# C.E. Amp. w/ Emitter Resistance (Quick/Simple)

Discussion in 'Homework Help' started by jegues, Apr 24, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
I've got quick question about the input resistance for this particular circuit.

The input resistance is looking in from the white node to the right of Rsig right?

Why doesn't his calculations for Rin include that 0.1MΩ resistor as well? Is there a mistake?

Can someone clarify?

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2. ### wmodavis Well-Known Member

Oct 23, 2010
737
150
I do not see any 0.1M Ohm resistor.

3. ### designnut Member

Apr 21, 2011
33
1
The .1 ma refers to an unknown current source. Did you mistake it for a resistor?

4. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
No, the 1M ohm resistor to the right first of the capacitor that couples the signal voltage.

5. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Still looking for some clarification on this one. My test is soon!

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Strictly speaking the 1MΩ should be included - maybe the person who wrote the solution thought it wouldn't change things by that much.

With the 1MΩ included Rin=50.5k||1M=48.07k which is about a 5% change and probably shouldn't be disregarded.

7. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Ah so that's the best way to deal with the 1MΩ being in there, find the input resistance to the right of it (say Rib), and combine that in parallel with the 1MΩ to get the actual input resistance(Rin).

Where, Rin = (Rib//1MΩ)

Thank you for clearing this up!