Q: feedback and input impedance in a emitter follower

--SNIP__

Sorry I haven't been able to get back to this for a while. Improve your calculations like this:

Pull up the collector 1 volt. Then you will have *three* delta currents to sum.

1. The current due to transistor action. This will be HFE times the current in the 4k transistor input resistance. The voltage at the junction of the bias resistors, the 1k input resistor, the 4k transistor input resistance and the 10k feedback resistor is applied to the 4k transistor input resistance and the current in the transistor input resistance is multiplied by the HFE of the transistor. The voltage at the junction of those 5 resistors will be:

(1k || 4k || 27k || 82k) / (10k + (1k || 4k || 27k || 82k))

It's just a simple voltage divider. Then the current in the 4k input resistance (HFE is 200) multiplied by HFE is:

((1k || 4k || 27k || 82k) / (10k + (1k || 4k || 27k || 82k))) / 4k * HFE

1. The current in the resistance seen looking from the collector into the

10k feedback resistor. This will be 10k in series with 4 resistors which are in parallel: 10k + (1k || 4k || 27k || 82k). Thus the current in the 10k feedback resistor is 1 / (10k + (1k || 4k || 27k || 82k)).

1. The current in the 4700 ohm load resistor which will be 1/4700 amps.

Take the reciprocal of the sum of these currents and get 257.797896 ohms, exactly what the nodal analysis gets.

In this case, just ignore the 1k input resistor.

Pull up the collector 1 volt. Then you will have *three* delta currents to sum.

1. The current due to transistor action. This will be HFE times the current in the 4k transistor input resistance. The voltage at the junction of the bias resistors, the 4k transistor input resistance and the 10k feedback resistor is applied to the 4k transistor input resistance and the current in the transistor input resistance is multiplied by the HFE of the transistor. The voltage at the junction of those 4 resistors will be:

(4k || 27k || 82k) / (10k + (4k || 27k || 82k))

It's just a simple voltage divider. Then the current in the 4k input resistance (HFE is 200) multiplied by HFE is:

((4k || 27k || 82k) / (10k + (4k || 27k || 82k))) / 4k * HFE

1. The current in the resistance seen looking from the collector into the

10k feedback resistor. This will be 10k in series with 3 resistors which are in parallel: 10k + (4k || 27k || 82k). Thus the current in the 10k feedback resistor is 1 / (10k + (4k || 27k || 82k)).

1. The current in the 4700 ohm load resistor which will be 1/4700 amps.

Take the reciprocal of the sum of these currents and get 78.05329 ohms, exactly what the nodal analysis gets.

Your analysis calculates the voltage gain of the circuit without the 1k input resistor, with the feedback resistor in place, and you get a gain of about -159. You then add the 1k input resistor and treat the whole thing like an inverting op amp circuit, but where the op amp has limited gain.

What if you first analyze the complete circuit, 1k input resistor included, but without the feedback resistor, and get an open loop gain. Then connect the feedback resistor and get the closed loop gain of the circuit with feedback. Don't you get a different feedback factor this way?

Why should that be?

Don't forget the 10k resistor ! That's another 1/10000 amps.

Now what do you make it ? Close to my answer by any chance ?

Graham

Correct.

Well you would. Let me think about that. I suspect it's not very relevant though.

'Loading' if indeed it's an issue.

Graham

I didn't forget it. Didn't you see No. 2? I quote (for the short circuited input) :

"2. The current in the resistance seen looking from the collector into the

10k feedback resistor. This will be 10k in series with 4 resistors which are in parallel: 10k + (1k || 4k || 27k || 82k). Thus the current in the 10k feedback resistor is 1 / (10k + (1k || 4k || 27k || 82k))."

Similarly for the open circuited input case.

If by "Now what do you make it ?", you think that I failed to take into account the current in the 10k feedback resistor, I explained just above that I *did* in fact take it into account, and therefore nothing changes. The output impedances are just as I show.

The relevance is that it's the identical circuit, analyzed in two different ways. Wouldn't you expect to get the same result from the two different analyses?

If two different analyses of the same circuit get different results, doesn't that suggest that at least one of the analyses is in error (by the difference in the two analyses if only one is in error)? Or, perhaps, they're both in error by amounts sufficient to account for the difference.

How shall a person determine which analysis is correct (or *most* correct)? And if the analyses are only approximate, how shall we determine just how much error they have?

I'm talking about the circuit as it stands, with no additional loading.