Driver circuit for medium loads

Low-voltage low-current to high-voltage high-current circuit.
The circuit is made out of three stages, two inverted voltage level shifters and a push-pull transistor circuit.

[@Enigma758]

The push-pull stage consists of two common collector transistors (NPN and PNP), so is the base resistor strictly necessary since there is load?

[@robertbolger2740]

Looks good. Do you have the specifications of the components, that could be added?

[@krakkus]

Any generic transistors would do the job, but these are the ones on the circuit.
PNP: 2N5401
NPN: 2N2222

[@jasonamosco318]

Does the circuit can make motor move forward and reverse by signal control from Arduino?

[@krakkus]

Hi, yes that is its purpose, but for that we need 2 of these pins/circuits. Pin 1 high and pin 2 low for one direction, and pin 1 low and 2 high for driving in the other direction. Like a h-bridge would do.

[@jasonamosco318]

​@@krakkusThank you

[@wolfgangvyhmeister1995]

Hi what's the difference between this and just using a sole transistor? This circuit uses some power while motor is off although it's very low

[@krakkus]

Hi, with a single NPN transistor. The output (emiter) voltage will not go above the base voltage, which is 3.3V for most microcontrollers. But it would give you more current. Also, this circuit connects to ground when off. Which a NPN also would not do. PNP would be better for voltage but inverted signal.
Two of these circuits will be able to drive a DC motor BOTH ways, like a h-bridge would do. Which is what I eventually would like to do.

[@Enigma758]

@@krakkus But I think the question likely applies to a single transistor in a common emitter configuration. In that case, it can exceed the base voltage.

[@davidmotley4207]

When the Q4 base is zero volt, why would Q4 turn on if its Emitter is floating. In order for the Q4 to turn on, its Emitter has to be higher by ~0.7v. In your test circuit, the motor is running when the Q1 base is ON and the current flows from Vcc through Q1 Emitter and through the load to ground. That means your load must be passive. So, the second output shown on scope is not true output but 0.75Vp-p since the scope provide the ground for Q1. The Q4 in your setup will never turn ON. Correct me if wrong.

[@krakkus]

Hi, thank you for your comment. I do not understand it completely. So I did a quick test on the circuit. Using the Q4 transistor.
ua-cam.com/video/coROebyYefI/v-deo.html
The motor is now connected to VCC and the other end to the output of the circuit/Q4, whenever the signal is low, the motor is running, which would suggest Q4 is conducting to ground.

[@davidmotley4207]

@@krakkus Thanks for checking. So, if you pull Q4 out, leaving a void, it should still work the way it did.

[@wolfgangvyhmeister1995]

You're right, I tested this in PROTO app from google play store and when deleting the Q4 transistor it does not affect the circuit in any way. In fact, Q4 does not receive any current when the load is turned off (motor in this case). And even when the motor is dissconected, because Q4 is a PNP transistor, it will not do anything as there is Q3 which is a NPN transistor.

[@davidmotley4207]

@@wolfgangvyhmeister1995 We are agreeing. For the next test, I would disconnect the Q4 collector and insert a PS of Vcc=-10v in such a way that negative connects to Q4 collector and positive to common ground. Technically, the motor should turn CW or CCW based on the input.