# Input impedance of an emitter follower

Discussion in 'Homework Help' started by umichfan1, Aug 30, 2012.

1. ### umichfan1 Thread Starter Member

Jun 16, 2012
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pp. 95-96 of "Student Manual for the Art of Electronics" (see attached) asks us to determine the input impedance of an emitter follower circuit (see p. 96 for the problem statement). The problem says to begin by measuring the voltage on either side of the 10k resistor. I did this, and the results are sketched in the other attachment. Basically, the signal on the far side of the 10k resistor (far from the AC source) is 20 mV below the signal on the near side (constant DC offset). So there is a 20 mV drop across the resistor. But how do I use this to calculate the input impedance? I thought maybe I should use (20mV)/(15V) = (10k)/(Z_in). This gives a value of 7.5 mega-ohms for Z_in. I'm skeptical that this is correct, though, because according to p. 66 of "The Art of Electronics," Z_in = (beta+1)*R_e. Here, R_e=3.3k, which implies that beta=2271. This is a very large value for beta, which I don't believe. But if this is not correct, then how do I calculate the input impedance given only the voltage drop over the 10k resistor?

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2. ### mlog Member

Feb 11, 2012
276
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I assume your waveform didn't look exactly as you sketched it. If you had 20 mV when the voltage crossed the zero line, then something's wrong. On the other hand, if you're trying to show that the delta is 20 mV at the peak, then OK.

If you measure the voltage across the 10k resistor, then you know the input current. If you divide the input voltage (presumably measured at probe 2) by the input current, you should get the input impedance.

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3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Like this .....

• ###### Input Resistance Measurement.pdf
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4. ### mlog Member

Feb 11, 2012
276
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You got a much different delta than he did, i.e. 200 mV vs. 20 mV, but I don't think he mentioned his input source voltage.

I just noticed something else. You added the bias resistors, 47k and 23k, and they dominate the input impedance.

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Yep - a practical emitter follower would of necessity have some biasing requirement.

But it's not too hard to work out the resistance "looking into" the base.

It's around 202kΩ.

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6. ### umichfan1 Thread Starter Member

Jun 16, 2012
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Thanks a lot for the help. I can see that I must have wired this up incorrectly, because the voltage difference is indeed 20mV both at the peak and when crossing the zero line (that is to say, it is a DC offset). Also, this DC voltage difference does not change when I vary the amplitude of the input signal--though it is attenuated when I decrease the magnitude of the power supply connected to the emitter resistor (at negative voltage relative to ground). Not sure what's going on--guess I've got some debugging to do. Thanks for the help!

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I was reminded that a technique called bootstrapping could be used to offset some of the loss in input resistance caused by the base biasing resistors. Requires two additional components as shown in the attachment.

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