mA = 29.09 \, mA$

This means that the current drawn from the power supply increases, but the effective current through the LED could be less because of the parallel configuration. If R3 has a significantly lower resistance, more current will be bypassed through R3, leading to a dimmer LED.

Connections

If this suits your circuit, your connections should be laid out as follows:

Connect R3:1 to D4:Anode.
Connect R3:2 to D4:Cathode.

If you can provide the exact values or more context regarding the power supply and other components, I would be able to offer more detailed insights or warnings.

Design Considerations

Use a parallel resistor with caution in LED circuits as it may affect the LED's brightness. Ensure the values align with your design goals.

11mo

Someone is typing...

What function does R5 have?

yudyud

@copilot How to calculate R4?

11mo

copilot

Calculating the value for resistor R4 in your 18650 battery charging circuit will depend on the specific function it plays in the circuit. Below are a few common scenarios involving resistors in battery charging circuits and how you would calculate their values:

Current Limiting Resistor: If R4 is used to limit current to the battery or an LED indicator, you can use Ohm's Law ( $V = I \cdot R$ ) to find the appropriate value. The resistor value can be calculated as: $R = \frac{V_{supply} - V_{load}}{I_{load}}$
Voltage Divider: If R4 is part of a voltage divider used to set a reference voltage for charging, the value can be calculated using the voltage divider formula: $V_{out} = V_{in} \cdot \frac{R4}{R4 + R_{other}}$
Feedback Resistor: If R4 is used in feedback to control a charging IC or transistor: $V_{sense} = I_{battery} \cdot R4$

Assuming typical scenarios:

Current Limiting Example:
- Supply Voltage ($V_{supply}$): 5V
- Load Voltage ($V_{load}$): 4.2V (Typical lithium-ion battery charging voltage)
- Desired Current ($I_{load}$): 1A
$R = \frac{5V - 4.2V}{1A} = 0.8\ \Omega$
Voltage Divider Example:
- Input Voltage ($V_{in}$): 5V
- Desired Output Voltage ($V_{out}$): 2.5V
- Assume $R_{other}$ is 1kΩ
$2.5V = 5V \cdot \frac{R4}{R4 + 1k\Omega}$ Solving for R4: $R4 = 1k\Omega$
Feedback Resistor Example:
- Desired Sense Voltage ($V_{sense}$): 0.6V
- Battery Current ($I_{battery}$): 1A
$R4 = \frac{V_{sense}}{I_{battery}} = \frac{0.6V}{1A} = 0.6\ \Omega$

Without specific details on the role of R4 in your circuit, you'll need to provide more information or identify its function in the schematic to calculate its precise value. Adjust the formula and assumptions accordingly.

11mo

yudyud

How to calculate R4 R5

11mo

copilot

To accurately calculate the values for resistors R4 and R5, we need to understand their specific roles in the circuit. From the information provided, it seems the resistors are part of "Net 15," which includes:

R4: (P2), R5: (P1) – Connected together at P2 and P1 respectively
C1: (P2) – Connected to the other ends of both R4 and R5
D1-TL431: (K) – Connected to other ends of both R4 and R5

Given this setup, it appears that R4 and R5 might be part of a voltage divider or control network that sets the reference voltage for a TL431 programmable shunt voltage regulator, common in battery charging circuits for accurate voltage regulation.

Calculating R4 and R5 in a Reference Voltage Setup

Assuming that you need to set a reference voltage for the TL431, the calculation involves determining the resistors to achieve the desired voltage.

Identify the desired reference voltage $V_{ref}$ for the TL431. The TL431 typically has a reference voltage of 2.495V.
Determine the input voltage $V_{in}$ at the top of the divider (likely the battery voltage or supply voltage).
Use the voltage divider formula to calculate R4 and R5. The formula for the voltage divider is: $V_{out} = V_{in} \cdot \frac{R5}{R4 + R5}$ To solve for R4 and R5, assume a ratio that gives the desired $V_{out} = V_{ref}$ .

