Hey Guys,

Quick question from a book I'm reading - picture attached for context.

The book states that in the push pull circuit in the attached diagram the amplified and inverted ac signal at the Q1 collector drives the bases of Q2 and Q3. On the positive half cycle, Q2 conducts and Q3 cuts off. On the negative half cycle, Q2 cuts off and Q3 conducts.

I'm confused about why the either transistor cuts off? I hope that question is specific enough.

Thanks,

Austin

 

Hello,

Have a look at the pages 3 and on in the following PDF:
http://educypedia.karadimov.info/library/lecture12.pdf

Bertus

 

In a push-pull circuit the output stage has little or no Dc bias current to minimize power dissipation. One transistor in the output conducts to provide power to the load on 1/2 of the cycle and the other transistor conducts to provide load power on the other 1/2 of the cycle. If the other transistor didn't cut off then some of the load current would instead go through that transistor, wasting power.

Why did you think they should both say on?

 

Hey, I have that same book! It kind of sucks at explaining basics. Tons of useful intermediate-level circuit fragments, though.

 

Hey Guys,

Thanks for the responses. Unfortunately I'm still a little confused. Let me explain better my interpretation of what's going on:

1. The ac signal Vin results in an ac signal at the collector of Q1 overlayed on a dc bias. The voltage at the Q1 collector can never be negative.
2. Because of the biasing diodes, there is always .7 volts across the be junction of Q2 and Q3, even with the ac fluction. Thus Q2 and Q3 are always biased into conduction.

Where is my thought process flawed?

 

Hey Guys,

Thanks for the responses. Unfortunately I'm still a little confused. Let me explain better my interpretation of what's going on:

1. The ac signal Vin results in an ac signal at the collector of Q1 overlayed on a dc bias. The voltage at the Q1 collector can never be negative.
2. Because of the biasing diodes, there is always .7 volts across the be junction of Q2 and Q3, even with the ac fluction. Thus Q2 and Q3 are always biased into conduction.

Where is my thought process flawed?

So far you are doing OK. The two diodes bias the output transistors so that they never fully turn off. This reduces what is called transition distortion. In some hi-fi amps this bias adjustment is very important to reduce distortion and to keep the amp from burning up. Too much bias and the amp runs hot. Too little bias and the distortion go up quickly.

OK?

 

A common misconception about a push pull amplifier is that the top transistor conducts when the waveform is above zero and the bottom transistor conducts when the waveform is below zero. This is a false analogy.

A better analogy is:
The top transistor conducts on the rising edge of the waveform and the bottom transistor conducts on the falling edge of the waveform. Transition distortion mainly occurs when the waveform transits from rising to falling, or falling to rising. This is where one transistor stops conducting and the other begins to conduct.

 

SkiBum, I think where you may be confused is that AC and DC aren't separate...the instantaneous voltage at the base of the transistors is the AC level plus the DC bias. And AC absolutely CAN be negative compared to the DC level it's riding on. Think of a virtual ground right between the two diodes, and look at each output transistor separately...as voltages goes up, it turns one transistor on and the other one off, as Les described. As it goes down, it turns the first transistor off and the second one on. The diodes mean that neither transistor turns COMPLETELY off (if they did, the part of the signal between -.7V and .7V would get lost), but they go as low as they can without ceasing to conduct.

 

I'm confused about why the either transistor cuts off? I hope that question is specific enough.

Look up the Classes of Amplifiers and the efficiency of each class.

Pure Class B amplification has "cross over distortion". It's been that way since vacuum tubes. To reduce that distortion you move the bias point away from class B and slightly closer to Class A. That class can be Class AB ... I've seen it called AB1 and AB2 as well, depending on where the bias was with respect to the class B bias point.

 

Les and to3metalcan,

Thanks for the help - I just figured it out. I was confused because I thought that the point between the transistors always had the same voltage as that between the two diodes. Realizing that those voltages fluctuate relative to each other, this all makes sense now.