I am a EE student from Denmark, in the middle of making a class D audio amplifier. In this matter, I have stumpled across a needed circuitry. I need a way to shut off my gate driver (essentially the output stage). The gate drivers I use have an active high disable pin. I will be running the logic on 5V.

In this sense, I for startes need to be able to use a switch turning off the gate-driver using the disable pin, whilst it turns on an LED.

enter image description here

The picture above shows my temporary try. I use an opamp, because I will eventually implement OT protection etc. Thus starting out with an op-amp, makes that addition easy.

Here I thought I would be able to feed the disable labelled wire to the disable pin, whilst it turns on the LED. However whilst simulating, I cannot get it to work properly. I cannot draw more than approximately 2 mA in the LED.

Do you guys have any suggestions, if so I am welcoming them gratefully.

It should be noted, that the opamp is a rail-to-rail output THS4222, and it is supplied with 5V.

Thanks in advance.


  • \$\begingroup\$ Ground the emitter of your transistor & move the LED & its resistor up to the collector instead. \$\endgroup\$
    – brhans
    May 11, 2016 at 21:16
  • \$\begingroup\$ 2nd, it's not really clear why you want an op-amp. If you want a comparator function, use a comparator, not an op amp. \$\endgroup\$
    – The Photon
    May 11, 2016 at 21:27
  • \$\begingroup\$ @ThePhoton I currently use an op-amp IC, and I have one gate unemployed, thus to spare an extra IC, I will for now use an op-amp. Not sure yet, if I wish hysteresis feedbacks etc. \$\endgroup\$ May 11, 2016 at 21:31

1 Answer 1



simulate this circuit – Schematic created using CircuitLab

Figure 1. Voltage follower. Figure 2. "Open collector" switch.

The way you have connected results in the situation in Figure 1. The base current flows through R1 which decreases the voltage at the base of Q1 to maybe about 3.6 V. The emitter will be 0.6 V less so should read about 3 V. You can check these with your multimeter. So, your circuit is losing voltage all the way.

You could make Figure 1 a little better by removing R1 so that it becomes an emitter follower. The emitter voltage would then be one diode voltage drop less than the output of OA1.

The normal way we do this is to ground the emitter and switch a load in the collector circuit as shown in Figure 2. The advantage is that with the 5 V drive from OA2 we can turn Q2 "hard on" giving minimum resistance between collector and emitter. This also has the advantage that we could use the transistor to switch a relay or lamp powered by a higher voltage - 12 V, for example.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.