An electrical connector acting as reusable interface to a conductor and creating a point where external circuits can be connected.Terminal- A net identifier used to create connectivity to other Net Portal with the same designator. #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
- 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
- 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
- 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
- 10uF Capacitor Aluminum Polymer 20% 16V SMD 5x5.3mm #forLedBlink #commonpartslibrary #capacitor #aluminumpolymer #radialcan
- Yellow 595nm LED Indication - Discrete 1.7V 1206 (3216 Metric) #forLedBlink
- Battery Holder (Open) Coin, 20.0mm 1 Cell SMD (SMT) Tab bate or batt #forLedBlink
U1
how would you make a lm393 delay pull up resistor circuit
10mo
To create a delay circuit using an LM393 comparator and incorporate a pull-up resistor, you would generally be looking at a setup where the comparator is used to trigger a delay to a stable output state. The LM393 is a dual differential comparator, which could be utilized in configuring a simple delay circuit, especially when combined with an RC timing network. Here’s a basic approach:
Components Needed:
- LM393 Comparator (assuming we use U4 from your project): This IC has two comparators; you can use one for the delay circuit.
- Resistor (R): This will be part of the RC delay network. Let's assume a value of 10kΩ (R16, for demonstration purposes).
- Capacitor (C): This, combined with the resistor, will create the RC time constant for your delay. Assuming a value of 100µF (C10, for example).
- Pull-up Resistor (R_pullup): This resistor is needed to pull the output to a high level when the comparator’s output is in the high impedance state. Let's assume a value of 10kΩ (R17).
-
Set Up the RC Network for Delay:
- Connect one end of R16 to U4:2IN- and the other end to the junction of C10:P1 and C10:P2 (where P2 is also connected to ground).
- Connect U4:2IN+ to a reference voltage (V_ref) that the input voltage (Vin) will be compared against. V_ref can be set using a voltage divider or a reference voltage source.
-
Pull-up Resistor:
- Connect R17:1 to U4:2OUT and R17:2 to your positive supply voltage (let's assume you're using 5V from your project).
-
Output:
- Directly take the output from the junction of U4:2OUT and R17:1. This node goes to the input of the following stage after the delay.
-
Input and Ground Connections:
- Vin should be applied to the RC network (R16:C10 junction).
- Ensure U4:GND (pin 4) is connected to your system ground.
- VCC of U4 (pin 8) should be connected to the same positive supply voltage as R17:2 (assumed to be 5V).
- When Vin rises above V_ref, the comparator output becomes open circuit (since LM393 outputs are open collector), and R17 pulls the output high.
- The RC network dictates how quickly the voltage at U4:2IN- rises, creating a delay before the output transitions states. The time delay (τ) can be approximated by the formula τ = R×C, where τ is in seconds if R is in ohms and C in farads.
- As Vin falls below V_ref, the comparator drives the output low, overcoming the pull-up resistor.
- I assumed you have a 5V supply available in your project for the pull-up resistor and that the general purpose of the circuit is to introduce a delay based on an input voltage exceeding a reference voltage.
- Values for R16 (10kΩ), C10 (100µF), and R17 (10kΩ) are placeholders. Based on your delay requirement, adjust R and C values accordingly. Smaller R/C values result in a shorter delay and vice versa.
If you need adjustments or have different requirements for the delay (such as a specific delay time or operating voltage), feel free to specify!
10mo
Someone is typing...
10
how would you make a negative dc offset to this amplifier
@copilot how would you make a negative dc offset to this amplifier
11mo
Someone is typing...
9
which lora version has the sx1276 semtech lora transceiver
11mo
The Semtech SX1276 LoRa transceiver is often associated with the LoRaWAN protocol version 1.0.2. However, you should check with your module or transceiver's datasheet or manufacturer's specification to confirm the specific version and compatibility.
11mo
Someone is typing...
4
tell me the difference between the hex inverter 74ls04 and the 74HC14D
11mo
Sure, both 74LS04 and 74HC14D are popularly used as logic inverters (also known as NOT gates), and they are a part of the 74-series logic gate ICs. However, there are significant differences between the two. Here they are:
-
Technology:
- The 74LS04 is a part of the Low-power Schottky (LS) family, which is a bipolar technology.
- The 74HC14D is a part of the High-speed CMOS (HC) family, which as the name implies is based on CMOS technology.
-
Input and Output Levels:
- The 74LS04, being based on LS technology, has a lower noise margin and a narrower range of acceptable input and output logic levels.
- Being a CMOS device, the 74HC14D has better noise immunity and a wider range for input and output logic levels.
-
Schmitt Trigger Feature:
- The "14" in 74HC14D indicates that it has a Schmitt trigger input, which provides hysteresis and allows the chip to smooth out the signal, making it more resistant to noise and ringing.
- The 74LS04 is just a plain inverter and does not have Schmitt trigger inputs.
-
Number of Gates:
- Both 74LS04 and 74HC14D contain six separate inverters.
-
Speed:
- The propagation delay time (a measure of speed) differs as well. The 74LS04 typically operates a bit slower than the 74HC14D.
-
Power Consumption:
- The 74HC14D, being a CMOS device, will consume much less power than the 74LS04 when not switching. The 74LS04 will consume more power, especially when switching frequently.
Overall, you'd want to use the 74HC14D if you need its Schmitt trigger capabilities (for example, to clean up a noisy signal), if you need its CMOS-level logic, or if low static power consumption is important. You might use the older 74LS04 if speed is not as critical, or if you need to interface with other LS-series parts.
11mo
Someone is typing...
3
What are some 74HC14D alternatives? because i just need one digital input, i am ...
11mo
Sure. The SN74AHC14QPWRQ1 you're using is a hex inverting Schmitt trigger, meaning it contains 6 individual inverting Schmitt triggers in a single package. If you're only in need of a single-channel device, you could use one of the following alternatives:
- 74HC1G14GW: It is a single gate, single input Schmitt trigger inverter made by NXP Semiconductors.