Example Calculation

Let's assume the input voltage $V_{in}$ is 4.2V (typical for charging a single Li-ion cell).
The reference voltage $V_{ref}$ for the TL431 is 2.495V.

Using the voltage divider formula:

2.495V = 4.2V \cdot \frac{R5}{R4 + R5}

Rearrange to find the ratio:

\frac{2.495V}{4.2V} = \frac{R5}{R4 + R5}

\frac{2.495}{4.2} = \frac{R5}{R4 + R5}

\frac{2.495}{4.2} = \frac{R5}{R4 + R5} \approx 0.594

So:

0.594(R4 + R5) = R5

0.594R4 + 0.594R5 = R5

0.594R4 = 0.406R5

R4 = \frac{0.406}{0.594}R5 \approx 0.683R5

If we choose R5 as 1kΩ (as a convenient value):

R4 \approx 0.683 \times 1k\Omega \approx 683\Omega

So, if R5 = 1kΩ, then R4 should be approximately 683Ω.

Conclusion

R4: 683Ω
R5: 1kΩ

These values set the TL431 reference voltage correctly for a 4.2V input to get a ~2.495V output.

Adjust these calculations based on the actual input voltage and the specific requirements of your regulation circuit. Let me know if further details or assumptions need tweaking for your exact setup!

11mo

yudyud

@copilot What function does R5 have?

11mo

copilot

You haven't specified the context or circuit in which R5 is used. Resistors can have various functions depending on their placement in the circuit, such as:

Current limiting
Voltage division
Setting bias points in amplifiers
Pull-up or pull-down resistor for logic circuits

If you can provide more details about the circuit, such as a partial schematic or description of the surrounding components and connections related to R5, I can provide a more accurate answer.

11mo

yudyud

Image

https://cdn2.flux.ai/flux-comment-files/files/commentFileId-dfd0f2e5e8de7ee8e325e1a0d11dc22c5a8411cd5fc5dd9078568489648743e2.png?GoogleAccessId=graviton-mvp%40appspot.gserviceaccount.com&Expires=4872988845&Signature=akQC0mr%2FjeH%2Bv7cg7F%2BzIDgktlh%2FxdEjTEBEASuDFIuqODhLjkjMrk4e%2FzuWl9cUusBPSPnROsKYFUjLKjZOgj9NHoBcvsINegic%2FauTAC303nh0wx8hoWSPT5mqoiYhsywOnxMCXmNKIpxPq47msTpZNgNKOc2GYnAcPNZpkFqyCbGLz9qNTOLuvPP83UWSkadjKNj%2F0rj1kVBG9oKyJTznuihqQZeLAUpXE9q7EBkXyUVxnBae2sPoTHpmewJvpXQvJ60Q1KaH4aV9LDvwbiikO9PL1cxmgcmepH2si3wM7l%2FOBsLmIXsLDdUMWp%2Fl9OtCOS%2FalTfvXnIhdu9OYg%3D%3D

11mo

copilot

Thank you for providing the schematic image. In this circuit, R5 is part of the voltage reference and current limit circuitry. It appears to serve the following functions:

Voltage Reference Creation:
- R5 is connected to the base of Q1, helping to create a voltage reference. Along with R4, it helps to set the biasing point of Q1.
Current Limiting:
- R5 forms a voltage divider with R4, which establishes the voltage at the base of Q1. It's crucial in determining the current flowing through the base-emitter junction of Q1, which in turn affects the current passing through the collector-emitter path of Q1.

The specific connections from the schematic are:

Connect $R5:2$ to the junction of $R4:2$ and $Q1:2$ (base of Q1).
Connect $R5:1$ to the ground.

Would you like more details on how this configuration can benefit the overall circuit operation?

11mo

yudyud

What function does R4 have?

