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. #portaljharwinbarrozo43.0M
- 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 #powerjharwinbarrozo11.4M
- 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 #CommonPartsLibraryjharwinbarrozo1.5M
- 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 #CommonPartsLibraryjharwinbarrozo1.5M
- 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 #genericsjharwinbarrozo14.4k
- 47 kOhms ±5% 0.125W, 1/8W Chip Resistor 0805 (2012 Metric) Automotive AEC-Q200 Thick Film #forLedBlinkjharwinbarrozo1.2M
- 10uF Capacitor Aluminum Polymer 20% 16V SMD 5x5.3mm #forLedBlink #commonpartslibrary #capacitor #aluminumpolymer #radialcanjharwinbarrozo1.2M
- Yellow 595nm LED Indication - Discrete 1.7V 1206 (3216 Metric) #forLedBlinkjharwinbarrozo1.1M
U1
how would you make a lm393 delay pull up resistor circuit
1y
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!
1y
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10
how would you make a negative dc offset to this amplifier
@copilot how would you make a negative dc offset to this amplifier
1y
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9
which lora version has the sx1276 semtech lora transceiver
1y
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.
1y
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4
tell me the difference between the hex inverter 74ls04 and the 74HC14D
1y
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.
1y
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3
What are some 74HC14D alternatives? because i just need one digital input, i am ...
1y
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.
1y
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1
How much voltage does the AC172 drops when connected to a 1A circtuit?
1y
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.
1y
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
1y
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.
1y
explain to me the last part, how do i tune the sensor?
1y
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.
1y
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5
can you tell me if this connection is alright for a operational apmplifier
1y
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.
1y
im tring to make an inversor amplifier wiht 100 of gain for a hall current sensor
1y
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.
1y
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8
3.3V
R1
Resistance
1k Ω
C1
Capacitance
470u F
D2
J6
R9
Resistance
1k Ω
Q1
VIN
C7
Capacitance
0.1u F
D3
R2
Resistance
1k Ω
ADC 2
R11
Resistance
1k Ω
D4
3A
3A
ADC 1
VIN
C6
Capacitance
0.1u F
ADC 3
J14
Q2
ADC
C2
Capacitance
470u F
U3
3.3V
R3
Resistance
5k Ω
J2
3.3V
AutentioNode
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Overlapping Copper • 189
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