- SN74LVC1G14DBVR: This Texas Instruments chip is also a single-gate Schmitt trigger inverter, with similar properties to the 74HC14.
Both alternatives are widely available and have similar electrical characteristics to the 74HC14D. Make sure to verify the specifications and packages for your specific use case.
However, please also keep in mind that even if you are not using all the pins on the SN74AHC14QPWRQ1, there is no harm in using it. It might be more economical or practical depending on available components and PCB space.
11mo
Someone is typing...
1
How much voltage does the AC172 drops when connected to a 1A circtuit?
11mo
I'm sorry, but I couldn't find an AC172 component in the project description you provided. Could you please confirm whether this component is present in your project? Moreover, AC172 seems to be a transistor rather than a voltage-dependent component. Usually, voltage drop is associated with components like resistors or diodes. Are you sure you want to know the voltage drop across AC172 or is there another specific component you're referring to?
Please provide more context so that I can assist you better.
11mo
sorry it was the AC712, a current sensor that would allow me to sense current through a cable using a hall effect, i want to implement it in my project but i was wondering how much voltage drops so i cause the minimum effect on the circuit where i would place the sensor
11mo
The AC712, more commonly referred to as ACS712, is a fully integrated, Hall effect-based linear current sensor IC from Allegro MicroSystems. It is very popular for its precision and sensitivity.
It's important to note that the ACS712 is a current sensor module, not something like a resistor that would introduce a voltage drop in your circuit. It's designed to cause minimal disruption to the circuit that it's measuring.
The ACS712 operates on the basis of the Hall effect, measuring both AC and DC currents. The sensor does not have to be in direct contact with any part of the circuit. Instead, the electrical current that the sensor measures generates a magnetic field, which the sensor then measures to determine the current.
These types of sensors generally have a very low resistance (a few milliohms), hence they introduce a very small voltage drop in the circuit they are measuring, typically in the millivolt range.
Before using it in your project, it's crucial to verify this with the specific datasheet of your model since there are different versions of the ACS712 sensor with different current measuring capabilities (5A, 20A, 30A).
Additionally, remember that to get accurate readings, you would need to tune the sensor properly based on the control voltage given out by the ACS712 for zero current flow.
11mo
explain to me the last part, how do i tune the sensor?
11mo
The ACS712 provides a analog voltage output that can be read by an analog-to-digital converter (ADC) channel on a microcontroller. This analog output voltage is proportional to the current flowing through the sensor.
When no current is passing through the sensor (known as the "zero current" condition), the sensor doesn't output a zero voltage. Instead, it outputs a mid-point voltage also called "Zero Current Output Voltage" ((V_{OUT(Q)})) which for ACS712 is typically around 2.5V. The output voltage then rises or falls proportionally around this "Zero Current Output Voltage" as the sensed current increases or decreases.
During the setup of the sensor in your circuit, you'll want to "tune" or "calibrate" your microcontroller's readings to know what output voltage of the sensor corresponds to zero current. This is usually done by reading the value of (V_{OUT(Q)}) from the sensor at powerup (when you are certain no current is flowing through the sensor), and storing that value as the "zero current" point. Once you've done this, the microcontroller can calculate the actual current by comparing future readings from the sensor to this calibrated zero point.
Remember that the ADC reading from the sensor will have to be scaled to the microcontroller's ADC resolution and reference voltage. For instance, if you are using a 10-bit ADC with 5V reference, then each bit is equivalent to ( \frac{5V}{2^{10}-1} \approx 4.89mV ). So, a zero current reading of 2.5V by the sensor would correspond to approximately 512bit on the ADC reading. This is the value that one would consider the zero-current point when calculating actual current from the sensor output.
There can be some variability due to factors like power supply voltage variation, component tolerances etc., so it's recommended to calibrate the sensor in the actual conditions in which the sensor will be used. Calibration should also be rechecked periodically, as the zero-point can sometimes drift over time.
11mo
Someone is typing...
5
can you tell me if this connection is alright for a operational apmplifier
11mo
For evaluating the connectivity of an operational amplifier (op-amp), typically, we look at how the inputs, outputs, power supply, and any feedback mechanisms are connected. However, you haven't specified which connection in your project you're referring to. Given the project details, I see that there's an LM358P op-amp designated as U2 with various pin assignments.
For a general overview, the LM358P is a dual operational amplifier. Here are some common practices for connecting such an op-amp in a circuit:
- Power Supply:
- Connect U2:4 (V-) to a ground reference.
- Connect U2:8 (V+) to a positive power supply. This can be a voltage within the range that the op-amp supports, typically 3V to 32V. In your schematic, this is probably connected to the VIN or 5V net for operation.
Based on the net connections provided in your project:
-
U2:8 (V+) is part of the net tagged
c791f3e3fe4e1370fae9b8ddf5c1a714, which is indeed connected to 5V and VIN, confirming that the power supply connection is correctly made to the op-amp. -
U2:4 (V-) is part of the net tagged
9632690c6cd6203abf7c1d9cd477fbcf, which needs to be connected to a ground reference for correct operation. Ensure this net connects to a ground in your project. -
Input and Output Connections:
- For a simple op-amp configuration (e.g., non-inverting amplifier), you would connect your input signal to U2:+IN1 (3), use U2:-IN1 (2) to set up feedback (often involving resistors to ground and back to this pin), and take your output from U2:OUT1 (1).
Without specific details on which connection you're referring to, I've made assumptions based on standard op-amp usage and your provided project description. If there's a specific connection you're unsure about, could you provide more details? For example, are you looking into creating a specific type of amplifier (e.g., non-inverting, inverting, differential)? This will help in providing a more tailored response.
