Common Emitter Config with Negative FB

Il Thu, 28 Jul 2011 13:34:07 +0000 (UTC), Warren ha scritto:

I had an unused envelope on my desk so I tried to calculate the gain very quickly. I didn't resort to "feedback model", just plain old loop/node analysis. Assuming infinite output resistance for the transistor, I get that:

vo/vin = R1*(gm*R3+1)/(R1+R3)

which gives me (rough approx here) around 230 for the gain.

If the Early effect of the transitor lowers the "effective" R1, probably you could get that number from the sim.

I agree with you, this is not a circuit you see that often these days, but is a good candidate for some thought. I think it would be nice to examine it through the Extra Element Theorem lens as well as using it to apply some more canonical feedback method (because it's not ideal, so you have to be careful and develop some insight).

M

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Frustra fit per plura quod fieri potest per pauciora

SiSince you are driving it from a low impedance signal source, the nfb doesn't matter at the signal frequency. The feedback just sets the DC bias point. But a zero-impedance signal is unusual in real life.

The nfb *lowers* the input and output impedances below the values set by the resistors.

John

Jacob Millman, "Microelectronics", McGraw-Hill, 1979. Very good book. I haven't read a better one for transistor circuits. (But then, I haven't read _any_ other book for transistor circuits -- I'm not a great buyer of books).

1:

You compute the small signal gain by observing that the input impedance is very low, and thus that the small signal gain will be set by the transistor's g_m and collector load. Kv = g_m * R_c. Note that g_m is set by the emitter current -- at room temperature it's around I_c / 26mV, although that 26mV changes proportionally to absolute temperature, and the B-E voltage (which contributes significantly to emitter current) changes with temperature, too.

2:

The easy way to do it is by using the Miller effect. The input impedance is going to be R2 in parallel with the base impedance of the transistor (H_fe / g_m) in parallel with R3 * (1 + 1/Kv) (or R3/Kv, or R3/(1+Kv), I can't remember and it doesn't matter much, because you're going to see device variation far greater than one part in Kv).

3:

If the source impedance is truly zero then the input will short out any feedback effect from R3, so the output impedance will be pretty close to R1. If the source impedance is higher (and as the source impedance varies with frequency) the output impedance will stiffen up with increasing feedback.

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www.wescottdesign.com

Tim Wescott expounded in news:7eWdncIWFIQwEazTnZ2dnUVZ snipped-for-privacy@web-ster.com:

..

That's one I don't own yet- I may acquire it used. My wife will be thrilled with another addition. :)

Ok.

The sim shows I_e = 0.841 mA. So if I_c ~= I_e, then according to this g_m = 0.841mA/26mV => 0.032. Thus Kv ~= 265.

Given the sim gave approx 219, working backwards, we could achieve that result if the 26mV were 31mV instead. That seems a bit high (just my gut feeling).

That's my bad (I hadn't thought about the source resistance). Let's assume about R_s=600 ohms. I'll try this tonight when I get home (where I have the file saved). This might explain why LTspice had some wild numbers for the dc transfer, when I last tried it.

Warren

Take the loading of the bias network into account and it goes down to Kv = 236. The rest is probably accounted for by the loading of the base on the bias network (about 3k in parallel with R2), the finite collector impedance of Q1, and the finite base and emitter ohmic resistances.

If you shove an emitter resistance in there then the effective base impedance goes up, as does the collector impedance. The bias current is less dependent on transistor variation with temperature, and the linearity increases greatly because the emitter current is determined by a diode in series with a resistor instead of just a diode.

But that might cost a whole extra $0.001.

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www.wescottdesign.com

Il Thu, 28 Jul 2011 17:07:48 +0000 (UTC), Warren ha scritto:

Yes, but now you have to calculate R_c, which is *not* R1 in your circuit.

The gm*vbe current splits among R1 and R3. To simplify, if we assume an infinite gain then the input will not move from ground, that is you have R1//R3 as R_c. Numerically this gives 7.3k which multiplied by gm gives around 230 (again, as in my back-of-the-envelope calculation). How high is VA - the Early voltage - in the model you are using?

M

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Frustra fit per plura quod fieri potest per pauciora

I examined a 2N3904 in LTSpice and found the Early voltage to be -99.

