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(Schematic taken from here)

I can't make heads or tails of the push-pull stage similar to the one below - I'd rather try to understand it before this goes to a prototype board.

Q1 will start to conduct as soon as its \$V_{BE}\$ is roughly 0.7V, but how will that happen when its emitter terminal is floating? May I assume Q3 and it's gate capacitance is discharged at the beginning which provides 0V to the node connecting the emitters of Q1 and Q2?

I guess my question is in large part about the circuit's initial conditions and my assumptions about them, but feel free to set me straight.

                          

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  • \$\begingroup\$ Remember that Vbe is the voltage BETWEEN the transistor's base and emitter, not the voltage between base and ground. \$\endgroup\$ Dec 6, 2020 at 19:35
  • \$\begingroup\$ @Peter Exactly, and this is the source of my confusion - how can the base of Q1 (or Q2) draw any current, i.e. start to conduct when its emitter is at an undefined potential? So far, all the articles about push-pull configurations I could find don't give a detailed explanation of how and why the BJTs will operate. May I assume that Q1's emitter is at \$0V\$ at t=0? \$\endgroup\$ Dec 7, 2020 at 19:56

4 Answers 4

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Each bipolar transistor, by itself, is a voltage follower. The upper transistor pulls up to Vin-0.6V, and the lower to Vin+0.6V.

With this kind of circuit it is impossible for both transistors to be on at the same time. Either the upper one is ‘pushing’ or the lower one is ‘pulling’, but never both. This is called Class B operation.

Let’s start with power on, Vin=0V, and the output (gate drive) is also 0V. Now, start increasing Vin. Nothing will happen until Vin is 0.6V, then the output will follow the upper transistor up, less 0.6V Vbe drop.

Go all the way up to Vin=10V. Output will be 9.4V because it followed the input up. The gate capacitor is now charged up to 9.4V.

Now, start decreasing the Vin voltage. Nothing will happen until Vin reaches… what? 0.6V below the output, or 8.8V. Vin had to swing -1.2V before anything happened. That’s the crossover ‘dead zone’ in Class B.

Keep Vin going down until Vin=0V. Then the output will follow the lower transistor down, and the final output voltage will be 0.6V.

You’ll note that the input has to swing at least 2*Vbe, or 1.2V, to change from ‘pulling’ to ‘pushing’ or vice-versa. While that may seem like a problem (and it is for audio), remember that our input is a square wave so we don’t care about this.


BONUS: simulate it here

enter image description here

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First of all, you can have a potential without any current flowing. Even if the emitter is "floating", it will still have a potential, notably, 1 diode drop different from the transistor base. Let's say, 0.7V. Technically, with no current flowing, it will be closer to the base potential, but that's a separate complication and it's irrelevant to understand how this circuit operates practically. You are on the right track with the rest of your understanding.

Consider this simplified model. I broke the operation down into two puedo-circuits: one for the off-to-on transition, and one for the on-to off transition. Since the bases of both bipolars are common, and they are complimentary, both bipolars can't conduct at the same time. So, as a simplification, only the conducting bipolar is depicted in its respective circuit. Another simplification is looking at the MOSFET as a capacitor because that is all the "driver" sees (strictly speaking, the capacitor model is a bit more complex then what I shown, but we are going to ignore the higher order effects as it adds unnecessary complications).

schematic

simulate this circuit – Schematic created using CircuitLab

In the first circuit, everything is off, PWM is low (0V) and MOSFET is off. Now, PWM goes high (VCC, or 12V in your case). Current starts flowing through the resistor and the MOSFET gate gets charged and begins to conduct. Current keeps flowing into the gate until it equals VCC. The MOSFET is now saturated and the drain to source channel is super low impedance - the MOSFET is fully on.

In circuit 2, the reverse happens. The PWM signal brings the base of Q2 to ground, forward-biasing the emmitter-base junction. The charge that was accumulated in the first circuit can now be shunted to ground. This cause the voltage across the MOSFET gate-source junction to fall to zero. Now the MOSFET is in it's off state where the drain-source channel is super high impedance.

When using a MOSFET as a switch (not an amplifier), you typically want to switch the transistor as fast as possible. And the reason you want to switch it fast as possible is to get the MOSFET out of its ohmic region as fast as possible to minimize heating up the junction - the transistor doesn't generate heat when there's no current through it (off state) and it doesn't generate heat when there's no voltage across it (on state). In the ohmic region in the middle of the switching transistion, you have both current flowing with a voltage dropped across it (P=V*I, P is heat).

At this point, you could argue: So what? It takes like a millisecond to turn the damn thing on or off. How much heat could that really make? And you know what? You'd be right, It's really not so bad; it's only a little bit of heat. You don't even need a heatsink. But, if you're turning that SOB on and off a thousand times a second; 20,000 times a second; 500k times a second, well then we're talking about real heat. This will matter in PWM circuits and especially power convertors that are hard-switching, where this transition heat is not only a problem, but no eats up the system efficiency.

So the whole point of this push-pull (partial) gate driver is to switch the MOSFET hard n' fast by moving a lot of current (charge) in and out of the gate so Vgs changes rapidly.

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I think you are right. When a Voltage is applied to the gate of Q1, a current can flow from VCC which will charge the gate capacitor of Q3 which will raise it the gates Voltage to VCC. The time this will take is depending on the value of the parasitic gate capacitance and the gate resistor R2.

Addition: When the PWM signal goes low again, the lower BJT will start conducting pulling the MOSFETs gate to ground by discharging its capacitor.

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    \$\begingroup\$ It is correct, this is a class B output stage which has the property that if no output current is drawn (when the gate capacitance has been charged) both transistors will be off and no current flows anymore. This happens "automatically". \$\endgroup\$ Dec 6, 2020 at 19:49
  • \$\begingroup\$ "When a Voltage is applied to the gate of Q1, a current can flow from VCC" - how, exactly? As far as I can see, the emitters of both BJTs are initially floating, so how can there ever be a |VBE| of 0,7V? My reasoning with Q3's gate capacitance providing virtual GND was only a guess, too much handwaving for my taste. \$\endgroup\$ Dec 8, 2020 at 1:39
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Just a visualization change will help..visualize the capacitor as impedance(resistor symbol connected to ground and the high frequency harmonics of pulse causing current to flow in it, in turn charging it)

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