I've been trying to design a control circuit to switch supply.

What I want to achieve?

I need 24V voltage for supplying an ultrasound sensor. There are two ways of gaining 24V: VIN is a physical 24V power supply and 24V_OUT is a boost converter placed in the other part of the circuit. I assume that when I plug any voltage to the VIN connector I obtain this voltage in the output decreased by the voltage drop on the D4 diode. For now, when I put 7.4V to the VIN input, I receive 1.8V in the output. I've checked the datasheet and threshold voltage of the Q3 transistor is typically 1.8V. Could you explain where had I made a mistake?

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It won't work because you are using an N MOSFET as high side switch without a proper driver.

The gate-to-source voltage must be positive and greater than Vth in order for the MOSFET to "turn on". But in your case, it is approximately -20.7 volts.

I would suggest using a P MOSFET instead. But remember typical P MOSFET Vgs_max ~ -20V, so you must protect the gate from overvoltage.

btw, I don't understand what you are going to achieve. The diode logic "automatically" choose the higher voltage source.

  • \$\begingroup\$ My main circuit can be supplied by either batteries or 24V power supply. Sensor is supplied just for measuring time and then it is switched off. What if I short R3 to VIN like it is suggested in the comment below? Would it work? \$\endgroup\$ – Y. Markov Dec 5 '18 at 12:44
  • \$\begingroup\$ If you do so, Vg = 24V, Vs ~ 24V => Vgs = 0V. Nothing happens. \$\endgroup\$ – Long Pham Dec 5 '18 at 12:49
  • \$\begingroup\$ So if I replace Q3 transistor by proper p-channel one would it work? I want to enable the output with a low state. I think that the R3 resistor should be connected to VIN \$\endgroup\$ – Y. Markov Dec 5 '18 at 13:38
  • \$\begingroup\$ Sure. There are several questions about using P MOSFET as high side switch on this site. I suggest you should have a read at them. \$\endgroup\$ – Long Pham Dec 5 '18 at 13:59
  • \$\begingroup\$ Long Pham thank yoy for your help and time. One more question: should I replace npn tranistor into pnp? \$\endgroup\$ – Y. Markov Dec 5 '18 at 14:00

If Q4 is off, you get 3.3V on Q3 gate. Q3 must have Vgs = 1.8V (approximately) to stay on. So, the source voltage is 3.3V - 1.8V = 1.5V (close to 1.8V you measure).

One partial solution would be to wire R3 to VIN instead of 3.3V, but you still would have a 1.8V voltage drop and get approximately 22.2V instead of 24V.


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