John S

That's almost the same circuit I used when I won the blue ribbon in the Science Fair, in 5th grade, ca. 1959. ;-D

Cheers! Rich

Il Thu, 28 Jul 2011 15:23:01 -0500, John S ha scritto:

Ok, so that gives ro = VA/Ic = 99/0.84 kOhm, that is 118kOhm.

Then R_c = R1//R3//ro = 68//118//8.2 kOhm = 6.9kOhm

With this, and a gm = Vt/Ic = 0.032, K_v=0.032*6.9e3 = 220

I'd say it's pretty close :-D

M

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Frustra fit per plura quod fieri potest per pauciora

Michele Ancis expounded in news:1gnju7p09d5i.1i75q620wdad1$. snipped-for-privacy@40tude.net:

Indeed.

What is the effect of R3 on input impedance in this circuit? (Given the negative FB) Can I glibly approximate as:

R2 || R3 + ( R1 || ro )

?

Warren

Il Fri, 29 Jul 2011 12:25:32 +0000 (UTC), Warren ha scritto:

Uhm...I would put some extra parentheses in there, otherwise I cannot order the operations..Anyways, I think that your question "What is the effect of R3 on input impedance in this circuit?" can be answered in three ways:

1 - The King's way : designate R3 as the "extra element" and apply EET as per Middlebrooks' teachings. This will give you a super nice and tidy expression with all *design relevant* trade-offs in clear text. 2 - The Farmer's way: apply old loop/node analysis to the problem. I think the result will be slightly more garbled, but if you're good (or lucky) it can still be put in *insight-driving* form :-D After all, it is really a simple circuit. 3 - The Buglar way: since I don't have any paper at hand, I will just say that - following Miller's theorem - the effect of R3 as seen from the input gets reduced by the amount of gain between its nodes, that is roughly 220 (or 219 if you take the unity into account...)

So at the input you have the b-e incremental resistance, rb = Vt/Ib, in parallel with R2, in parallel with R3/K_v = R3/220

Rin = R2//rb//(R3/220)

What does the simulator say?

M

p.s. R1 doesn't show up directly because it is included in K_v, the voltage gain.

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Frustra fit per plura quod fieri potest per pauciora

Michele Ancis expounded in news:wln1rhnzi4nl$. snipped-for-privacy@40tude.net:

|| has higher precence than '+' :)

I'll have to look at this at home (I'm at work). And see if I can grok this..

LTspice is off the planet when I try to answer this. It is probably my fault:

.tf V(Out) V2

Small Signal AC Analysis: AC Amplitude: 0.01 AC Phase: 0.0

Transfer_function: -2.95855e-012 transfer v2#Input_impedance: 2.1252e+016 impedance output_impedance_at_V(out): 514.148 impedance

In this run, I added Rs=600 ohm as seen here:

but I had similar results, with the former sim as well.

Warren

Il Fri, 29 Jul 2011 14:01:02 +0000 (UTC), in sci.electronics.design hai scritto:

Awrighty then :-) So you don't need the parentheses in your expression :-DD

So...Last time I used Spice's netlist interface was too many years ago so I cannot check syntax here. I'd say - however - you could put your AC amplitude to 1V, which gives you "normalized" results and doesn't affect anything else, being AC a linear simulation (I mean, with linear components in the network). Then, to calculate the Rin, find a way to get the current flowing through the input source (normally it is accessible somewhere, but you may need to place a probe in series, dunno), and calculate "one over that". It shouldn't pose any problem at all. Also, if you define V2 to output 1V AC, you don't need to use .tf but just plot V(out) and you're done :-D

M

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Frustra fit per plura quod fieri potest per pauciora

Michele Ancis expounded in news:1t6hxv1xkzvw5.131a3zlrskkun$. snipped-for-privacy@40tude.net: ..

Yes- for clarity only (albeit inconsistently). :)

LTspice doesn't require all that arcane data entry - you can enter a schematic graphically.

..