11mo

copilot

To determine the function of R4 in the circuit, let's analyze its placement in the schematic:

Location: R4 is connected between the base of transistor Q1 and the/an emitter of transistor Q4 (though it looks as if it has voltage from the same node as LED1 and LED2 through a resistor).
Function: R4 looks to serve a couple of purposes:
- Base Current Limiting: It limits the base current to Q1, ensuring that the transistor does not get damaged by excessive current.
- Biasing: It's part of the biasing network that sets the operating point of Q1, ensuring Q1 is in the correct region $($ likely saturation or active region, depending on the circuit design $)$ .

To dive deeper:

Since R4 connects to the base of Q1, it helps control the switching behavior of Q1 by regulating the current flowing into the base.
This resistor helps in stabilizing the transistor's operation and prevents excessive base current which could lead to thermal runaway.

Would you like to analyze any other component's function?

11mo

yudyud

How to calculate the value of R4 and R5

11mo

copilot

To calculate the values of resistors R4 and R5, we'll make several assumptions about the transistor Q1's requirements and the operating conditions of the circuit:

Transistor Parameters:
- Assume the base-emitter voltage drop $(V_{BE})$ of Q1 is approximately 0.7V (typical for silicon BJTs).
- Assume the desired base current $I_B$ for Q1.
- Assume the collector current $I_C$ of Q1.
Voltage Supply:
- Let's assume the voltage at the junction of LEDs and R4 is around 5V (assumed typical forward voltage drop for LEDs and resistor configuration).

To calculate the resistor values:

Calculating R4:

R4 will determine the base current of Q1.

Determine the Base Current $I_B$ : $I_B = \frac{I_C}{\beta}$ where $\beta$ (h_FE) is the current gain of the transistor. Assume $\beta = 100$ and $I_C = 1A$ (or according to the circuit's load requirement).

$I_B = \frac{1A}{100} = 0.01A = 10mA$
Calculate R4: The voltage drop across R4 (let's call it $V_{R4}$ ) is the supply voltage minus the base-emitter voltage drop.

$V_{R4} = V_{supply} - V_{BE}$ Assuming $V_{supply} = 5V$ and $V_{BE} = 0.7V$ ,

$V_{R4} = 5V - 0.7V = 4.3V$

Now, calculate R4 using Ohm's Law:

$R4 = \frac{V_{R4}}{I_B}$ $R4 = \frac{4.3V}{0.01A} = 430 \Omega$

Calculating R5:

R5 assists in biasing and can stabilize the transistor's operation.

Assume Desired Stability Current $I_R5$ Through R5: Assume the current through R5 ( $I_R5$ ) is small compared to the base current. Let's assume $I_R5 = 10\% \times I_B = 1mA$
Voltage Drop Across R5: The voltage drop across R5 is the same as $V_{R4}$ .
Calculate R5:

$R5 = \frac{V_{R5}}{I_R5}$ $R5 = \frac{4.3V}{0.001A} = 4.3k\Omega$

Summary of Calculated Values:

$R4 = 430 \Omega$
$R5 = 4.3k\Omega$

Note these are theoretical calculations and should be adjusted based on the actual circuit requirements and design constraints. If more precise calculations are required, consider measuring the actual voltage drops and currents in the circuit.

Let me know if you'd like to modify these assumptions or have further details to refine the calculations!

11mo

yudyud

Nhưng mà tôi tính R4 = R5 = 1k. Hãy tihs lại đi Vsupply = 12.3V

11mo

copilot

Sure, let's recalculate the values for R4 and R5 given that the supply voltage (

V_{supply}

) is 12.3V, and we want both resistors to be 1kΩ. Here's how you can verify the biasing:

Verifying Voltages and Currents with R4 = R5 = 1kΩ

Assumptions:
- $V_{BE}$ of Q1 is 0.7V.
- $\beta$ (h_FE) of Q1 is 100.
- Base current $I_B$ is set by the resistor network.

Calculations:

Let's find the currents and voltages assuming R4 and R5 are both 1kΩ.

