groundgroundingpush-pulltransistorsvirtual-ground
In this simple "push-pull" circuit, there are +V, -V and ground connections.
up until now, i always considered ground to be the minus side of the battery.
but that is also the way i understand -V, so i don't see how current would flow through the PNP transistor from ground to the -V
If i am using this circuit to drive a simple DC motor forwards(push) and backwards(pull) with a battery
what would the -V and ground connections mean ?
1.) The transistors M4 and M5 act like a floating voltage source. M3 is a current source and therefore the voltage across M4 and M5 is constant.
M6 and M3 form a common-source stage that allows to move the floating voltage source up and down. This eliminates the region where neither M1 nor M2 is conducting, therefore distortion is reduced.
2.) All these transistors should be in saturation. M4, M5 will always be in saturation since they are diode-connected.
3.) The dimensioning is usually done for the quiescent condition where the output signal is zero. For this case M4/M1 and M5/M2 act like current mirrors.
The ratio determines the quiescent current.
The problem is that your feedback signal is taken from the wrong place.
Figure 1. Corrected feedback loop. (Red = remove, green = add.)
In your original layout the feedback signal is monitoring the output of the opamp but has no idea what's happening at the load which is where it matters. If the circuit is working the voltage at RL2 would be the same as that on pin 3. I think that if you measure it you will find that it is out by 0.7, the emitter-base voltage drop of Q2.
Instead, take the feedback from the junction of Q1 and Q2's emitters. The opamp will now correct until that point is at the same potential as pin 3.
Note that this arrangement can be problematic in some situations (push-pull audio amplifiers, for example, because the to switch from sourcing base current for Q1 to sinking base current from Q2 the opamp has to suddenly switch from +0.7 V to -0.7 V. To solve this a biasing arrangement is used to pass some "quiescent current" to keep both transistors slightly on near this point. It may not matter in you application.
Best Answer
When using a 'split rail' like this, ground will be the mid terminal of the battery or power supply. So if you had a +12v and a -12v supply, that would need two 12v batteries, connected in series, with their mid point taken to be ground.
This split rail connection allows you to generate both positive and negative voltages into the load. Your motor would connect between ground and \$V_E\$. This would allow it to be driven in either direction.