Q: feedback and input impedance in a emitter follower

even bother

I'll analyse the circuit for you if you like.

Initially without feedback

First off I want to know the hfe for the BC547A. Interesting since Philips doesn't list a BC547A so I'l use the BC547B !

200 min 290 typ 400 max @ 2mA which is close enough to our needs.

Note that the other h parameters you want aren't listed - nor have I ever seen them in any datasheet.

The base bias voltage is 12*27/(27+82) V = 2.97V from a source resistance of 27*82/27+82 k = 20.3 k

Vbe can be estimated as 0.55V.

You can make a nice sinulataneous equation if you like to calcualate the next but but I'll approximate again. Assuming minimum loading of the bias circuit the emiiter voltage will be (

2.97 - 0.55 )V = 2.42V

Which gives an Ie of 2.42 / 1.8 mA = 1.34mA

So Ib = 1.34mA/200 = 6.7uA

What's 6.7uA in 20.3k ? it's 136mV so I'm not really going to bother going round the loop again. You could if you wanted to by making the emitter voltage 2.85V.

The dynamic emitter impedance re will be 26/1.34 Ohms = 20 Ohms.

Making the input impedance of the transistor itself 20*200 = 4 k Ohms

So the total input Z is 4k in parallel with 20.3 k and all in series with 1k. =

4.34k

The ac gain will be ~ 4.7k / 20 = 235 which suggests there was something wrong with Peter Andersen's method.

Closing the feedback loop adds a futher 10k in paralle with the 4k7 collector load so the Avol drops to 10k || 4k7 / re = 159x

This is now an inverting configuration with a virtual earth at the node comprised of R1, R2, R6 and R5.

The actual impedance here will be R5 / Avol = 10k / 159 = 63 ohms. So the input impedance including R6 will be 1.063 k Ohms.

If the gain block had 'infinite gain' the voltage gain would be R5/R6 which is

10x. Since Avol is 159x it's going to be less by 10/10+159 x = 6%

So I make the closed loop gain 9.4x

Repeat as required.

OTOH you can simply see the dominating effect of hfe*re in the first instance. In the second you can see it's nearly parameter proof ( but it'll be noisy )

A far beter configuration by far here would be shunt derived series applied feedback.

Graham

--SNIP--

doesn't list a

them in any

They're in the Vishay data sheet available at:

Figure 8 in that data sheet shows how they vary with collector current.

but but I'll

voltage will be (

round the loop

4.34k

with Peter

load so the Avol

comprised of R1, R2,

input impedance

10x. Since Avol

In the second

feedback.

doesn't list a

seen them in any

That is *very* unusual / untypical however.

I've never seen any semi maker quote these before.

Graham

doesn't list a

seen them in any

Look in the big brown Motorola "Small-Signal Transistor Data" book.

The venerable 2N2222, the 2N3903 family and others have h parameters. A number of the higher frequency devices have admittance parameters.

The National Semiconductor data book gives h parameters for the 2N2907,

2N3019, 2N3904, 2N5086, and 2N5088.

The Fairchild Semiconductor data book gives h parameters for the 2N2695,

2N3072, 2N3120, 2N3502, 2N3930 and quite a few more.

My old General Electric data book gives h parameters for the 2N43 (germanium), 2N461, 2N524, 2N4424, 2N3858, 2N5232, 2N3392, 2N5827 and LOTS more.

I can't speak for the European manufacturers, but here in the states it's fairly common.

How would you calculate the output impedance for the open loop case and the feedback case?

share

doesn't list a

seen them in any

From that era I only have Motorola's European Supplement First Edition.

I do see those numbers for the BC147 ( similar to BC547 ) but not for much in the way of much later devices. Simply because it's not needed !

In any case what's the point of having hie specified at 2 mA when you want it at

1.3mA in this case ?

Interestingly the BC107A hie is quoted as 6.3k Ohms @ 2 mA which looks wrong to me !

Who still even uses these devices aside from the 2N3903,4 etc series which seems to be unkillable ?

My 1967 RCA databook doesn't seem to have much other than hfe at a quick glance.

