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for Q5 what is the max voltage recomennded to come into the anode
ril3y

ril3y

if we supplied 3v3 into the annode what resistor would we need to limit the current to 50ma to protect the led?
1y
ril3y

ril3y

@copilot for Q5 what is the max voltage recomennded to come into the anode
1y

copilot

The component designator Q5 is an optocoupler with the part number EL817(B)-F. The EL817 series optocouplers are designed to isolate electrical signals between two circuits by using light. The maximum forward voltage (V_F) of the LED inside the optocoupler is an important parameter to consider when determining the maximum voltage that can be applied to the anode.
The EL817 datasheet typically specifies a forward voltage (V_F) of about 1.2V at the recommended forward current (I_F). However, the absolute maximum forward current (I_Fmax) is a critical parameter that determines the maximum voltage that can be safely applied. For the EL817(B)-F, the absolute maximum forward current is 50mA.
Given a typical forward voltage (V_F) of 1.2V and assuming you'll be operating the LED within its recommended current range (not exceeding the absolute maximum ratings), the voltage at the anode should ideally not exceed this typical forward voltage by a significant amount when the recommended forward current is applied.
However, the actual maximum voltage you can apply at the anode is also influenced by the current-limiting resistor you connect in series with the LED. In practice, to calculate the maximum voltage (V_max) that can be applied to the anode, you need to consider the source voltage (V_S), the LED's forward voltage (V_F), and the forward current (I_F) you plan to apply through the resistor (R_series):
Vmax=VF+(IF×Rseries)V_{max} = V_F + (I_F \times R_{series})
Since you haven't specified the value of the series resistor or the forward current you intend to use, I'll assume a typical forward current of 20mA (0.02A) for calculation purposes. Assuming you're using a current-limiting resistor calculated for a 5V source (which is common in many circuits):
Rseries=VS−VFIFR_{series} = \frac{V_S - V_F}{I_F}
Plugging in the typical values for a 5V source:
Rseries=5V−1.2V0.02A=190ΩR_{series} = \frac{5V - 1.2V}{0.02A} = 190 \Omega
Given this scenario, the maximum voltage you apply to the anode would be around the forward voltage of 1.2V plus whatever voltage drop occurs across the resistor due to the forward current.
Keep in mind, the applied voltage can actually be higher than 1.2V as long as you ensure the current through the LED doesn’t exceed the maximum forward current by carefully selecting your series resistor value based on the source voltage. Always refer to the specific datasheet for the most accurate and application-specific recommendations.
1y
ril3y

ril3y

if we supplied 3v3 into the annode what resistor would we need to limit the current to 50ma to protect the led?
1y

copilot

To limit the current to 50mA (0.05A) through the LED in the optocoupler (Q5) when applying a supply voltage of 3.3V, we can use Ohm's Law ((V = IR)) and the formula for determining the value of a current-limiting resistor ((R)):
[R = \frac{V_{S} - V_{F}}{I}]
Where:
  • (V_{S}) is the supply voltage (3.3V in this case),
  • (V_{F}) is the forward voltage of the LED (typically around 1.2V for the EL817),
  • (I) is the desired current (0.05A).
Plugging in the numbers:
[R = \frac{3.3V - 1.2V}{0.05A} = 42\Omega]
However, using a resistor value to drive the LED at its absolute maximum forward current is generally not recommended for continuous operation as it can significantly reduce the LED's lifespan and reliability. It is typically better to operate the LED well below its maximum rating to ensure longevity and stable performance—common practice suggests using 70-80% of the maximum rating as a safer operating point.
But directly addressing your question with the assumption of needing to limit the current to 50mA with a 3.3V source, a (42\Omega) resistor would technically achieve that current limit. However, for reliability, considering a lower current might be preferable.
1y
ril3y

ril3y

ok what would 80% resistor need to be
1y

copilot

If we aim to operate at 80% of the 50mA maximum recommended current for the LED within Q5, we first calculate 80% of 50mA:
[I = 50mA \times 0.80 = 40mA = 0.04A]
Next, we calculate the required resistor value to achieve this current level with a 3.3V supply and a forward voltage ((V_F)) of 1.2V for the LED:
[R = \frac{V_S - V_F}{I} = \frac{3.3V - 1.2V}{0.04A}]
[R = \frac{2.1V}{0.04A} = 52.5\Omega]
Therefore, a resistor value of (52.5\Omega) would be required to limit the current to 40mA, which is 80% of the absolute maximum rating of 50mA for the optocoupler LED. Since it's unlikely you'll find a resistor with exactly (52.5\Omega), you would typically select the closest higher standard value resistor available to ensure the current does not exceed 40mA. The closest higher standard value is (56\Omega).
1y
please explain what Q5 does.
ril3y