11mo
im tring to make an inversor amplifier wiht 100 of gain for a hall current sensor
11mo
For an inverting amplifier configuration using your LM358P op-amp (U2) with a gain of -100 for interfacing with a Hall current sensor, you need two resistors. The gain of an inverting amplifier is given by the formula
Gain=−RinRf
where:
- (R_f) is the feedback resistor connected from the op-amp's output to its inverting input.
- (R_{in}) is the input resistor connected from the inverting input to the signal source.
Given that you want a gain of -100, you could choose a 100kΩ resistor for (R_f) and a 1kΩ resistor for (R_{in}). These values are quite typical and should be easily available. Let's assume R13 (100kΩ) could act as your (R_f) and R11 (1kΩ) as your (R_{in}) based on the components listed in your project description.
Assuming U2:+IN2 is not used and connected to ground through a resistor to set the bias correctly (this could be a typical connection setup not specified but is recommended for unused op-amp units to ensure stability).
Steps to Configure U2 as an Inverting Amplifier:
-
Input Connections:
- Connect the output of the Hall current sensor (let's assume it outputs its measurement through a designated net, say
SCT1:P1) to R11:P1. - Connect R11:P2 to U2:-IN1.
- Connect the output of the Hall current sensor (let's assume it outputs its measurement through a designated net, say
-
Feedback Connections:
- Connect R13:P1 to U2:OUT1.
- Connect R13:P2 to U2:-IN1.
-
Grounding and Power Supply:
- Ensure U2:+IN1 is connected to ground to properly bias the unused op-amp unit (if not using dual characteristics). This can be through a typical value resistor, say a 10kΩ, to ground but it's not expressly listed in your items.
- Connect U2:V- to GND.
- Connect U2:V+ to a suitable positive power supply, e.g., 5V from your power net designation.
-
Output:
- The output of your inverting amplifier will be available at U2:OUT1.
-
Unused Op-Amp Configuration (for LM358P which is a dual op-amp):
- It's a good practice to configure the unused op-amp in a stable configuration to avoid oscillation or interference.
- For the LM358P's unused section (+IN2 and -IN2), connect +IN2 to ground and OUT2 directly to -IN2 which sets it as a voltage follower that won’t oscillate.
Given your existing components:
- Connect SCT1:P1 to R11:P1.
- Connect R11:P2 to U2:-IN1, R13:P2.
- Connect R13:P1 to U2:OUT1.
Keep in mind, the actual connection to SCT1:P1 or the Hall sensor's output pin might need to be adjusted based on your specific sensor and its wiring situation in your schematic. Also, ensure the feedback and input resistors meet the precise requirements (in value and tolerance) needed for your application to achieve the desired gain accurately.
11mo
Someone is typing...
8
Vref
R7
Resistance
3.3k Ω
3.3V
R1
Resistance
1k Ω
D2
C1
Capacitance
470u F
J6
2IN+
R9
Resistance
1k Ω
Q1
VIN
D3
R2
Resistance
1k Ω
D3
ADC 2
R11
Resistance
1k Ω
D1
3A
3A
ADC 1
3.3V
VIN
ADC 3
Vref
Q2
ADC
4IN+
C2
Capacitance
470u F
R8
Resistance
3.3k Ω
3.3V
R3
Resistance
5k Ω
3.3V
3.3V
C7
Capacitance
0.1u F
J8
D4
C6
Capacitance
0.1u F
J14
J1
U3
J2
AutentioNode
Controls
Properties
Availability & Pricing
| Distributor | Qty 1 | |
|---|---|---|
| Digi-Key | $4.91–$12.91 | |
| LCSC | $5.95–$6.31 | |
| Mouser | $8.10–$13.39 | |
Assets
Simulation
WHAT IS THE MAX VOLTAGE THE CD4025 ADMITS\_
how to make a signal widht extender to make a 0.7v peak 2ms signal transform it ...
10mo
To extend a 0.7V peak, 2ms pulse signal to a 10ms signal using a diode (such as the 1N5818, assuming the 1N4818 was a typo), a capacitor, and a resistor, you can employ a simple RC (resistor-capacitor) circuit to achieve pulse stretching. The diode will prevent the capacitor from discharging quickly back into the signal source. Here's a basic approach:
Components Required:
- Diode: 1N5818 (a Schottky barrier diode, chosen for its low forward voltage drop).
- Resistor (R): To determine the resistor value, we use the RC time constant equation.
- Capacitor (C): To set the desired pulse width extension.
-
Theoretical Background:
- The RC time constant ((\tau)) is defined as (\tau = R \cdot C), where:
- (\tau) is the time constant (in seconds),
- (R) is the resistance (in ohms),
- (C) is the capacitance (in farads).
- A pulse is considered to have returned to 63.2% of its initial value after one time constant.
- For a signal to be extended to approximately 10ms ((T)), you may aim for a time constant that matches or slightly exceeds this target to ensure the voltage across the capacitor does not fall below the desired level within the 10ms duration.
- The RC time constant ((\tau)) is defined as (\tau = R \cdot C), where:
-
Choosing Component Values:
- Assuming a simplistic model where we aim for a (\tau) (time constant) equal to or slightly longer than 10ms to maintain the signal.
- Without specific constraints on the resistor or capacitor values, let's target a (\tau) of around 10ms. As an example, you might choose (R = 100k\Omega) and (C = 100nF), hence (\tau = 100k\Omega \cdot 100nF = 10ms).
-
Circuit Connection:
- Connect the anode of the diode (1N5818) to the signal source. This ensures that the signal can charge the capacitor but not discharge it back into the source.
- Connect the cathode of the diode to one end of the resistor (R) and to one terminal of the capacitor (C). This node is where your extended signal is obtained.
- The other end of the resistor is connected to ground.
- The other terminal of the capacitor is also connected to ground.
- The extended signal is taken across the capacitor, which will gradually charge up to the peak voltage of the incoming pulse and then slowly discharge through the resistor when the pulse ends, thereby extending the signal duration.