Maybe in the 60s and 70s but not now and not typically for modern devices either.

Graham

In the open loop case the output Z is 4.7k in parallel with the collector resistance. Since the collector resistance is very high indeed, it's essentially negligible.

So for all practical purposes it's 4.7k.

In this case the feedback R lowers the collector load to 3.2k. This value is then divided by the feedback factor 159/9.4. which gives ~ 190 Ohms.

Incidentally, looking at the value of hie in a couple of other early data sheets it's not toleranced ( unlike hfe ) which makes it completely meaningless !

You would actually *introduce* errors by using these numbers for hie ! It also enourage rote thinking instead of a proper analysis.

Graham

prefer

The OP was asking how to calculate Zin with and without the feedback connected. Without, and assuming a beta of 500 (not at all unusual these days) the input impedance at the base is ballpark 13K, and depends strongly on beta.

But I think his point wasn't what the value is, but how to calculate it. My point was that you don't need an overall formula to get the answer; if you know the 26*B/Ie thing, you can get everything else step-by-step.

John

a

share

Philips doesn't list a

seen them in any

the way of much

at 1.3mA in this

That's what Figure 8 in the Vishay BC547 data sheet is for. It shows the variation of the parameters with collector current. Almost every transistor data sheet that gives h parameters also has graphs showing the variation of the parameters with collector current.

me !

On this page:

the second choice down is from Micro Electronics. On the third page of the data sheet they show h parameters for 3 gain groups. Look at group C; the typical hie is 8.7k @ 2 mA. But look at the hfe range. It's pretty clear that hie rises with increasing hfe. In fact the rule of thumb formula for input resistance of a transistor (hFE * 26 / Ie) gives a fair approximation to hie, but hie is measured with an AC short on the collector which can cause it to be somewhat different than the DC formula value.

The 2N3903/2N3904 data sheet:

shows for the 2N3904 (at 1 mA) a 10 to 1 variation in hie, but only a 4 to

1 variation in hfe, so there's something going on that the formula hFE * 26 / Ie doesn't capture. This means that the DC formula (hFE * 26 / Ie) for transistor input resistance can be in substantial error, and open circuit performance may be considerably different than the DC formula would predict. Of course, if there's enough feedback the final circuit performance may not be much affected.

But I think it's safe to say that hie and hfe track at least in the sense that they increase or decrease together in different transistors. You won't find the largest hie associated with the smallest hfe in a given transistor.

The 2N3903 data sheet also has a graph showing the variation of the h parameters with collector current. If you look at the corresponding graph for other transistors which specify h parameters, you will note that the line for hie versus current is usually a near straight line. The BC107 and BC547 sheets show it ruler straight. The 2N3903 shows a slight curvature and the Motorola 2N2222 is somewhat S shaped. To the extent that hfe doesn't vary much with collector current, the line will probably be straighter. But in all cases, hie is varying just about inversely with collector current, which is what I would expect.

seems to be unkillable

either.

There aren't many modern (meaning recently designed), low-frequency, low-power, general purpose bipolar transistors are there? If you need a low-cost, general purpose little bipolar you're probably going to be using a 2N3903, 2N4401, 2N2222, 2N2907, etc., or some European equivalent. About the only modern low-power, low-frequency transistors are part of an integrated circuit.

Your analysis has failed here. The correct value (with hfe of 200 and transistor input impedance of 4000 Ohms, which is what you used in your example analysis) is about 78 Ohms. You can verify this with Spice.

I think your condemnation is exaggerated. If you calculate the input resistance with the DC formula (HFE * 26 / Ie), using typical HFE, you also get an untoleranced number. You have to look at the tolerance on HFE to get an approximate tolerance on the input resistance.

The designer experienced in the use of h parameters knows that hie at a given collector current is just about proportional to hfe, so from the tolerance on hfe one can get a idea of the tolerance on hie. But, as I pointed in another post, it's only approximate. The 2N3904 data sheet gives a 4 to 1 variance in hfe, but a 10 to 1 variance in hie, so using HFE

• 26 / Ie as an approximation to hie can be substantially in error.