Voltage at the base of Q1:

If the supply voltage is 12.3V:

Voltage drop across R4 and R5 would directly affect the base voltage, let's denote this voltage (V_R):
Using voltage dividers:

V_{base} = V_{supply} \times \frac{R5}{R4 + R5}

Since

R4 = R5

V_{base} = 12.3V \times \frac{1kΩ}{1kΩ + 1kΩ} = 12.3V \times \frac{1}{2} = 6.15V

Determining Base Current

I_B

The current through R5 (from base to ground through R5) sets the base voltage, so:

I_B = \frac{V_{base} - V_{BE}}{R4}

I_B = \frac{6.15V - 0.7V}{1kΩ} = \frac{5.45V}{1kΩ} = 5.45mA

With current gain

\beta

of 100, the collector current

I_C

is:

I_C = \beta \times I_B = 100 \times 5.45mA = 545mA

Summary:

Voltage at Base of Q1: $V_{base} = 6.15V$
Base Current $I_B$ : $I_B = 5.45mA$
Collector Current $I_C$ : $I_C = 545mA$

Thus, with R4 = R5 = 1kΩ and supply voltage of 12.3V, your base-emitter configuration for Q1 results in significant operation currents that can be verified or adjusted based on load and requirements.

Let me know if you have any further questions!

11mo

yudyud

Q1 Q6 is a darlington combination

11mo

copilot

A Darlington pair (or combination) amplifies the current by configuring two transistors in such a way that the current amplified by the first is amplified further by the second. This gives the combination a very high current gain (

\beta

Given that Q1 and Q6 form a Darlington pair, the calculations would be adjusted accordingly. Here’s how:

Darlington Pair Characteristics:

The effective base-emitter voltage $(V_{BE})$ for a Darlington pair is approximately $1.4V$ (since it passes through two base-emitter junctions, each dropping around $0.7V$ ).

Recalculated Values for R4 and R5:

Assumptions:
- $V_{BE}$ of the Darlington pair is approximately $1.4V$ .
- Overall current gain $\beta$ of a Darlington pair is roughly the product of the gains of individual transistors (very high).

Verifying Voltages and Currents with R4 = R5 = 1kΩ: Voltage at the base of Q1 (Q1 + Q6 combo):

If the supply voltage is 12.3V:

V_{base} = V_{supply} \times \frac{R5}{R4 + R5}

Since

R4 = R5

V_{base} = 12.3V \times \frac{1kΩ}{1kΩ + 1kΩ} = 12.3V \times \frac{1}{2} = 6.15V

Determining Base Current

I_B

of Darlington Pair:

Since the effective base-emitter voltage drop of a Darlington pair is 1.4V:

I_B = \frac{V_{base} - V_{BE}}{R4}

I_B = \frac{6.15V - 1.4V}{1kΩ} = \frac{4.75V}{1kΩ} = 4.75mA

Given the Darlington pair has a high current gain, the collector current

I_C

could be very high: It is safe to assume the overall gain,

\beta \approx \beta_1 \times \beta_2

. For high current gain, the value could be approximately

10000

. Practically, we can consider typical range efficiency.

I_C \approx \beta \times I_B \approx 10000 \times 4.75mA = 47500mA$$ => 47.5A. ``` However, in real circuits, the power limitations of the transistors and other practical aspects will limit this. Therefore, realistic current with additional limiting elements would be in a safe range, like $$I_C = 300mA$$. ### Summary 1. Voltage at Base of the Darlington pair: $$ V_{base} = 6.15V $$ 2. Base Current $$I_B$$: $$I_B = 4.75mA$$ 3. Collector Current $$I_C$$: $$I_C = 47.5Amps$$ The practical draw will stabilize to functional limits based on the connection being correct. Would you like further adjustments or component visual?

11mo

yudyud

Recalculate

11mo

copilot

Sure, let's recalculate the values more realistically considering Q1 and Q6 form a Darlington pair and using the given supply voltage of 12.3V.

Darlington Pair Review:

Darlington Pair Effective $V_{BE}$ : Approximately 1.4V.
Assume High Beta ( $\beta$ ): The current gain of a Darlington pair is very high, any value above 1000 is typical.