ril3y

@copilot please explain what Q5 does.
1y

copilot

Q5, designated as a part of your project, is an optocoupler (also known as an opto-isolator) with the specific part number EL817(B)-F. In the context of your project, here's what Q5 does:
  • Function: Optocouplers like Q5 are used to transmit signals between two isolated circuits by using light to modulate the signal from input to output. They help in preventing high voltages from affecting the system receiving the signal.
  • Pin Configuration:
    • Pin 1: Anode of the LED inside the optocoupler.
    • Pin 2: Cathode of the LED.
    • Pin 3: Emitter of the output transistor.
    • Pin 4: Collector of the output transistor.
  • Operation: When a current flows through the LED (Anode to Cathode), it emits light. This light falls on the phototransistor, causing it to turn on. Thus, the signal on the input side (LED side) gets transferred to the output side (transistor side) without electrical connection, providing electrical isolation between the input and output.
  • Role in Your Project: Based on the net connections in your project:
    • The Anode (Pin 1) is connected to the 1.2V net (inferred from net b437f0efabc186114c927ad45b7cceba), possibly serving as the control signal input.
    • The Cathode (Pin 2) is linked to the 48V net through R20 and R21 (inferred from net b437f0efabc186114c927ad45b7cceba), which is likely part of biasing or enabling circuit for the optocoupler.
    • The Emitter (Pin 3) connects to the common ground net, ensuring the lower side of the output transistor is grounded.
    • The Collector (Pin 4) to J6 pin 2 (inferred from net 576617262850c4f68bba55b259a36e2c), indicating the output of Q5 is used to interface with an external device or system, potentially to signal a control logic level based on the presence of the input signal.
The exact operation detail and the reason for using Q5 in this way would depend on further context regarding the role of the connected nets and components, such as the specific function Q5 is intended to perform within the overall circuit logic, such as signaling, switching, or isolation purposes.
1y
5
if 12v is present at the gate of q3 and 48v is at the drain of q3 will allo q3 t...
ril3y

ril3y

@copilot if 12v is present at the gate of q3 and 48v is at the drain of q3 will allo q3 to start conducting current though its drain to its source at 48v?
1y

copilot

Based on your description, you're inquiring about the behavior of Q3 (an IRFR120NTRPBF MOSFET) when a 12V signal is applied to its gate (G) and a 48V potential is present at its drain (D), with respect to its source (S) conducting at 48V.
For a MOSFET like the IRFR120NTRPBF, which is an N-channel MOSFET, the key parameter to understand in this scenario is its threshold voltage ((V_{th}))—the minimum gate-to-source voltage ((V_{GS})) required to turn on the MOSFET and allow current to flow from the drain to the source effectively.
If the gate-to-source voltage ((V_{GS})) is greater than the threshold voltage ((V_{th})), the MOSFET will enter the "on" state and conduct current from the drain to the source. However, the actual behavior also depends on how the MOSFET is incorporated into the circuit, particularly regarding how the source voltage ((V_S)) and gate voltage ((V_G)) are determined.
In your case, applying 12V to the gate and having 48V at the drain:
  1. The source voltage ((V_S)) is crucial but not specified. For the MOSFET to conduct, (V_{GS}) (the voltage difference between the gate and the source) needs to exceed the (V_{th}).
  2. If we assume the source is connected to a common ground (0V) and the MOSFET's (V_{th}) is well below 12V—common (V_{th}) values for power MOSFETs range from 2V to 4V—applying 12V between the gate and source would indeed turn the MOSFET on.
  3. In this "on" state, assuming the source is at ground level, the MOSFET would attempt to pull the drain towards the source voltage, enabling current flow if a complete path (circuit) exists for the current. The voltage at the drain would not be directly 'pulled down' to ground because the MOSFET acts as a variable resistor in this state, and actual drain-to-source voltage ((V_{DS})) would depend on the resistance in the conducting (on) state ((R_{DS(on)})) and the load connected to the circuit.
To sum up, if 12V is present at the gate relative to the source (ensuring (V_{GS} > V_{th})) and 48V is at the drain relative to the source, Q3 will indeed start conducting, allowing current to flow from the drain to the source, underlining the importance of the relative voltages to ensure proper operation.
1y
11
should i put r4 on the cathode ground side to avoid the voltage dropping too low...
ril3y