How would using the data sheet numbers for typical hie be more guilty of introducing errors than using the typical values for hfe? Anytime you use "typical" parameter values, you introduce error. That's why a full analysis with all parameters can be a good thing. You can vary the parameters over the expected tolerance range, re-run the analysis and see how much things change. You speak of "experience". Experience can teach one how much typical variation in the individual h parameters ought to be expected even if the tolerances aren't on the data sheet.

I have long known of and used the method your example analysis demonstrates, but I cannot agree that it is any more "proper" than a full nodal analysis. The full nodal method doesn't give results that are any less accurate than the simple method.

When I want a quick "back of the napkin" analysis, I use the simple method myself. A "proper" analysis should take into account *all* parameters that can affect circuit performance. It's a fact that *all* of the h parameters have an effect on circuit performance, and I would say that it's "rote" thinking to believe that hre and hoe need never be taken into account. It may be true in some particular circuit, but how would a person know if they don't do the analysis? If a person does a full analysis when first encountering a particular circuit and determines that, for *this* circuit, hre and hoe may be ignored, that fact becomes part of one's experience. But that doesn't mean that some other circuit shares that same property. And if circuit reliability is important, a full analysis should be done even if *similar* circuits from the past showed little sensitivity to some parameter.

have a

share

Philips doesn't list a

ever seen them in any

the way of much

at 1.3mA in this

Not the Motorola ones I saw !

to me !

Which is what I use.

The point is that a BC107A will never have an hie of 6.3k @ 2mA.

I'd like to see a proof of that.

Of course to be totally acurate you should also include rbb.

Of course.

I'd expect so !

seems to be unkillable

glance.

either.

I use none of those 2N parts. Of the European parts I tend to use the BC546,7,8,9,550 family and their complements for low level, MPSA42,43,92,93 for higher voltages and selected Japanese parts for very low noise.

I've never needed anything other than hfe out of the h parameters to use them just fine.

Just how different I wonder are the BC107, 147 , 182, 337, 547 ?

Graham

I'm suspicious of this.

Please explain how this is.

Which my method does automatically.

I'd like to see a reason for that. It sounds wrong to me.

Graham

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.

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.

Does that make sense?

John

is

I've always understood that negative feedback lowered the output impedance by the feedback factor. Maybe that's an oversimplification ? I'm puzzled how it could be lowered by more though.

The quiescent current was in fact ~ 1.34mA. Re was ~ 20 ohms.

Putting those numbers in gives 213 ohms which is quite close to my ~ 190 ohms.

That sounds fine as far as it goes but is unrepresentative of real operation. I didn't consider this example.

Sure but it seems my simple method is actually quite accurate ! Note that your

286 ohms was based on the wrong Ic. When I adjusted for Ic = 1.34mA I made that number 213 ohms.

Graham

I don't know what to tell you except that if you do a full nodal analysis, that's the result you get. If you simulate the circuit in Spice, you should also get the same result, although I had to adjust the hfe in the transistor model to be exactly 200.

Or, you could try this:

Imagine an ideal circuit element, a current-controlled current source, with a current gain of 200 (currents are all directed into a node), oriented input to output from left to right. Such an element would have a short at the input, and infinite output impedance. Put a 4000 Ohm resistor in series with the input (left end) to represent the 4000 Ohm input impedance of the transistor. Place a 4700 Ohm load resistor from the output of the current source to ground. Also put a 27k and 82k to ground from the left end of the 4000 Ohm resistor at the input to the controlled current source. The left end of the 4000 Ohm resistor is to be the input node. This arrangement constitutes an amplifier similar in its major characteristics to the transistor amplifier we've been discussing, minus the 1000 Ohm input resistor which is in the transistor circuit.

Set up the circuit equations in whatever mode you favor, and solve for voltage gain, input impedance and output impedance.