Recalculating Resistor Values: Voltage at the base of the Darlington pair:

Using voltage dividers:

V_{base} = V_{supply} \times \frac{R5}{R4 + R5}

Since

R4 = R5 = 1kΩ

V_{base} = 12.3V \times \frac{1kΩ}{1kΩ + 1kΩ} = 12.3V \times \frac{1}{2} = 6.15V

Determining Base Current

I_B

For the Darlington pair, the effective

V_{BE}

is approximately 1.4V:

I_B = \frac{V_{base} - V_{BE}}{R4}

I_B = \frac{6.15V - 1.4V}{1kΩ} = \frac{4.75V}{1kΩ} = 4.75mA

Determining Collector Current

I_C

The current gain of a Darlington pair is very high. Using a conservative estimate:

\beta_{Darlington} \approx 1000

So:

I_C = \beta \times I_B = 1000 \times 4.75mA = 4750mA = 4.75A

This high current suggests that any practical implementation would have a current limiting mechanism to match actual load requirements.

Summary of Recalculated Values:

Voltage at Base of Darlington Pair: $V_{base} = 6.15V$
Base Current $I_B$ : $I_B = 4.75mA$
Collector Current $I_C$ : $I_C = 4.75A$ (in an ideal scenario, with the actual current dependent on loading conditions and circuit limits).

This shows well that R4 and R5 values with the given supply voltage would stabilize your circuit at a significant current gain. Would you like assistance in setting up experimental limits or other components?

11mo

yudyud

What is the effect of this resistor R5 and How to calculate the value of R5?

11mo

copilot

Resistor R5 plays an essential role in defining the operating point of the Darlington pair

Q1

and

Q6

. Specifically, R5 influences the base voltage and subsequently the base current of the transistors.

Role of R5:

Voltage Division:
- R5 forms a voltage divider with R4, which sets the base voltage of the Darlington pair. This base voltage $V_{base}$ is critical for the proper biasing of the transistors.
Bias Stabilization:
- By forming a potential divider, R5 helps stabilize the operating point of the Darlington pair against variations in transistor parameters and temperature changes.

Calculating the Value of R5:

Given:

Supply voltage $V_{supply}$ = 12.3V
Desired $V_{base}$ for the Darlington pair
Base-emitter voltage drop $V_{BE}$ of the Darlington pair is approximately 1.4V.

Steps:

Determine Desired $V_{base}$ :
- For the Darlington pair to operate properly, $V_{base}$ should be such that $V_{BE}$ is achieved.
- Let's assume the ideal base voltage we want is slightly above the combined $V_{BE}$ , say 1.5V for safety.
Voltage Division Calculation:
- Voltage at node between R4 and R5: $V_{base} = V_{supply} \times \frac{R5}{R4 + R5}$
- Rearrange to solve for R5: $R5 = \frac{V_{base} \times (R4 + R5)}{V_{supply}}$ $R5 = \frac{V_{base} \times R4}{V_{supply} - V_{base}}$
Choosing a Suitable $V_{base}$ :
- An adequately higher base value, suitable here, will be around: $1.5V$ approximately for detailed biasing.
Plug in the values:
- $R4 = 1k\Omega$
- $V_{supply} = 12.3V$
- $V_{base} = 6.15V$ (optimal between max supply nodes)
Calculating:

$R5 = \frac{6.15V \cdot 1kΩ}{(12.3V - 6.15V)} = \frac{6150}{6.15} \approx 1kΩ$

By our assumption:

Determine Current Path: You'll design to ensure proper base current for driver stability

This ensures the base current sets from calculated divider ensuring right saturation (Operation).

Practical Selection of R5:

Slight variation (real-case tuning) may still consider a value around

1kΩ

ensuring enough leeway for circuit stabilization:

Verification: The above calculation stabilizes considering practical licensors ensuring transistor maintains steady electron transfer for optimal current transduction.

If specific load conditions for lower stabilizing reconfiguration, practical tweaking may slightly vary the above balance for enhancing operable limits.

Would you want further assistance in optimal placement and calibration arrangements?

11mo

Someone is typing...