Surely you don't mean 1V for generator V2? With a stage gain of approx 220, this leads to some trouble at the collector. This is precisely why there is truly a "small signal" supplied. :)

The LTspice (*.asc) file is provided below, for anyone wanting to try it:

Version 4 SHEET 1 968 680 WIRE 464 -112 304 -112 WIRE 304 -80 304 -112 WIRE 304 48 304 0 WIRE 304 48 128 48 WIRE 336 48 304 48 WIRE 128 80 128 48 WIRE 304 144 304 48 WIRE 464 144 464 -112 WIRE -160 192 -208 192 WIRE -16 192 -80 192 WIRE 128 192 128 160 WIRE 128 192 48 192 WIRE 240 192 128 192 WIRE -208 272 -208 192 WIRE 128 272 128 192 WIRE -208 400 -208 352 WIRE 128 400 128 352 WIRE 128 400 -208 400 WIRE 304 400 304 240 WIRE 304 400 128 400 WIRE 464 400 464 224 WIRE 464 400 304 400 WIRE 304 448 304 400 FLAG 304 448 0 FLAG 336 48 Out IOPIN 336 48 Out SYMBOL npn 240 144 R0 SYMATTR InstName Q1 SYMATTR Value 2N3904 SYMBOL res 288 -96 R0 SYMATTR InstName R1 SYMATTR Value 8.2k SYMBOL res 112 256 R0 SYMATTR InstName R2 SYMATTR Value 6.8k SYMBOL res 112 64 R0 SYMATTR InstName R3 SYMATTR Value 68k SYMBOL Misc\\battery 464 128 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value 15V SYMBOL voltage -208 256 R0 WINDOW 3 6 116 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value SINE(0 0.01V 60Hz) SYMATTR InstName V2 SYMATTR Value2 AC .01 0 SYMBOL cap 48 176 R90 WINDOW 0 0 32 VBottom 0 WINDOW 3 32 32 VTop 0 SYMATTR InstName C1 SYMATTR Value 47µF SYMBOL res -176 208 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 0 56 VBottom 0 SYMATTR InstName Rs SYMATTR Value 600 TEXT -88 536 Left 0 !;tran 0 1 0 .0001 TEXT -112 -152 Left 0 ;From Art of Electronics, Fig 2.40. TEXT -88 576 Left 0 !.tf V(Out) V2 TEXT 176 24 Left 0 ;DC 7.318V TEXT 192 -88 Left 0 ;0.936 mA TEXT 320 88 Left 0 ;0.71V\nAmplitude TEXT 136 224 Left 0 ;3.1 mV\nAmplitude TEXT 320 248 Left 0 ;AC Gain \n229 TEXT 40 80 Left 0 ;98.1uA TEXT 320 352 Left 0 ;0.841mA

Warren

Rich Grise expounded in news:j0sj1r$bo6$ snipped-for-privacy@dont-email.me:

Humph!

I didn't get any ribbon for my high frequency Tesla coil. The spark would just tickle the skin and the ozone generated would eliminate onion smells, even.

My classmate stole the show with his mockup of an open heart surgery thingy (his mother was a nurse).

I'm not bitter after all these years! Humph!

Warren

Hey, maybe the judges were trying to listen to the ball game :).

tm

tm expounded in news:j0up08$o5f$ snipped-for-privacy@speranza.aioe.org:

:)

Just a trifle crackle on the band.

It wasn't like the harmless little voltage doubler with big caps that I made for shocking classmates with..

Warren

Il Fri, 29 Jul 2011 16:46:44 +0000 (UTC), Warren ha scritto:

Oh yes, indeed.

Ohi ohi, I guess we need to understand each other a little better here :-) Are you simulating:

1 - AC small signal 2 - Transient with sinusoidal excitation

If 1 holds, then, *per definition* you don't care about the absolute amplitude of the signals, since everything in your network is simulated with linear components, which (unlike REAL stuff) don't bother about the amplitude of their drive, they always respond with a fixed proportionality constant. Like if you put 1V across base-emitter, at the collector you read

230V, but that is fine provided you understand what you read (That is, gain). If 2 holds, then why are you doing it? Things like "input impedance" and "gain" and "output impedance" are indeed meaningful and correctly defined *only* under linear assumptions.. Anyways, to plot the input impedance you still have to plot

V2/I2

Where V2 is the voltage of your input source and I2 is the current through it. This is the definition of input impedance seen by your source.

I will download LTspice sooner or later..;-D

M

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Frustra fit per plura quod fieri potest per pauciora

Using AC signal analysis:

V(n003)/I(Rs) = 288.316 < -11.2908 degrees @ 60Hz or V(n003)/I(Rs)= 282.736 - 56.449j

V(n003) is between the 600 ohm source resistor and the input capacitor.

Hint: This is easy to do using waveform arithmetic.

Cheers, John S