- Connect the diode so it conducts when the signal is high, allowing the capacitor to charge quickly and block the discharge path back to the signal source.
- The RC network then controls the discharge time, extending the signal duration to your target of 10ms.
- Depending on the exact requirements (such as how definitively the signal needs to hold its peak or how rapidly it can decay), you might need to adjust (R) and (C).
This setup is simplistic and designed for educational purposes. The actual performance can be influenced by various factors, including the diode's forward voltage drop, the initial charging speed of the capacitor, and how the signal's shape affects the decay pattern over time. Testing and iteration may be necessary for precise applications.
10mo
what is the best comparator op amp with 4 amps, like a LM393 but with 3 instead of 2
10mo
For a comparator with more than the two comparators found in an LM393 but with similar functionality, you would typically look for quad comparator ICs. Unfortunately, there's no direct "LM393 with 3 comparators" because the common configurations manufacturers opt for are either dual (like the LM393) or quad (four comparators in a single package).
A good example of a quad comparator that could meet the needs of someone looking for something "like an LM393 but with more comparators" is the LM339. The LM339 is a quad comparator, providing four independent voltage comparators in a single package. This makes it suitable for applications requiring multiple comparison operations without the need to use multiple dual-comparator ICs.
Key Features of LM339:
- Quad Comparators: It contains four independent voltage comparators.
- Supply Voltage Range: Operates from a single power supply over a wide range of voltages, or can operate from dual supplies.
- Input Offset Voltage: Typically low, which enhances comparison accuracy.
- Output Type: Open-collector outputs that can be connected to achieve wired-AND relationships.
- Package Options: Available in various packages, suitable for different types of circuit boards and design preferences.
- If you're looking for more than two comparators like in LM393 but less than four, using a quad comparator like the LM339 and simply not using one of the comparators would be the practical route. It's common in electronics design to select a component with slightly more functionality than currently needed to provide flexibility for future modifications or expansions of the circuit without a significant increase in cost or complexity.
If the LM339 doesn't meet your specific needs or you're looking for comparators with advanced features (like faster response times, lower power consumption, etc.), there are other quad comparator options available, such as:
- LT1394: High speed, low power quad comparator.
- MAX339: Precision, quad comparator with different voltage reference options.
When selecting a comparator, it's crucial to consider factors beyond the number of comparators per package, including the power supply requirements, input offset voltage, response time, and packaging type, to ensure it fits well within the design constraints and application requirements.
10mo
can i replace the schotchy diode with a cable
9mo
Replacing a Schottky diode with a simple wire (cable) in a circuit can profoundly affect the circuit's functionality, especially if the diode is being used for its rectifying properties, voltage drop characteristics, or fast switching capabilities. Here's a look at scenarios where a diode might be used and the implications of replacing it with a wire:
-
Power Rectification: Schottky diodes are often used in power supplies for their low forward voltage drop and fast switching. Replacing it with a wire would remove the rectification capability, potentially allowing AC components to pass through or causing short circuits in DC applications.
-
Voltage Clamping/Protection: Schottky diodes are employed in circuits to prevent reverse voltage conditions. A wire replacement would eliminate this protection, possibly leading to damage from reverse currents.
-
Logic Circuits or Signal Rectification: In digital circuits, Schottky diodes can be used for logic level shifting or preventing logic inversion. Replacing them with a wire could introduce errors in signal interpretation or unintended signal inversion.
-
Or-ing Power Supplies: Diodes are used in circuits to combine power from multiple sources without the sources back-feeding into each other. A wire would not prevent back-feeding, which could lead to damage or malfunction.
-
Solar Panels: Diodes prevent the back-flow of current into solar panels during low light conditions. A wire cannot provide this form of protection, potentially reducing efficiency or damaging the panels.
-
Switching Speeds: Schottky diodes are favored for their fast switching speeds in RF applications and power rectifiers. A wire cannot replicate this property and could lead to performance issues in high-speed circuits.
While replacing a diode with a wire might seem like a simplification or a way to avoid voltage drops, it typically results in circuit malfunction, potential damage to components, or complete failure of the circuit to perform its intended function. Diodes serve crucial roles in directing current flow, protecting against reverse currents, rectifying signals, and more. Removing them without a carefully considered alternative that can fulfill the same role will compromise circuit integrity.
If the goal is merely to bypass a particular point in the circuit (ignoring the diode's protective or rectifying functions), using a wire might work, but this should be done with a thorough understanding of the circuit's operation and the consequences of such a modification.
9mo
i use the schotchy for the polaritazation of the supply, so when i invert the negative and positive it doesnt affect the circuit
9mo
Using a Schottky diode for protection against reverse polarity is a common and effective method to safeguard your circuit. The key advantage of a Schottky diode in such applications is its low forward voltage drop, allowing more of the supply voltage to be used by the circuit while still blocking reverse current effectively.
The Role of the Schottky Diode in Reverse Polarity Protection:
-
Forward Connection: When the power supply is correctly connected (positive to positive, negative to negative), the Schottky diode is forward-biased. It allows current to flow with a minimal voltage drop (typically between 0.2V to 0.45V for Schottky diodes), ensuring the device receives power with slight loss.
-
Reverse Connection: If the power connections are reversed (positive to negative, negative to positive), the Schottky diode is reverse-biased. It blocks the current from flowing, protecting sensitive components from being damaged by reverse polarity.
Replacing the Schottky diode with a wire (or direct connection) in this context would remove the circuit's protection against reverse polarity. If the supply voltage were accidentally inverted, the direct connection would allow the reverse current to flow through the circuit, potentially damaging or destroying integrated circuits, transistors, and other components sensitive to polarity.
Alternative Methods for Reverse Polarity Protection:
If there's a specific reason you're considering removing the diode—perhaps related to the voltage drop or other concerns—there are alternative methods to protect against reverse polarity without significant voltage drop:
-
P-Channel MOSFETs: An efficient way to provide reverse polarity protection with a very low voltage drop. The source is connected to the power supply, the drain to the load, and the gate to ground through a resistor. When the polarity is correct, the MOSFET turns on, providing a path with a very low on-resistance. If the polarity is reversed, the MOSFET remains off, blocking the current flow.
