Q: feedback and input impedance in a emitter follower

[Eeyore]

What bugs me about that approach is that it makes the closed-loop Zout
be dominated by the collector load resistor value.

Well yes !

Change that to,
say, 1K, and the resulting output impedance would decline to 52 ohms.
So putting about 1.5K across the existing collector load resistor (ie,
across the 190 ohm Zout) reduces Zout from 190 to 52, which is sorta
strange.

No. Reducing the collector load also reduces Avol so the feedback factor reduces and
the output Z stays virtually unchanged.

Or replace the collector load resistor with a big inductor, and this
technique computes Zout of about 590 ohms. Fishy ditto.

How did you compute that ?

I think the safe thing to do is assume some delta-v at the collector,
and then compute all the contributions to delta-i, and divide.

See above.

Graham

 

[The Phantom]

On Fri, 27 Oct 2006 13:18:54 -0700, John Larkin

--SNIP__

I liked all your math up to the Zout calc, which didn't make sense to
me. Try this:

Assume high beta and that the input is shorted. Hook a signal
generator to the output and wiggle the collector voltage. If you pull
the collector up 1 unit (let's say 1 volt for argument, ignoring
linearity), the base voltage goes up about 1/11 V. Assuming 1 mA bias,
Re = 26, that induces (1/26)*(1/11) A of collector current. So the
collector itself looks like about 286 ohms.

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

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)).

3. 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.

But if you leave the amp input open, it's different. At beta=500,
pulling up Vc by "one volt" dumps about 43 uA into the base, giving 21
mA in the collector, so the transistor looks like 46 ohms. I've
ignored the loading of the base bias resistors, so the real value will
be higher, so probably the Spice 78 ohms is reasonable.

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


2. 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)).

3. 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.

 

[The Phantom]

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?

 

[Eeyore]

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

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

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

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

Graham

 

[Eeyore]

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.
Correct.


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?

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

Why should that be?

'Loading' if indeed it's an issue.

Graham

 

[The Phantom]

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

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.

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

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 Phantom]

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

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?

'Loading' if indeed it's an issue.

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