Why is the voltage at this node equal to 12.3V with the input voltage being 15V?

yudyud

@copilot Why is the voltage at this node equal to 12.3V with the input voltage being 15V?

11mo

copilot

There can be several reasons for a voltage drop from 15V to 12.3V at a particular node in your circuit. Let’s go through some potential causes:

Voltage Drop Causes

Resistive Drop: If there is a resistor or a series of resistive components between the input voltage and the node in question, there will be a voltage drop due to the current flowing through them.

$V_{drop} = I \cdot R$
Diode Forward Voltage: If there is a diode in the path, it will have a forward voltage drop (typically between 0.7V for silicon diodes and 0.3V for Schottky diodes).
Load on the Node: If there is a significant load (resistor or other components) connected to that node, it can draw current and cause a voltage drop.
Voltage Regulators: If there is a linear or switching regulator connected between the input voltage and the node, it can regulate the voltage down to 12.3V. Regulators can be designed to have specific output voltages quite commonly found in standard IC parts.
Parasitic Elements: PCB trace resistance or inductance could also contribute to a small voltage drop.

Steps to Troubleshoot

Check Circuit Components: Verify the components connected between the 15V input and the node. Look for resistors, diodes, or voltage regulators.
Measure Current: Measure the current through the path to identify if resistive drops are significant.
Verify Connections: Ensure that there are no loose connections or faulty components causing unintended drops.
Component Specifications: Verify the specifications of any regulators or converters to see if 12.3V is a plausible regulated output for those components.

Example Calculation

If there is a resistor of

R = 2.7 \, \Omega

and a current

I = 1 \, A

flowing through it, the voltage drop would be:

V_{drop} = I \cdot R = 1\,A \cdot 2.7\,\Omega = 2.7\,V

This would drop a 15V input to:

V_{node} = 15\,V - 2.7\,V = 12.3\,V

Summary

To determine the exact reason for the voltage drop in your circuit, you will need to carefully check each of these potential causes in the circuit design and physical layout. If you provide more details about the components and connections, I can give a more precise analysis. If you're using the ESP32 or a voltage regulator, remember that the ESP32 operates at 3.3V and any regulators will need to step down from your input voltage accordingly.

11mo

Someone is typing...

What is the effect of this resistor?

yudyud

@copilot What is the effect of this resistor?

11mo

copilot

To analyze the effect of a resistor in a circuit, we need to know which specific resistor you're referring to. In your 18650 battery charging circuit, you have several resistors (e.g.,

R_1

R_2

R_3

, etc.). Could you please specify the designator of the resistor you're interested in? For example,

R_1

R_{10}

, etc.

Alternatively, if you want to understand the general role of certain resistors in the context of the 18650 battery charging circuit, please let me know, and I'll explain based on their typical functions.

11mo

copilot

Sure, let's analyze the effect of resistor

R_8

From the provided netlist,

R_8

is connected as follows:

$R_8: P_1$ is connected to Net 5, which also includes $Q_1: E$ , $R_4: P_1$ , $R_{13}: P_1$ , $R_6: P_2$ , $R_{25}: P_1$ , $L_5: Pin_1$ , $L_1:1$ , $R_1: P_1$ .
$R_8: P_2$ is connected to Net 2, which includes $Q_4: E$ , $F_2: P_2$ , $D_{13}: cathode$ .

In a typical battery charging circuit,

R_8

(which has a resistance of 4.7 ohms) is likely serving one or more of the following roles:

Current Limiting: $R_8$ could be used to limit the current flowing through part of the circuit. Given that it's connected to $D_{13}$ (a diode, likely for reverse voltage protection) and fuse $F_2$ , limiting current is necessary to protect components or maintain a safe charging current for the 18650 battery.
Voltage Drop: By Ohm's Law, $V = IR$ , $R_8$ will cause a voltage drop proportional to the current passing through it. This can be important in setting the voltage levels at different points in the charging circuit.
Balancing and Filtering: In some designs, resistors are used in combination with capacitors to filter or stabilize the voltage. Given the placement of $R_8$ in Net 2 and with capacitors available in the circuit, this could also be a secondary function.

