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Where is a single stage CE BJT amplifier with a bypass capacitor supposed to clip with a normal 4-resistor biasing circuit? Like I seem to be able to get the upper peak of the output voltage all the way to VCC before it clips without a bypass. With a bypass, the gain is actually a lot larger, but clips a lot sooner, as in before the upper output voltage peak reaches VCC. I get the clipping without a Bypass is caused by it reaching the voltage of the supply, but what about with a bypass, what causes it to clip before reaching the supply. I see something similar with CC and CB amplifiers. Though maybe it's all for similar reasons.

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  • \$\begingroup\$ It would help if you'd provided a schematic. But I think I have an idea about what you need to understand. You aren't seeing "clipping" when you add the bypass capacitor, but something else. \$\endgroup\$ – jonk May 8 '18 at 19:10
  • \$\begingroup\$ I'll provide a short answer. If you have questions, feel free to ask them. \$\endgroup\$ – jonk May 8 '18 at 20:53
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Assuming that what you did is place a capacitor across the emitter resistor, then what you are seeing is the effects of very strongly variable gain. Suppose your collector resistor is \$3.3\:\text{k}\Omega\$ and the quiescent current is \$1\:\text{mA}\$. With a well-applied bypass capacitor across the emitter resistor (the value isn't important right now), your effective dynamic emitter resistance depends upon the collector current, which itself of course depends on the collector voltage.

The dynamic resistance at \$I_\text{C}=1\:\text{mA}\$ will be about \$26\:\Omega\$. But as the collector voltage approaches the upper rail voltage, this current diminishes towards zero, of course. Let's say it diminishes towards \$I_\text{C}=100\:\mu\text{A}\$. Then at this point the dynamic resistance will be \$260\:\Omega\$. And at the other end let's say that the collector current increases to as much as \$I_\text{C}=2\:\text{mA}\$. Then here the dynamic resistance will be about \$13\:\Omega\$. This means the gain will vary from about \$\frac{3.3\:\text{k}\Omega}{260\:\Omega}\approx 12.7\$ to \$\frac{3.3\:\text{k}\Omega}{13\:\Omega}\approx 250\$.

So during each cycle you will see quite a variation in the gain, with the lowest gain occurring when the collector voltage is at its highest voltage and the highest gain occurring when the collector voltage is at its lowest voltage during the cycle.

The effect of this widely varying gain is that the upper part of the curve will look more "mushy" (gradually rounded) and the lower part of the curve will look sharper and more pointed. This may look to you very much like clipping because the tops of the curve look a lot like that. But it is NOT that.

With enough signal applied, you can still force the output voltage all the way towards the upper rail. But then it is looking a lot more like a square wave. Long, long before that point and well underneath the voltage rail, you will see that "nearly clipped" looking signal.

Here's an example:

enter image description here

The blue line is the quiescent point for the amplifier. The green curve is the amplified signal. You can see that the tops do not reach the \$10\:\text{V}\$ rail. So there is no clipping taking place. This is variable gain in action.

Usually, if this kind of amplifier is applied (and it won't be so badly designed), there will be some global negative feedback applied to correct the distortion while still getting access to higher but variable gain.

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