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what to do in this circuit to trip the relay?
shivam2002

shivam2002

@copilot what to do in this circuit to trip the relay?
1y

copilot

To trip the relay $K1$ in your circuit, we need to create a situation where the relay coil is energized. This can be achieved by sending a signal through the relay control circuit, which typically involves switching a transistor that controls the current through the relay coil. Since the relay in question is SRD-12VDC-SL-C, it requires a 12V supply across its coil to operate.
Based on the project components and connections provided, here's a step-by-step guide to trip the relay $K1$:
  1. Ensure Power Supply: Verify that $V1$ (12V supply) is properly connected and powered, as this voltage is needed to energize the relay's coil.
  2. Transistor Switching Mechanism: To control the relay, we can use $Q1$ or $Q2$ as a switch. For simplicity, let's use $Q2$ (PC817, an optocoupler) in this example. When the LED inside $Q2$ is forward-biased (current flows from ANODE to CATHODE), it will turn on the phototransistor segment, thus allowing current to flow from COLLECTOR to EMITTER, making it possible to control the relay.
    To energize the relay:
    • Activate $Q2$: Apply a signal to the ANODE of $Q2$ through $R2$. This can be done via a GPIO pin from $U1$ (for example, $U1:D8$) set to a high state. Ensure this pin is configured as an output in your Arduino sketch.
    • Connecting $U1:D8$ to $R2:P1$ and ensuring $R2$ is connected to $Q2:ANODE$ (which is already done in your circuit) is the key step here.
Here are the detailed steps summarized:
  • Ensure $V1$ is connected and supplying 12V.
  • Connect $U1:D8$ to $R2:P1$ (already done).
  • Issue a command from the Arduino Nano ($U1$) script to set $D8$ HIGH, which will forward bias the LED in $Q2$, activate the phototransistor, and allow current to flow from $V1$ through $Q2:COLLECTOR$, $Q2:EMITTER$, to energize $K1$'s coil and trip the relay.
This arrangement provides a way to digitally control high-power devices connected to the relay, isolating the microcontroller from the higher voltage circuit controlled by the relay.
1y
shivam2002

shivam2002

is there any issue in this circuit? ensure all the voltage and current requirements are properly satisfied with the use of resistors in this circuit.
1y

copilot

To evaluate the voltage and current requirements and examine potential issues with the use of resistors in the circuit, let's analyze the key components and their connections considering their specifications and typical operating conditions.
1. Relay $K1$ (SRD-12VDC-SL-C)
  • Voltage and Current: The relay requires a 12V supply to operate, which is provided by $V1$. The typical coil current is around 37.5mA at 12V. The existing arrangement through $Q2$ should adequately control the relay coil without an additional resistor directly on the relay coil circuit.
2. Optocoupler $Q2$ (PC817X1CSP9F)
The optocoupler plays a crucial role in isolating the low voltage control signal from the high voltage relay coil circuit.
  • LED Forward Voltage and Current: Typically, the forward voltage of an optocoupler LED is about 1.2 to 1.4V, and the forward current recommendation is around 10 to 20mA for proper operation. However, there's no detailed specification provided for this exact part, so typical values are assumed.
3. Resistor $R2$ for $Q2$ LED
Given that $U1$ (Arduino Nano) operates at 5V on its digital pins:
  • Using the calculations with an assumed forward voltage of $V_F = 1.2V$ and a target forward current of $I_F = 10mA$ for the optocoupler's LED: R=Vsource−VFIF=5V−1.2V10mA=380ΩR = \frac{V_{source} - V_F}{I_F} = \frac{5V - 1.2V}{10mA} = 380\Omega
However, a 330$\Omega$ resistor is used in the project ($R2$), which will slightly increase the current, but this is still within safe limits for typical LEDs.
4. LED1 ($WP7113SYD$)
This LED has a forward voltage of about 2V (as specified in the project) and can handle a forward current up to 30mA. It is critical to ensure this LED is not overdriven.
  • Current Limiting for LED1: Assuming the LED is powered directly from a 12V source ($V1$) with no current-limiting resistor detailed in the path, which is a significant oversight since it can lead to exceeding the maximum forward current and potentially damage the LED.
To properly size a resistor for LED1: R=Vsource−VLEDILED=12V−2V20mA=500ΩR = \frac{V_{source} - V_{LED}}{I_{LED}} = \frac{12V - 2V}{20mA} = 500\Omega A current value of 20mA is chosen to ensure bright illumination without approaching the absolute maximum rating.
Potential Issue Identified
  • Missing Resistor for LED1: There is a critical need for a current-limiting resistor in series with $LED1$ to prevent it from being overdriven by the 12V source. An approximate value of 500$\Omega$ would be appropriate.
Summary
  • The circuit's control path for the relay, via the optocoupler, appears correctly designed.
  • There is a potential oversight with $LED1$, where a current-limiting resistor needs to be added to prevent damage to the LED.
  • All other components and nets should function correctly based on the available information and typical values assumed for calculation purposes.
Before implementing any changes, always verify the exact specifications of your components, as the calculations here are based on typical values and assumptions.
1y
Q1
V2
K1
R2
Resistance
330 Ω
D1
V1
R1
Resistance
1k Ω
LED1
Q2
U1
U2
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Strong Olive Interocitor

Strong Olive Interocitor
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Created
Last updated by shivam2002
1 Contributor(s)
shivam2002

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DistributorQty 1
Digi-Key$1.35–$1.92
LCSC$2.32–$2.46
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