ril3y

@copilot should i put r4 on the cathode ground side to avoid the voltage dropping too low to turn on the q1 full?
1y

copilot

Placing R4 on the cathode side of Q1 in your circuit is a strategic move if you're trying to control the voltage seen by Q1's gate with respect to its source (for a FET) or base (for a BJT). Based on your description, it looks like you're considering the scenario where, in the absence of R4, the voltage at the cathode of Q1 could potentially drop too low, affecting the turn-on capability of the transistor.
If Q1 is a MOSFET, the gate-to-source voltage ($V_{GS}$) determines whether the transistor turns on fully (enters saturation) or not. If Q1 is a BJT, then it’s the base-emitter voltage ($V_{BE}$) that’s critical for turning the transistor on. In both cases, you want to ensure that the voltage is sufficient to fully turn on the transistor.
For a MOSFET, typically a $V_{GS}$ of 10V is more than enough to turn most power MOSFETs fully on, though many can turn on with less. For a BJT, a $V_{BE}$ above approximately 0.7V is generally needed for full turn-on.
Placing a resistor (like R4) in the correct position to ensure the voltage drop across the transistor's control terminal (gate or base) and its reference (source or emitter) is within an acceptable range to ensure full turn-on is a good practice. This is especially critical in circuits with high currents, where even small resistance values can produce significant voltage drops due to Ohm's Law ($V=IR$).
Given this, if your intention with R4 is to drop a specific voltage to control the operation of Q1 more precisely, then ensure you calculate the necessary resistance value based on the expected current through that path and the desired voltage drop. Make sure this configuration doesn't drop the voltage too low, as that would counteract your intention to turn Q1 on fully. Essentially, your decision should hinge on a careful balance between ensuring enough voltage to turn Q1 on while also managing any undesirable effects of having the resistor in place.
Remember, depending on whether Q1 is a MOSFET or BJT, your approach might need to slightly differ since their operation and requirements for "fully on" differ.
1y
7
do i need r4 at all? precharge to the anode is 3v3
ril3y

ril3y

@copilot do i need r4 at all? precharge to the anode is 3v3
1y

copilot

Given that the precharge to the anode is at 3.3V, let's review the role of resistor R4R4 in your circuit.
  • R4 Connection: Connects from precharge to Q1Q1's anode
  • Purpose of R4: Typically, a resistor in this position would be used to limit the current flowing into the anode of an optocoupler (Q1Q1, in this case). This is essential for protecting the LED inside the optocoupler from excessive current which could otherwise lead to its damage.
Given the context that the precharge voltage is 3.3V, whether you need R4R4 or not depends on the forward voltage and maximum forward current of Q1Q1's internal LED.
  • Assumption: Assuming Q1Q1 is a typical optocoupler with a forward voltage of approximately 1.2V to 1.4V and a desired forward current of 10mA for proper operation.
The value of R4R4 can be calculated using Ohm's Law:
R=Vsource−VfIfR = \frac{V_{source} - V_{f}}{I_{f}}
Where:
  • Vsource=3.3VV_{source} = 3.3V (the precharge voltage)
  • Vf≈1.2VV_{f} \approx 1.2V (forward voltage of the LED inside the optocoupler)
  • If=10mA=0.01AI_{f} = 10mA = 0.01A (desired forward current)
R=3.3V−1.2V0.01A=210ΩR = \frac{3.3V - 1.2V}{0.01A} = 210\Omega
Considering this calculation, if the actual value of R4R4 around this calculated value (210Ω), it suggests that the resistor's purpose is to limit the current to a safe level for the optocoupler's internal LED, indicating that R4R4 is indeed necessary.
However, if the precharge directly to the anode is consistently at 3.3V, and the current needs to be limited to prevent exceeding the optocoupler's rating, then yes, R4R4 would still be necessary to ensure the safe operation of Q1Q1. Skipping this resistor could risk damaging the optocoupler by allowing too much current to flow through its internal LED. Always refer to the specific components' datasheets for accurate parameters.
Do check Q1Q1's datasheet to confirm the forward voltage (VfV_f) and the recommended forward current (IfI_f) for more precise calculation or validation of this assumption.
1y
ril3y

