modified 6 years ago

Dead zone reduction with VI feedback topology

00:53:18

The 50 ohm is the output impedance of the voltage source. To see the effect of feedback, just close those switches. 

published 6 years ago

wty

6 years ago

the dead zone is not reduced by this circuit. Closing the switches reduces the voltage on the base, and the transistor is always in cutoff region when all switches are closed. The voltage on the load is caused by the feedback resistors. You can prove this by removing the power source.

hurz

6 years ago

@wty, right. Only this cant reduce the dead zone. I think best would be to explain first what this "dead zone" is all about before you try to fix it.

KeenZa

6 years ago

@wty Thank you very much for your clarifying. I completely forgot about the operation of transistors which made this circuit become a voltage divider.

KeenZa

6 years ago

@hurz I think the dead zone of a device is the range of input in which we cannot use to turn on it. In this case, i think i have to bias those transistor to a forward active region.

hurz

6 years ago

yes, its this what makes the cross over destortion in class B push pull stages. Right, you need a biasing to avoid/suppress this.

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