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This circuit is going to be in a production environment. I want to turn a relay on and off using a P-channel MOSFET. This action of turning on and off is not a periodic signal at a high frequency, hence a MOSFET driver is not needed.

When the controller is set high (activated), it pulls the output to ground.

R3 is added to limit the current into the gate. R2 is added to pull up the gate to 12V, so it is not floating. Diode added is a flyback diode.

The internal circuitry contains a 3V voltage that is pulled through R1.

Is this circuit sufficient for a production design? What standards do you guys follow when designing a similar circuit?

schematic

simulate this circuit – Schematic created using CircuitLab

Also by the looks of it the IRF9530 FET will not work because the Vgs is to high, are my calculation coorect?

schematic

simulate this circuit

Edit 1:

schematic

simulate this circuit

Edit 2:

enter image description here

Edit 3:

A redundant pull-up is added so even if the control internal circuit is not plugged in, the load is still OFF.

enter image description here

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  • \$\begingroup\$ Others have pointed out issues with your circuit. I would like to ask you why you want to use a magnetic relay. Magnetic relays have a relatively high energy cost, and therefore, I would not suggest using one unless you need to use one. So, what exactly is the load that the relay will be switching? Could a solid state device, such as a mosfet serve your purpose instead? \$\endgroup\$ Dec 28, 2021 at 20:30
  • \$\begingroup\$ @MathKeepsMeBusy I have to use the Relay, no other option. \$\endgroup\$
    – JoeyB
    Dec 28, 2021 at 21:06

2 Answers 2

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If the circuit is as shown it will not work at all, (or it will pull the 3V supply up to 12V).

There will be about 9V on the coil with switch open and about 11V on the coil with switch closed. The MOSFET will never turn off or on completely. Remember Vgs is what controls the MOSFET.

You should probably be using an N-channel MOSFET (or two MOSFETs) depending on what you are trying to do.

DC relays do not have an inrush current, so not sure where your numbers are coming from- they sound more like an AC relay specification. Powering an AC relay from the nominal AC voltage in DC will likely lead to the acrid smell of disappointment.


Generically, you probably want to do something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Would not high side switching solve these problems? \$\endgroup\$
    – JoeyB
    Dec 28, 2021 at 13:54
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    \$\begingroup\$ You will now get 9V with switch open and almost 12V with it closed. Still not ideal since the relay will never drop out. \$\endgroup\$ Dec 28, 2021 at 14:03
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    \$\begingroup\$ @JoeyB - You are missing the point entirely. The problem lies with the existence of R1. As long as that is there, the voltage from +12 to the gate of the FET (Vgs) cannot fall below 9 volts, and the FET will always be on. \$\endgroup\$ Dec 28, 2021 at 15:40
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    \$\begingroup\$ Yes, this should work. An advantage of the transistor (as well as giving you inversion) is that the MOSFET gate is being driven from 12V so you don't need to use a logic-level MOSFET, just one that can withstand (say) +/-20V Vgs to be safe with +12. \$\endgroup\$ Dec 28, 2021 at 20:51
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    \$\begingroup\$ Thanks. I will still accept your answer, since it was your idea that lead to the final solution \$\endgroup\$
    – JoeyB
    Dec 28, 2021 at 21:05
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I think it is worth pointing out that the solution pointed out is still very much incomplete. The reason for this is that the circuit pointed out by Spehro and also the circuit in my Edit 2 is based on the assumption that the NPN BJT is operating in the Cut-off and Saturation regions.

If you operate the BJT in the active region then Vce can swing between 0V and 12V, which means the FET can always be "ON" if you do not use the correct biasing resistors

Hence we take the following NPN BJT ZXTN2038FTA, datasheet here.

Now back to the circuit at hand. We require Vce to be 0V (saturation region), hence Ic (collector current) will be 1.2mA (12V/10kohm).

Therefore from the datasheet we can find the values for beta and Vbe.

enter image description here

enter image description here

Hence, we get the following values, assuming we are using the BJT at a temperature of 25:

beta@Ic=1.2mA = 220 (approx.)

Vbe@Ic=1.2mA = 0.6V (approx.)

Therefore,

Ib = 1.2mA/220 = 5.45uA

Rb (base resistance) = (2.8 - 0.6)/5.45uA = 404 kohm

Hence, in order to ensure the BJT remains in the saturated region, Rb must be less than 404 Kohm

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