Specifically, within your circuit:

Connect $R_8: P_2$ to $Q_4: E$ , $F_2: P_2$ , and $D_{13}: cathode$ .
Connect $R_8: P_1$ to $Q_1: E$ , $R_4: P_1$ , $R_{13}: P_1$ , $R_6: P_2$ , $R_{25}: P_1$ , $L_5: Pin_1$ , $L_1:1$ , $R_1: P_1$ .

Would you like a more detailed explanation on any specific role?

11mo

yudyud

11mo

copilot

Let's examine the role of

R_8

in your circuit.

Analysis of $R_8$

Designator: $R_8$
Specifications:
- Resistance: 4.7Ω
- Power Rating: 3W
Connections:
- $R_8\colon P2$ to $Net\ 2$
- $R_8\colon P1$ to $Net\ 5$

Context in the Circuit

Net 2 contains:

$Q_4\colon E$
$F_2\colon P2$
$D_{13}\colon\ cathode$

Net 5 contains:

$R_{13}\colon P1$
$R_6\colon P2$
$R_{25}\colon P1$
$R_1\colon P1$
$Q_1\colon E$
$L_5\colon Pin_1$
$L_1\colon 1$
$R_4\colon P1$

From this context,

R_8

appears to be in the path from

Q_4\colon E

(emitter of transistor

Q_4

) to the common connection points of resistors

R_{13}, R_6, R_{25}, R_1

, and the emitter of

Q_1

, etc.

Possible Functions:

Current Limiting: $R_8$ could be used to limit the current through the emitter of $Q_4$ and other connected components.
Voltage Division: By creating a voltage drop across $R_8$ , it can help set biasing conditions for the connected transistors, particularly $Q_4$ and $Q_1$ .
Power Dissipation: With a power rating of 3W, $R_8$ can also handle significant power dissipation, indicating it's meant to manage higher currents in the circuit.

Given its low resistance (4.7Ω) and high power rating (3W),

R_8

is likely providing a small but crucial voltage drop and current control essential for the safe and proper operation of the charging circuit. This helps to regulate and stabilize components, ensuring they operate within safe limits.

11mo

Someone is typing...

Capacitance

0.1u F

R26

Resistance

220k Ω

Resistance

1k Ω

R18

Resistance

56k Ω

Resistance

3k3 Ω

R25

Resistance

3k3 Ω

Resistance

3k3 Ω

PowerNet1

Resistance

1k Ω

Resistance

39k Ω

R12

Resistance

1k Ω

D3-TL431

Resistance

56k Ω

R27

Resistance

56k Ω

R10

Resistance

56k Ω

Capacitance

0.1u F

Capacitance

0.1u F

Resistance

1k Ω

D1-TL431

Capacitance

0.1u F

D6-TL431

D2-TL431

Resistance

39k Ω

D12-LED1

Resistance

4.7 Ω

D13

D4-LED2

R13

Resistance

1k Ω

R11

Resistance

1k Ω

R20

Resistance

1k Ω

R17

Resistance

39k Ω

R21

Resistance

1k Ω

End of Life

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	Designators	Net
	F2:1, L6:1	Net 1
	D2-TL431:3, R11:2	Net 18
	R10:2, R17:1	Net 26
	Q7:3, R17:1	Net 26
	L5:1, R25:1	Net 5
	Q2:2, Q7:3	Net 26
	R4:1, R1:1	Net 5
	Q1:3, L5:1	Net 5
	Q1:2, Q6:3	Net 24
	R2:2, Q6:3	Net 24
	Q2:3, R9:1	Net 24
	R9:2, D2-TL431:1	Net 13
	L8:1, L7:1	Net 25
	R20:1, D2-TL431:2	Net 19
	D3-TL431:1, R17:2	Net 14
	L2:1, L3:1	Net 30
	R7:1, D12-LED1:1	Net 10
	R2:1, R1:2	Net 12

yudyud / 18650 battery charging circuit

Schematic

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