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What I understand is: in discrete amplifier circuits, the aim of the DC analysis is to find the DC operating point. (I get this part)

And the texts I read so far mentions that AC analysis is done to obtain the Voltage gain, input and output impedance, and frequency response.

My confusion is that: why do we need AC analysis but not DC analysis to find the input and output impedance? Why do we ignore the DC part of an input signal when finding input or output impedance?

Is there a way to clarify/understand this in a simple fashion? Would be nice to consider a BJT common emitter amplifier with a resistor at its emitter such as:

enter image description here

enter image description here

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  • \$\begingroup\$ Note that the DC input and output resistances of your amplifier are both infinite. \$\endgroup\$ – Olin Lathrop Nov 19 '16 at 23:17
  • \$\begingroup\$ @OlinLathrop so is there something called DC input and output impedance? And can you tell me why they are infinite? There is still current flowing through base even there is no input signal. What is going on there? \$\endgroup\$ – floppy380 Nov 19 '16 at 23:19
  • \$\begingroup\$ Capacitors are open circuits at DC, meaning they have infinite resistance. \$\endgroup\$ – Olin Lathrop Nov 19 '16 at 23:22
  • \$\begingroup\$ @OlinLathrop Some examples don't have caps when deriving impedance. I ruined my question by these caps. I should have added an example without capacitors. What if C1 wasn't there? Would we still ignore the DC component and make AC analysis when finding the input output impedance? \$\endgroup\$ – floppy380 Nov 19 '16 at 23:26
  • \$\begingroup\$ if C1 wasn't there, then what is it replaced with? an open circuit? a short circuit? \$\endgroup\$ – robert bristow-johnson Nov 19 '16 at 23:47
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The amplifiers you are speaking of are audio, that is, they only concern themselves with AC. As such, they are ordinarily AC coupled at both input and output with capacitors. So their DC input impedance is expected to be infinite, and their DC output impedance infinite.

EDIT - You added your schematic while I was composing this. Note the input and output capacitors. END EDIT

For amplifiers which are expected to include DC, DC input impedance must be specified, and likewise output impedance. You'll see both of those addressed in the data sheets of operational amplifiers. For inputs, you'll see specifications for input offset voltage, input offset current, and input bias current. For outputs, you'll commonly see the maximum voltage swing available for one or more load resistors, which is directly related to output impedance (although the presence of loop gain complicates the matter).

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  • \$\begingroup\$ I should have added an example without a capacitors. What if C1 wasnt there? Would we still ignore the DC component when finding the input output impedance? \$\endgroup\$ – floppy380 Nov 19 '16 at 23:25
  • \$\begingroup\$ @doncarlos - The circuit you show is not usable as a DC amplifier except over a small range, as varying the DC inputs changes the operating point. So nobody uses it as a DC amplifier, and nobody cares about its impedance in this case. \$\endgroup\$ – WhatRoughBeast Nov 19 '16 at 23:28
  • \$\begingroup\$ I think I get your point. But just to clarify one more thing: Imagine I remove C1(short the input to the base) and apply a constant small DC signal in my first circuit's input. And in this case wouldn't I obtain the same input output impedance as in AC analysis case? \$\endgroup\$ – floppy380 Nov 19 '16 at 23:34
  • \$\begingroup\$ @doncarlos - To a first approximation, the input impedance will be the Thevenin input equivalent - that is, the parallel resistance of R1 and R2. In general, you set the operating point by assuming the base current is much smaller than the bias resistor current, so this approximation is valid. \$\endgroup\$ – WhatRoughBeast Nov 20 '16 at 0:23
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After you eliminate C1 you can determine DC input resistance this way:
1) Determine DC base voltage.
2) Apply a DC source having exactly the same voltage as base voltage.
The current drawn from this DC source will be zero. At this point, you could say that input resistance is infinite (V/I is infinite). This is dissimilar to a result of AC analysis.
AC analysis applies a changing voltage or current to your circuit. Input impedance is determined from delta V / delta I.
So you could raise the DC source voltage a small amount (say 10mv). Remeasure the current drawn from this source - it will be larger than zero as before, let us say it turns out to be +100uA. Input impedance is 10mV/100uA, or 100 ohms. But one must consider what to do with the transistor emitter. It should be grounded (as you have properly done with the AC schematic analysis). But this cannot be done for DC because it disturbs transistor biasing. The best we can do is to keep emitter voltage the same as it was before the 10mV rise in base voltage. The result for input resistance should correspond to results from AC analysis.
You must use small deltas to do this properly - in AC analysis, your circuit is assumed to be linear, and this is approximated by small disturbances to your source.

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  • \$\begingroup\$ What is the difference between current gain in AC and current gain in DC? \$\endgroup\$ – floppy380 Nov 20 '16 at 0:42
  • \$\begingroup\$ @doncarlos Have edited my answer to keep emitter voltage constant. Current gain is usually considered to be output current to RL only, divided by input current through Rg. The delta method still applies, with the caveat of keeping emitter voltage constant. If you insist on keeping all capacitors active, then DC current gain is zero, as is DC voltage gain. Am unsure what you really mean by DC gain vs. AC gain. \$\endgroup\$ – glen_geek Nov 20 '16 at 2:16

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