You shouldn't need to solve the equations to determine input and output impedance without feedback, but doing so will verify that you have the equations set up correctly. The left end of the 4000 Ohm input resistor is equivalent to the base of the transistor. The impedance at this point is

4700 || 27k || 82k, or 3341.8868 Ohms, since the input impedance of the controlled current source is zero. The output impedance of this arrangement is, of course, just 4700 Ohms, the load resistance. The voltage gain is -235.000.

Now place a 10k Ohm resistor from the output of the controlled source to the left end of the 4000 Ohm input resistor. Solve the equations again, and you should get a voltage gain of -159.544, an input impedance of

61.1484 Ohms, and an output impedance of 78.0533 Ohms (input open circuited for the output impedance calculation).

Notice that the gain reduction factor is 1.473, the input impedance is reduced by a factor of 54.652, and the output impedance is reduced by a factor of 60.215. This is all intended to represent the transistor amplifier without the 1k Ohm input resistor. Where you went wrong was in using the feedback factor for the complete amplifier, 1k input resistor included, which you calculated as 159/9.4, and assuming the output impedance would be reduced by that factor. A much closer result would have been obtained by taking the factor by which the input impedance at the base of the transistor was reduced, 3.34k/63 = 53 using your numbers, and dividing the 4700 Ohm load resistor by that factor, giving 88.68 Ohms, much closer to the correct 78.0533 Ohms, but still in error by 13.6%.

Notice that it is only approximately correct to assume that the input impedance and output impedance reduction factors, 54.652 and 60.215 (of the amplifier without the 1k Ohm input resistor), are the same, and neither of them is close to the gain reduction factor of 1.473.

These results correspond to the transistor amplifier with the 1k Ohm input resistor left open at the left end, and the base of the transistor considered to be the input (and left open when calculating output impedance). If a voltage source were connected to the 1k input resistor in the full transistor amplifier, then that would be the AC equivalent of shorting the left end to ground. In the controlled current source model, this effect can be had by simply connecting a 1k resistor from the input to ground. Then the output impedance becomes 257.8 Ohms.

I also noticed that the output impedance of the full amplifier (with 1k input resistor) is not much stabilized by the feedback. If the transistor parameters are allowed to assume the following values:

HFE transistor input impedance

50 1000 100 2000 200 4000

the closed loop gains are hardly changed, but the output impedance varies quite a bit:

HFE output impedance output impedance with input shorted with input open

50 379 215 100 299 125 200 258 78

I don't understand what you mean. If you only use the typical value of HFE, then you haven't taken into account the tolerances.

But if you redo the analysis with the minimum and maximum HFE from the data sheet, then you *have* taken the tolerances into account. Is that what you mean, that you redo the analysis with max and min HFE?

If that is what you mean, then the same thing can be done when using h parameters, and is just as "automatic".

You'll notice the same thing to a greater or lesser degree whenever h parameters are given with tolerances (for all the silicon transistors I checked, anyway), so I doubt that it's wrong. I think the reason is the internal feedback in the transistor, the same effect that causes hre to have a non-zero value.

Your calculation of the 1/11 ratio has ignored the input impedance of the transistor (because you assumed HFE high), and that's how HFE gets into the picture. If HFE is low then the 1/11 ratio has to be changed. But Eeyor's computation of output impedance uses the feedback factor which doesn't change much with HFE from, say, 50 to 200 and therefore indicates that the output impedance doesn't change with HFE, which as far as I can tell, isn't true (whether the input is open or shorted).

I have a feeling that Eeyor is going to say that he meant the output impedance with the input AC shorted, whereas I had an open circuited input in mind.

In my most recent long posting, I give an analysis of output impedance variation with HFE, and I think there is still a problem with Eeyor's analysis even if we agree that the input will be AC shorted to ground.

If HFE is reduced to 50, and we redo Eeyor's analysis, the emitter current is slightly reduced, perhaps to 1.17 mA instead of 1.34 mA. The following changes occur (cutting and pasting the relevant part of his analysis):

-------------------------

"The dynamic emitter impedance re will be 26/1.34 Ohms = 20 Ohms. 1.17 22 Making the input impedance of the transistor itself 20*200 = 4 k Ohms 22*50 = 1.1k So the total input Z is 4k in parallel with 20.3 k and all in series with

1k. = 4.34k 1.04k The ac gain will be ~ 4.7k / 20 = 235 which suggests there was something 4.7k / 22 = 214 wrong with Peter Anderson's method.