-
Crowbar Circuit: A more complex solution, it involves using a voltage sensing circuit that triggers a thyristor (SCR) to create a short circuit when incorrect polarity is detected. This action usually results in a blown fuse, disconnecting the power supply to protect the circuit. It’s effective but sacrifices the fuse each time a reverse polarity occurs.
-
Specialized ICs: There are integrated circuits designed specifically for reverse polarity protection, which effectively control the current flow without significant voltage drops. Examples include the LTC4364 among others. These provide protection with additional features like overvoltage protection but at a higher cost.
For effective reverse polarity protection that also allows for minimal voltage drop across the protective element, replacing the Schottky diode with a wire is not advisable. Consider maintaining the diode or implementing an alternative method like a P-Channel MOSFET for similar effectiveness with minimal voltage drop. Always assess the specific needs and limitations of your circuit to choose the most suitable protection method.
9mo
@copilot how to make a signal widht extender to make a 0.7v peak 2ms signal tran...
@copilot how to make a signal widht extender to make a 0.7v peak 2ms signal transform it into a 10ms signal with a 1N4818 .a capacitor and a resistor
10mo
hola
how can i amplify the signal of a current sensor, beacuse the signal maintains t...
11mo
Signal amplification in current sensors can be achieved using Op-Amps (Operational amplifiers) or specific current sense amplifiers. Op-amps amplify the voltage difference between its two inputs. The gain, or amplification factor, is dependent on the feedback resistor network in the circuit.
Here are some suggestions:
-
Using a non-inverting operational amplifier: The sensor output goes to the positive input of the operational amplifier, and a voltage divider network acts as a sensor for feedback to the negative input of the operational amplifier. You can adjust the ratio of resistance in your voltage divider to adjust the gain of your amplifier.Here is a simple connection schematic:
- Connect the sensor output to the non-inverting input (+) of the Op-Amp (designator of the Op-amp: Ux, pin: IN+).
- Connect the invert input (-) of the Op-Amp to the junction of two resistors (Rx and Ry). This is your voltage divider acting as the sensor for feedback.
- Connect one end of resistors Rx and Ry together. Connect the other end of Rx to the Op-amp output and the other end of Ry to the ground.
- Connect your sensor signal to the non-inverting terminal of Ux.
-
Using a specific current sense amplifier: There are ICs specifically designed to measure and amplify small sense voltages across a current sensing resistor (also known as shunt resistor). They also provide isolation between input and output in many cases. Current sense amplifiers such as Texas Instruments' INA180, INA219 or MAX4172 by Maxim Integrated are good choices.
Please note that these are general techniques to amplify sensor signals. You should consider the specific type of your current sensor, the sensor output range, and the desired output range to select the appropriate amplifying mechanism and configuration.
11mo
so there i would get a broader difference between the signal so when current goes to the wire the signal differeitiates more ?
11mo
Yes, by amplifying your signal, you essentially increase the sensitivity of your measurement. This allows small changes in the sensed current to be represented as larger changes in the output signal, making it easier to measure and interpret these changes.
For example, if your sensor's output is say, 0.1 V/A and you want an output of 1 V/A, you would use an amplifier with a gain of 10. So a 0.1 A change in current, which would normally give a signal of 0.01 V, will now give a signal of 0.1 V after amplification.
Just remember, the gain does not change the resolution of the sensor, it just magnifies the change in signal for a given change in current. Be careful not to set the gain so high that your amplifier saturates (reaches its maximum output voltage), as once this happens, you can no longer accurately measure the current.
11mo
and wwhat op amp do u reccomend
11mo
The choice of an op-amp depends on several factors, namely the supply voltage, current, gain, bandwidth, input/output impedance, noise level, and cost. Some commonly used op-amps perfect for signal amplification include:
-
Texas Instruments LM358: This is a dual op-amp, meaning it has two independent op-amps in one package. It works well for a supply voltage as low as 3V and up to +32V.
-
Texas Instruments LM741: This is a single, general-purpose op-amp with a wide supply voltage range. It's known for its versatility.
-
Microchip's MCP6002: This dual op-amp is designed for low power and low voltage applications. It's operational from 1.8V to 6V, making it suitable for battery-powered devices.
-
Analog Devices' AD822: This one is a FET input op-amp, which means it has very high input impedances and low bias currents. It's excellent if you're worried about the op-amp loading down your sensor's output.
Bear in mind that to achieve high gain, you might require a rail-to-rail op-amp, i.e., an op-amp that can output voltage all the way to the supply rails. Some op-amps cannot do this and have a maximum output voltage a couple of volts less than the supply voltage. This is particularly important for amplifiers running on a low supply voltage.
You have to decide based on your specific needs and constraints, including available power supply voltage, current, gain, bandwidth requirement, and budget.
11mo
how can i extend the signal after an opamp for 3 seconds, like when the voltage is gretaer than 3 volts, maintain that voltage for 2 seconds
10mo
To hold or "extend" an output signal at a specific level beyond the actual duration of the input signal—especially after an event where the voltage surpasses a threshold like 3 volts, and you wish to maintain this state for an added duration (e.g., 3 seconds)—you are essentially looking into designing a circuit that incorporates a time delay or a "hold" feature. This can be achieved using a combination of analog and digital components. Here, a simple approach combining analog design with a bit of digital logic is described, focusing on a concept that's easier to prototype with standard components.
Sample Circuit Concept
-
Comparator: This checks if the output of the Op-Amp is greater than 3V. It changes its output state when the Op-Amp's output crosses 3V.
