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Following on from my last question Cascaded BJT Amplifier Load - I'm trying to understand what it is (via math or otherwise) that causes the input signal to appear malformed when the output of the signal starts 'clipping'

Looking at my circuit:

enter image description here

Am I right in saying that for the output at the collector (V(n002)):

  • The clipping at the high end is because it cannot swing above Vcc?
  • the clipping at the low end is because the voltage at the collector (V(n002)) cannot go below the voltage at the transistors emitter (3.18V)?

Then my main question is - Why is it that the limits of the biasing/BJT cause a change to the input signal when measured at the base of the BJT (V(n005)?

Many Thanks

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  • \$\begingroup\$ Yes, the positive clipping acquires because the BJT's is cut-off (Ic =0A) and the negative clipping is when the BJT is entering the saturation region (Vc = VB). And I don't understand your question. Can you elaborate it more? \$\endgroup\$ – G36 Aug 16 '20 at 13:52
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  • Yes
  • Yes

I'm assuming that you're trying to create a linear amplifier. But perhaps you're investigating behaviour of an overdriven amplifier.

You are driving the base with an input signal too large for linear operation where the transistor has current gain.

The base biasing resistors R1, R2 weakly establish base voltage because their values are large. When the transistor is over-driven, the base-emitter junction pulls a lot more current on positive peaks, because during these positive peaks the transistor has no current gain.
Try looking at base current in LTspice:current at base of transistor after 100 seconds
The difference between I(R1) and I(R2) is base current: about 10uA. Base bias current of R1, R2 is only about 25uA with no input signal. When over-driven by the signal generator, current into the base peaks at 44uA - this current dominates the static bias current of 25uA, and causes base voltage to change from its static value ( where you have no AC input).

If you decrease input voltage from 0.1V peak to 0.01V peak, base current becomes more sinusoidal shape, with average value near 2.5uA. Base bias currents from R1, R2 (about 25uA) now dominate over transistor base current. The transistor is now working within (mostly) its linear range, and for the full duration of the input signal cycle, its current gain is nearly constant @ hfe=100.

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Why is it that the limits of the biasing/BJT cause a change to the input signal when measured at the base of the BJT

If you try and "brute-force" the collector down to the emitter voltage, the normally reverse-biased BJT base-collector region becomes forward biased and the current gain plummets. After all, the base-emitter region is forward biased and, if collector drops close to emitter then that region is also forward biased thus ruining hFE.

The effect of this is that your base signal is now supplying a lot of the current into the collector and emitter and hence it clips due to the series resistance you have in the base circuit (Rsig).

The clipping at the high end is because it cannot swing above Vcc?

Correct.

the clipping at the low end is because the voltage at the collector (V(n002)) cannot go below the voltage at the transistors emitter

Correct.

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