Closing the feedback loop adds a father 10k in parallel with the 4k7 collector load so the Avon drops to 10k || 4k7 / re = 159x 145x This is now an inverting configuration with a virtual earth at the node comprised of R1, R2, R6 and R5.

The actual impedance here will be R5 / Avon = 10k / 159 = 63 ohms. So the 145 = 69 input impedance including R6 will be 1.063 k Ohms. 1.069 k If the gain block had 'infinite gain' the voltage gain would be R5/R6 which is 10x. Since Avon is 159x it's going to be less by 10/10+159 x = 6% 145x 10/10+145 x = 6.45% So I make the closed loop gain 9.4x 9.35x"

---------------------------------------------------------------

So then his output impedance calculation would be:

"In this case the feedback R lowers the collector load to 3.2k. This value is then divided by the feedback factor 159/9.4. which gives ~ 190 Ohms." 145/9.35 206 Ohms.

But the actual output impedance (with input shorted and HFE = 50) is about 379.56 Ohms.

And as you can see by the small changes in his analysis, changing HFE from 200 to 50 has not much effect on the open loop gain, 4700/re, because re in this approximation only depends on the emitter current, which doesn't change much with HFE. And connecting the 10k to the collector gives about the same gain for almost any reasonable HFE for the same reason, so the calculated feedback factor hardly changes.

So how does this analysis account for the fact that the actual output impedance (with input either shorted or open) changes substantially with changes in HFE? And by "actual output impedance", I mean the impedance calculated using network equations. I verified the results obtained by the network solution in Spice with HFE of 50, 100 and 200 and with the input both open and shorted, and got very good agreement.

I allowed HFE to be 10,000,000 in the network solution, and the input impedance of the transistor proportionately larger, and the output impedance with the input shorted approaches a limit of 214.8 Ohms as HFE gets really large. So the classical formula for feedback amplifier output impedance works ok if the gain is really high, but it is in substantial error for HFE = 50, it would seem.

I'm thinking I'll have to look through the transistor drawer and see if I can find some suitable parts to actually build up this circuit. Does this mean that I don't trust Spice?

Of course. What's the point in measuring something with the input open ? You can't use it like that ! As the input Z is ~ 1k then it makes sense that the source would be low Z hence I replaced it with a short.

I can't see the point of that. The circuit serves no function with the input open.

Remeber that the impedance at the virtual earth node was ~ 60 ohms with typical hfe so this won't actually vary much.

I realised just as I was going out to the pub that I had actually made a slight mistake. I should have used Avol/11 not Avol/10 for the feedback factor.

This makes my ~ 190 ohms output Z actually ~ 210 ohms which is damn close to the

213 ohms found by using JL's method.

Graham

can't

would

open.

It serves no function with the output open, either. If you don't connect some kind of load, why have an amplifier, eh? And, yet you calculate the output impedance with no load connected. You can treat the complete amplifier as a two port with input and output either open or shorted and achieve a complete characterization. That characterization can then be used to calculate the output (or input) with any load on the other port.

hfe

the 213

What about the fact that the output impedance (with input shorted and with feedback) varies a lot with HFE, and your method gives almost no indication of that? With HFE of 50, your method still gives around 200 Ohms, but the true output impedance is about 379 Ohms.

Actually, I got that value of 379 Ohms by assuming that as the HFE went down to 50, the transistor input impedance went down to 1000 Ohms. But if the emitter current for HFE=50 drops to about 1.17 mA, then the HFE * 26 / Ie method would give an input impedance of around 1100 Ohms, and the output impedance would be about 397 Ohms.

And if the amplifier had an input resistor of 2k Ohms instead of 1k Ohms, with an HFE of 50, the error in output impedance calculated by your method would be even greater. Why is the error so large for output impedance when the calculation for closed loop input impedance and gain are rather close?

and

is

the

be

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

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

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.

John