-
Monostable Multivibrator (Timer): A device like the 555 timer in monostable mode can be configured to have an output high period of 3 seconds when triggered by the comparator's output. This means once the voltage exceeds 3V and triggers the 555 timer, its output will go high and stay high for the predetermined time (3 seconds in this case) irrespective of the comparator's state post-triggering.
-
Comparator Setup:
- The Op-Amp's output is connected to one of the comparator's inputs.
- The other comparator input is set to 3V (using a voltage divider or reference).
- When the Op-Amp's signal exceeds 3V, the comparator's output switches state.
-
555 Timer Setup in Monostable Mode:
- The comparator's output is connected to the trigger input of the 555 timer.
- Configure the 555 timer in monostable mode with an RC time constant to achieve a 3-second high signal upon triggering. The duration ((T)) is determined by the formula (T = 1.1 \times R \times C), where (R) and (C) are the resistance and capacitance values.
-
Output:
- The output of the 555 timer will now represent the extended signal. When the comparator detects a voltage greater than 3V, it triggers the 555 timer whose output remains high for 3 seconds.
- Precision and variability: The accuracy of the timing depends on the precision of your resistor and capacitor for the 555 timer. Temperature and supply voltage fluctuations can affect the timing accuracy.
- Comparator choice: Choose a comparator with a response time faster than the minimal signal duration you expect to detect.
- Voltage levels: Ensure that the voltage levels are compatible between the comparator, the 555 timer, and the subsequent stages to avoid any logic level mismatch.
This approach provides a basic framework but may require adjustments based on specific application constraints such as power supply levels, the exact duration requirement, ambient conditions, and the need for customization of the time extension period.
10mo
what 555 would u reccomend for this situation
10mo
For general-purpose applications where you need a reliable timer with a wide range of operating voltages, the NE555 timer IC is a solid choice. It's arguably the most popular timer IC and is suitable for various timing and pulse generation applications, including the one you described.
Here’s why the NE555 is recommended:
- Voltage Range: The NE555 can operate between 4.5V to 15V, making it versatile for both 5V and higher voltage applications.
- Output Current: It can drive up to 200mA, allowing it to directly drive small relays or LEDs if needed for signal indication.
- Temperature Stability: Its timing remains relatively stable under varying temperatures.
- Availability and Cost: The NE555 is widely available and very affordable, making it an excellent choice for prototyping.
For your specific need to maintain an output voltage for 3 seconds after detecting a signal greater than 3 volts:
- You'll configure the NE555 in the monostable mode where it will output a high signal for a precise duration upon receiving a trigger.
- Calculate the RC time constant to achieve the 3-second delay. Given the formula for the delay time in monostable mode is approximately (T = 1.1 \cdot R \cdot C), you can choose values of R and C that satisfy your timing requirement. For example, using a 100kΩ resistor (R) and a 27µF capacitor (C) will give you around 2.97 seconds, which is close to your desired 3-second delay.
Make sure to use a pull-up resistor on the trigger pin to ensure it remains high until triggered to low, as the 555 timer trigger is active-low.
Additionally, if your application is sensitive to supply voltage variations or requires high precision, consider the CMOS versions like LMC555 or ICM7555, which have similar functionalities but with lower power consumption and sometimes better precision due to their CMOS construction.
10mo
i want to use an 1N4148 diode and a capacitor to replace the 555, what tould be the formula that tells me the time by the capacitor and the resistance for the diode
10mo
To create a simple delay circuit using a diode like the 1N4148 and a capacitor, without an active timing component like a 555 timer, one common approach is to use the RC (resistor-capacitor) charging circuit concept. This kind of circuit, however, would not strictly "replace" a 555 in terms of functionality but could provide a delay in the signal's response time based on the RC time constant.
In such a setup, you can use the diode to control the direction of current flow, ensuring the capacitor charges through one path and discharges through another, or to prevent backward current flow that could affect the charging time.
Circuit Description:
- Charging Path: When power is applied, the capacitor starts charging through a resistor.
- Timing Calculation: The time it takes for the capacitor to charge up to a certain voltage can be approximated with the RC time constant formula.
The RC time constant, denoted as τ (tau), gives an idea of how quickly the capacitor charges up to about 63.2% of the supply voltage when charging through a resistor. The formula is:
τ=R×C
Where:
- ( R ) is the resistance in ohms (Ω),
- ( C ) is the capacitance in farads (F),
- ( \tau ) (tau) is the time constant in seconds (s).
To estimate how long it takes for the capacitor to charge close to the full supply voltage, you can use a multiple of the time constant ( \tau ). A common rule of thumb is that the capacitor will charge up to about 99.3% of the full voltage in ( 5 \tau ) (5 times the RC time constant).
Example:
If you want a delay of approximately 3 seconds (( T = 3 ) s) and you have chosen a 47µF capacitor (( C = 47 \times 10^{-6} ) F), you can calculate the required resistor value (( R )) for the charging path using the time constant formula targeted for ( 5\tau ) (since you're interested in near-full charge, not just 63.2%):
T=5τ=5RC
Solving for ( R ):
R=5CT
Substituting ( T = 3 ) s and ( C = 47 \times 10^{-6} ) F:
R≈5×47×10−63≈12.77kΩ
For practical purposes, you might choose a standard resistor value of 12kΩ or 13kΩ.
Note:
- The diode (1N4148) in this circuit is typically for direction control or to shape the discharge path rather than altering the timing, hence it doesn't figure into the timing calculation precisely.
- This approach provides a less precise control over timing compared to a 555 timer circuit.
- The real-world behavior of the circuit can vary from the ideal calculation due to factors like the diode's forward voltage drop and the leakage current of the capacitor.
Experimentation and adjustment might be necessary based on actual components and the specific requirements of your application.
10mo
i have that circuit to extend the signal time, where should i connect my digital signal to, the exit of the capacitor_
10mo
how can i amplify the signal of a current sensor, beacuse the signal maintains t...
how can i amplify the signal of a current sensor, beacuse the signal maintains the same even tough it has to be circulating more current through a wire
11mo
What is the SN74HC14N?
