question about npn clipping from art of electronics book

AnalogKid

Joined Aug 1, 2013
8,106
In order for the output voltage to go lower and lower, the transistor has to conduct less and less. It is the emitter resistor that "pulls" the output down to -10 V. However, the load is another 1 K resistor to ground. When the transistor is completely turned off, the two resistors form a simple voltage divider between GND and -10 V. If the two resistors are of equal value, the middle node is half the total voltage, -5 V. If the load were 111 ohms, the clipping level would be -1 V (-0.999 V).

ak
 
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wayneh

Joined Sep 9, 2010
16,102
The text says it clips at an output voltage of -5V, not input. But neither makes sense to me.

Gaaa, AK beat me to it and even got it right. I had missed he ground symbol which is not the same as -10V. I wish they'd use 0V instead.
 

Thread Starter

tpny

Joined May 6, 2012
216
I also didn't see the ground symbol. second you on using 0v insteadl

Hi ak, thanks for illuminating duly. But you said if load were 111 ohms, the clipping would be 1v, did you mean -1v? thank you!!
 

wayneh

Joined Sep 9, 2010
16,102
I don't. 0 V and GND are not interchangeable symbols or concepts.
You're right, but I think in the example the important thing we both missed was that the point marked with the ground symbol was at 0V. That's our bad, but explicit labelling of that point as 0V might have prevented the miscommunication.
 

wayneh

Joined Sep 9, 2010
16,102
**even**?

And, in this unschooled, wild west internet age, they *even* got the ground symbol pointing in the correct direction. even.

ak
All I meant by that is that you were both faster and correct, not just one or the other. I could have used "also".
 
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