ril3y

is 3v3 in precharge enough to fully turn on q1?
1y

copilot

To determine if 3.3V is sufficient to fully turn on Q1Q1, an optocoupler in this case, we need to consider two main parameters of the optocoupler's internal LED: the forward voltage (VfV_{f}) and the current required for proper operation.
The forward voltage (VfV_{f}) is the voltage drop across the LED inside the optocoupler when it's conducting. For many optocouplers, VfV_{f} typically ranges from about 1.1V to 1.4V, but this value can vary based on the specific device and manufacturer. The datasheet for Q1Q1 should provide this exact value.
Assuming that Q1Q1's internal LED has a VfV_{f} within the common range mentioned, and considering that most optocouplers can be effectively turned on (i.e., get into active operation mode) with current as low as a few milliamperes (mA), a supply voltage of 3.3V is generally enough to turn on the optocoupler, given appropriate current limiting resistance to protect the LED.
From the earlier calculations with resistor R4R4, we used a rough estimation to find a resistance value that would allow around 10mA of current (a common value for many LEDs to be well within operating range) through the optocoupler's LED with a supply voltage of 3.3V:
R=Vsource−VfIf=3.3V−1.2V0.01A=210ΩR = \frac{V_{source} - V_{f}}{I_{f}} = \frac{3.3V - 1.2V}{0.01A} = 210\Omega
This calculation suggests that with a suitable limiting resistor (like R4R4 calculated as 210Ω, or close depending on optimal design and safety margins), the 3.3V supply voltage is indeed enough to fully turn on Q1Q1's internal LED.
However, it's critical to review Q1Q1's datasheet for the exact forward voltage and recommended forward current to make precise calculations and ensure that 3.3V will sufficiently operate the device within its safe operational parameters. Remember, turning on the optocoupler's LED is one part, but the ability to transfer the signal effectively through the optocoupler depends on the current through the LED and the characteristics of the specific optocoupler being used.
1y
8
for U1 there needs to be a cap connected to VOUT but the datasheet doesnt say th...
ril3y

ril3y

@copilot for U1 there needs to be a cap connected to VOUT but the datasheet doesnt say the value needed can you help figure this out. datasheet is here: https://www.lcsc.com/datasheet/lcsc_datasheet_2304140030_Allegro-MicroSystems--LLC-ACS70331EOLCTR-005U3_C2155535.pdf
1y