11mo
The SN74HC14N is a Hex Schmitt-Trigger Inverter IC manufactured by Texas Instruments. It's a digital logic component that includes six independent inverters each with a Schmitt trigger on its input.
Here are some of its key features:
-
Each circuit functions as an independent inverter, but with a Schmitt trigger action on the input. This provides the inverter with a hysteresis characteristic (there's a specific "on" and "off" threshold) which makes the IC useful for handling noisy signals.
-
It operates with a wide supply voltage range from 2V to 6V enabling it to be used in low voltage applications
-
Operates in a wide range of frequency up to 80 MHz which makes it suited for high speed applications
-
It has high noise immunity due to Schmitt-trigger action at all inputs
-
It is fabricated with CMOS technology, which offers the benefits of wide supply voltage range, low power consumption and high noise immunity
The Schmitt trigger functionality makes the SN74HC14N ideal for a variety of applications, including wave shaping, signal conditioning and noise removal.
11mo
Hi! I’m copilot. Think of me like an experienced electrical engineer that can he...
Hi! I’m copilot. Think of me like an experienced electrical engineer that can help you brainstorm ideas, debug problems, research components, and even make wiring connections for you. Ask me questions like:
- @copilot I'm designing a PCB for a medical device that measures heart rate and temperature. Can you give me the list of components I will need?
- @copilot How would I connect these parts to make the LED flash at 1kHz?
- @copilot can you choose 4 digital pins on the ATMega328P-AU that I have here to use as GPIO given that I am already using some pins for reset, the external clock, UART, and I2C.
Reach me at any time by tagging me on comments or in the chat with my handle, @copilot.
11mo
Someone is typing...
Passive Components
Resistor Power Rating
BetaChecks that the power traveling across a given resistor is under its rating with 50% headroom.
Learn morePull Up/Pull Down Presence
BetaChecks that IC pins that require pull up or pull down resistors have them.
Learn moreCapacitor Voltage Rating
BetaChecks that capacitors have appropriate voltage ratings to prevent overheating and ensure reliable circuit operation.
Manufacturing
Parts Available
BetaChecks that all parts have a part number and are available from distributors.
Design Rule Checks
Overlapping Copper • 189
Des | Net | Des | Net | |
|---|---|---|---|---|
Pad | — | P2 | GND | |
Pad | — | Trace aa64 | Net 17 | |
Pad | — | Trace b1e8 | Net 17 | |
P2 | GND | Pad | — | |
Pad | — | Trace ddcb | Net 17 | |
P1 | Net 17 | Trace feae | Net 7 | |
Trace 2839 | Net 17 | A | — | |
A | — | Trace 8b22 | Net 17 | |
Pin_4 | — | Trace 8917 | Net 17 | |
Pin_4 | — | Trace 797a | Net 17 | |
Pin_4 | — | Trace a729 | Net 17 | |
Pin_3 | — | Trace a29d | Net 17 | |
Pin_3 | — | Trace 41ff | Net 17 | |
Pin_2 | — | Trace 3f99 | Net 17 | |
Pin_1 | — | Trace 7030 | Net 17 | |
Pin_1 | — | Trace e72d | Net 17 | |
P2 | GND | Trace cf68 | Net 17 | |
P2 | GND | Trace bb65 | Net 17 | |
P2 | GND | Pad | — | |
P2 | GND | Trace 5e9a | Net 17 | |
P2 | GND | Trace 50fa | Net 17 | |
P1 | Net 27 | Trace fb9e | Net 14 | |
P1 | Net 27 | Trace a494 | Net 14 | |
P1 | Net 27 | Trace 56e0 | Net 14 | |
P2 | Net 14 | Trace 5198 | Net 27 | |
P1 | Net 3 | Trace fa74 | Net 23 | |
P2 | GND | Trace b1e8 | Net 17 | |
P2 | GND | Trace ddcb | Net 17 | |
～ | GND | Trace b647 | Net 17 | |
P2 | Net 16 | Trace dd91 | Net 7 | |
Trace 2807 | GND | P1 | Net 7 | |
Trace 2a13 | Net 28 | A | Net 15 | |
K | Net 28 | Trace 7dbf | Net 14 | |
K | Net 28 | Trace c8ea | Net 14 | |
K | Net 28 | Trace dd36 | Net 14 | |
Trace 678c | Net 10 | 2 | — | |
Trace 39b9 | Net 17 | 3 | GND | |
Trace 472d | Net 26 | 1 | Net 23 | |
Trace 1f8a | Net 26 | 1 | Net 23 | |
Trace 3896 | Net 33 | K | Net 3 | |
K | Net 3 | Trace d9f7 | Net 33 | |
P2 | — | Trace a494 | Net 14 | |
Trace 1b4b | Net 13 | 33 | — | |
Trace 2309 | Net 26 | 13 | — | |
37 | — | Trace eb9e | Net 1 | |
Pin_8 | — | 34 | Net 12 | |
5V | Net 20 | Pin_2 | — | |
12 | — | Trace a0ea | Net 29 | |
Pin_4 | — | 36 | Net 1 | |
GND | — | Trace b69b | Net 20 | |
Pin_2 | — | 3V3 | Net 26 | |
Trace 203a | Net 17 | GND | — | |
GND | — | Trace 81b0 | Net 17 | |
GND | — | Trace 8319 | Net 17 | |
Trace 61ab | Net 17 | GND | — | |
Trace 3ff6 | Net 17 | GND | — | |
Trace 27ec | Net 17 | GND | — | |
GND | — | Trace 894f | Net 17 | |
Trace 6188 | Net 17 | GND | — | |
Trace 623c | Net 1 | 38 | — | |
Trace 186e | Net 13 | 32 | — | |
Trace 1a53 | Net 13 | 32 | — | |
Trace 33bb | Net 26 | 3V3 | — | |
Pin_9 | — | 35 | Net 21 | |
Pin_7 | — | 39 | Net 8 | |
Trace 33bb | Net 26 | Pin_2 | — | |
Trace 203a | Net 17 | Pin_1 | — | |
Pin_1 | — | Trace 81b0 | Net 17 | |
Pin_1 | — | Trace 8319 | Net 17 | |
Trace 61ab | Net 17 | Pin_1 | — | |
Trace 3ff6 | Net 17 | Pin_1 | — | |
Trace 27ec | Net 17 | Pin_1 | — | |
Pin_1 | — | Trace 894f | Net 17 | |
Trace 6188 | Net 17 | Pin_1 | — | |
Pin_16 | — | Trace a0ea | Net 29 | |
Trace 2231 | Net 1 | Pin_4 | — | |
Trace 186e | Net 13 | Pin_10 | — | |
Trace 1a53 | Net 13 | Pin_10 | — | |
Trace 1b4b | Net 13 | Pin_11 | — | |
Trace 623c | Net 1 | Pin_6 | — | |
Trace 2309 | Net 26 | Pin_17 | — | |
Pin_5 | — | Trace eb9e | Net 1 | |
Trace 727b | Net 20 | Pin_2 | — | |
Pin_1 | — | Trace b69b | Net 20 | |
VCC+ | — | Trace d462 | Net 26 | |
Trace 3d78 | Net 10 | NON-INVERTING_INPUT_3 | — | |
NON-INVERTING_INPUT4 | — | Trace dfb2 | Net 10 | |
Trace 6dc7 | Net 3 | NON-INVERTING_INPUT2 | — | |
Trace 092b | Net 29 | OUTPUT1 | — | |
Trace 0e9e | Net 17 | VCC- | — | |
VCC- | — | Trace da21 | Net 10 | |
NON-INVERTING_INPUT_1 | Net 22 | Trace abc0 | Net 33 | |
Trace 4ebc | Net 31 | OUTPUT4 | — | |
INVERTING_INPUT1 | — | Trace 95a3 | Net 10 | |
Trace 027c | Net 10 | INVERTING_INPUT3 | — | |
INVERTING_INPUT2 | — | Trace eae7 | Net 10 | |
INVERTING_INPUT2 | — | Trace b12b | Net 10 | |
OUTPUT3 | — | Trace ad4d | Net 18 | |
Trace 8027 | Net 18 | OUTPUT3 | — | |
P2 | GND | Trace aa64 | Net 17 | |
Trace 797a | Net 17 | P1 | — | |
P1 | — | Trace a729 | Net 17 | |
Trace 278f | Net 17 | P2 | — | |
Trace 66c3 | Net 17 | P2 | — | |
P2 | GND | Trace e563 | Net 20 | |
P2 | GND | Trace da5f | Net 20 | |
Pin_1 | — | Trace d495 | Net 1 | |
Trace a6e1 | Net 1 | Pin_1 | — | |
Trace 5dc2 | Net 1 | Pin_3 | — | |
Trace 9dcf | Net 1 | Pin_3 | — | |
Trace 1bf0 | Net 1 | Pin_6 | — | |
Trace 357a | Net 1 | Pin_6 | — | |
Pin_6 | — | Trace d438 | Net 1 | |
Trace 17de | Net 1 | Pin_4 | — | |
Pin_4 | — | Trace f231 | Net 1 | |
Trace 938f | Net 1 | Pin_2 | — | |
Pin_2 | — | Trace ed62 | Net 1 | |
Pin_2 | — | Trace ff99 | Net 1 | |
Pin_2 | — | Trace fa24 | Net 1 | |
Trace 5075 | Net 1 | Pin_5 | — | |
Pin_5 | — | Trace c213 | Net 1 | |
Trace 68cd | Net 17 | ～ | GND | |
Trace 334d | Net 17 | ～ | GND | |
Trace 4cc4 | Net 17 | ～ | GND | |
Trace 71d6 | Net 17 | ～ | GND | |
Pad | — | P2 | GND | |
Trace 3d3a | Net 17 | P2 | GND | |
Trace 96e7 | Net 17 | P2 | GND | |
P2 | GND | Trace e186 | Net 17 | |
Trace 6d30 | Net 33 | P1 | Net 3 | |
Trace 7954 | Net 33 | P1 | Net 3 | |
Pad | — | P1 | Net 3 | |
P1 | Net 3 | Trace d9f7 | Net 33 | |
P1 | GND | Trace fc06 | Net 20 | |
Trace 2d35 | Net 20 | P1 | GND | |
Trace 1e4d | Net 20 | P1 | GND | |
P1 | Net 26 | Trace e8c3 | Net 18 | |
Trace 858a | Net 26 | P2 | Net 30 | |
Trace 9e25 | Net 26 | P2 | Net 30 | |
P2 | Net 30 | Trace fda5 | Net 26 | |
Trace 5fd3 | Net 26 | P2 | Net 30 | |
Trace 1795 | Net 17 | GND | GND | |
GND | GND | Trace f2fa | Net 17 | |
K | Net 33 | Trace fa74 | Net 23 | |
Trace ad7e | Net 20 | P2 | Net 7 | |
Trace 3f99 | Net 17 | P1 | — | |
Trace a981 | Net 17 | P2 | — | |
Trace 5e9a | Net 17 | P2 | — | |
Trace 444c | Net 17 | P2 | — | |
Trace 004f | Net 23 | Trace 576a | Net 26 | |
Trace 310d | Net 26 | Trace 38c4 | Net 23 | |
Trace 1fcc | Net 26 | Trace 4bd1 | Net 23 | |
Trace 472d | Net 26 | Trace f28c | Net 23 | |
Trace 1f8a | Net 26 | Trace f28c | Net 23 | |
Trace 1fcc | Net 26 | Trace f28c | Net 23 | |
juan2704
copilot