copilot

The Allegro ACS70331 mentions that it's a high sensitivity, galvanically isolated linear current sensor IC which operates on the principle of the Hall effect. However, the datasheet you've linked requires an account to access, and I cannot directly review it as of my last training cut-off in September 2023. Generally, I can provide you with some guidance on how to select an appropriate capacitor value for VOUT in applications like this one.
For Hall effect sensors like the U1 (ACS70331), the capacitor on the VOUT pin typically serves two functions:
  1. Filtering: Reduce the noise on the sensor output signal, which can come from the power supply or be induced by electromagnetic interference in the environment.
  2. Stability: Ensure the output signal is stable without oscillations or abrupt changes that might occur due to the internal dynamics of the sensor or due to the layout of the PCB.
The selection of the capacitor value depends on several factors:
  • Bandwidth Requirement: A smaller capacitor value will provide higher bandwidth, meaning the sensor can accurately follow faster changes in the current it measures. This might be desirable in fast-switching applications.
  • Noise Filtering: A larger capacitor will provide better filtering of noise but will slow down the response of the sensor, which might not be suitable for applications requiring quick detection of current changes.
Given these considerations, and lacking specific advice from the datasheet, I would recommend starting with a commonly used value in signal filtering applications, which is a 0.1uF ceramic capacitor for digital applications. This should provide a decent balance between noise filtering and maintaining a suitable response time for many applications.
If your application is sensitive to noise (requiring smooth DC measurements, for instance), you might also consider adding a larger capacitor in parallel (e.g., 10uF) to further smooth out any low-frequency voltage variations on the output.
So your connection would be:
  • Connect C_(appropriatedesignator)(appropriate designator):P1 to U1:VOUT
  • Connect C_(appropriatedesignator)(appropriate designator):P2 to a ground point, ideally close to U1:GND for return path minimization.
Remember, this recommendation is a good starting point, but you might need to adjust based on your application specifics or after observing the performance in prototype testing. It would also be beneficial to consult the datasheet or application notes from Allegro MicroSystems for any recommended practices specific to this sensor.
1y
12
L1
Inductance
2.2u H
LED1
PGND
BOOT Selector Mode
GND
Note 3
Line 1
RB Resistor Selection
Line 2
3.3v - R1 33.2k, R2 10k
Line 3
5v - R1 54.9k, R2 10k
R2
Resistance
220 Ω
12V
RESET
CANBUS INPUT
R1
Resistance
220 Ω
BOOT1
3V3
R12
Resistance
110 Ω
GND
3V3
ANALOG PINS
Line 1
PA0~PA7 are all 0~3v3 12bit ADCs
3V3
R15
Resistance
33.2k Ω
PGND
CONTACTOR
RESET
Y2
PRECHARGE
C6
Capacitance
.1u F
LED4
LED_VAL2
GND
C2
Capacitance
1u F
C4
Capacitance
.1u F
Q5
3V3
VOLTAGE_SENSE
Note 4
Line 1
20 to 68 uF
LED_VAL1
GND
R5
Resistance
10k Ω
C1
Capacitance
10u F
GND
C12
Capacitance
.1u F
KEY INPUT
J6
LED_VAL1
C3
Capacitance
.1u F
48V
R7
Resistance
390 Ω
Q2
3V3
U2
Manufacturer Part Number
STM32F103C8T6
CANTX
12V
Note 2
Line 1
3v3 @ 1A
C9
Capacitance
10u F
R9
Resistance
10k Ω
LED_VAL4
R10
Resistance
10k Ω
GND
VOLTAGE_SENSE
LED_VAL2
CONTACTOR
LED3
GND
PGND
PGND
J4
LED2
KEY_INPUT-48v
IC1
Manufacturer Part Number
TPS561208DDCR
R13
Resistance
10k Ω
J3
MOTOR+
Q3
RESET
3V3
LED_VAL4
3V3
CANTX
C8
Capacitance
.1u F
R8
Resistance
10k Ω
Logic Key
Line 1
When 12V is high,
Line 2
TTL is low
C11
Capacitance
22u F
GND
RESET CIRCUIT
CONTACTOR_SW
Q1
3V3
Out to Motor Controller
CANRX
CONTACTOR
12V
SWCLK
R21
Resistance
715 Ω
J5
C7
Capacitance
22u F
3V3
LED_VAL5
Decoupling Caps
J2
Current Monitoring Section
48V
C13
Capacitance
22p F
PGND
R16
Resistance
1k Ω
confirmed
confirmed
GND
CONTACTOR CIRCUIT
BOOT0
C14
Capacitance
22p F
Key Switch Input
Line 1
48V comes into P1 which goes though the opto anode then back to p2 (-48v gnd) from the cathode. This will drive the START_TTL_IN low
CONTACTOR_SW
CANRX
LED_VAL3
LED5
GND
LED_VAL3
GND
12V
R4
Resistance
110 Ω
R6
Resistance
220 Ω
PRECHARGE
MOTOR+
Q4
End of Life
C5
Capacitance
.1u F
GND
PGND
PGND
CL Calculations
Line 1
(CX+strayC)/2
Line 2
(39pf + 5)/2=22pf
BOOT1
J1
U1
Not Recommended for New Designs
Part Type
Sensor
Manufacturer Part Number
ACS70331EOLCTR-005U3
GND
R17
Resistance
1k Ω
confirmed
confirmed
R14
Resistance
10k Ω
48V
R3
Resistance
220 Ω
R7 is a power resistor
GND
SWCLK
GND
C10
Capacitance
10u F
R18
Resistance
56 Ω
R20
Resistance
10k Ω
D1
My Logic
Line 1
Q2 should start conducting which would allow for Q1 to switch 48V into IP+ pin. This will come out IP- pin on the current monitor (u3). If Q2 does not conduct it should make Q1 open as well and no current should be conducting through Q1.
BOOT0
GND
R19
Resistance
4.7k Ω
SWDIO
Minimal Design Schematic
Line 1
http://www.haoyuelectronics.com/Attachment/STM32F103C8T6-DEV-BOARD/STM32F103C8T6-DEV-BOARD-SCH.pdf
SWD Interface
GND
SWDIO
3v3 output
START_TTL_IN
GND
GND
GND
GND
LED_VAL5
R11
Resistance
220 Ω
  • Terminal
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Inspect

current_limiter 9251

current_limiter 9251
Description

Created
Last updated by ril3y
1 Contributor(s)
ril3y

Controls

Properties

Availability & Pricing

DistributorQty 1
Digi-Key$6.92–$9.65
LCSC$3.00–$4.34
Mouser$8.33–$8.96

Assets

    Simulation

    µs

    Simulation Speed