# Input impedance of a BJT in CE config

• posted

I'm designing a very simple (dummy I'd say) linear amp with a biasing network like the one shown in this image:

Ignore the values reported in the image. I'm interested in understanding the input impedance seen at the base by the AC source, using the \pi model of the BJT.

For AC analysis DC sources are shutdown (become grounds), the capacitor near the emitter shortcircuits the parallel resistor, and all the other caps caps are shortcircuited hence letting AC through.

The impedance is the ratio between voltage and current, so in my case,

r_\pi = v_{BE} / i_B

Unfortunaltely v_BE and i_B are related to the exponential relationship between current and voltage of a diode.

Looking around, I found mention to formulas like

r_\pi = V_T / i_B

where V_T= kT/q, that is the thermal voltage of the BE junction which accounts for the so called "ohmic effects" of the diode model.

Unfortunately, if both the above formulas are correct it looks like V_{BE} = V_T. If this is the case, then my input source signal (V_s in the image) is not varying the V_{BE} voltage which is only driven by thermal noise. Is this a correct guess?

Can you explain me the procedure to derive the input impedance in AC, given the knowledge of the voltage V_s and its series resistance R_s.

Thanks, RM

• posted

VS and its series resistance do not influence the _circuit_ input impedance.

Does your homework require an accurate calculation of AC input impedance good up to substantial frequency ?>:-} ...Jim Thompson

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| James E.Thompson                                 |    mens     |
| Analog Innovations                               |     et      | 
• posted

NOTATION upper case: DC current/voltages (bias), lower case: AC current/voltages (signals),

Well, my goal is to dimension R_C in such a way to ensure that the range of the input signal i_B guarantees exact full swinging of the output signal V_{CE} in the admitted datasheet dynamic.

For the purpose, I need to know the range of i_B.

In order to know the max range of i_B I need to know the source and its series resistance and the input impedance of the transistor, which will define how much current will be injected in the base, and therefore the range of i_B.

Precisely, given a range of frequencies of my input signal v_s, I need to know the minimum BJT base AC input impedance in that range, which will admit the maximum current i_B into the base.

Once I know the range of i_B, I can dimension R_C so to admit exactly that range and guarantee that the maximum i_B value causes V_CE to get near saturation, and the minimum i_B value causes V_CE to reach V_CC.

Am I misunderstanding on this?

• posted

NNTP-Posting-Date: Thu, 26 Sep 2013 11:41:25 -0500 From: Jim Thompson Newsgroups: alt.binaries.schematics.electronic Subject: Biasing Question from S.E.D - BiasQuestionManfrinSED.pdf Date: Thu, 26 Sep 2013 09:41:24 -0700 Message-ID:

If you can't access A.B.S.E, let me know and I'll post to my website.

NOTE: I'm moving into our new house, so it may be a day before I respond again. ...Jim Thompson

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| James E.Thompson                                 |    mens     |
| Analog Innovations                               |     et      | 
• posted

You still have to consider the bias resistors. which look like 30k // 130k (//= in parallel) Then there's the 1.5k For just the transistor I think Zin looks like beta times the 1.2k emitter resistor..(?) But I should check in Art of electronics by H&H to be sure, has your copy arrived yet? They explain all this stuff very well...

• posted

In comparing those two resistances you're forgetting that it's an AC model, and letting the DC voltages creep in.

The AC impedance of the base, when the emitter is shorted, is V_t/i_B, where V_t = kT/q (so note that it changes with temperature!), and i_B is the --> DC

• posted

It appears to be homework. ...Jim Thompson

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| James E.Thompson                                 |    mens     |
| Analog Innovations                               |     et      | 
• posted

I'm going to read more about it in the documentation (btw no I'm still waiting for that book to ship). R

• posted

The book is available in its entirety online.

• posted

Yes, 2nd ed. is available in a "non-profit" way .. I'm reading through it right now.. page 76 says

"C is chosen so that all frequencies of interest are passed by the high-pass filter it forms in combination with the parallel resistance of the base biasing resistors (the impedance looking into the base itself will usually be much larger because of the way the base resistors are chosen, and it can be ignored);"

This is unexpected.. if the emitter is bypassed by C_E, the impedance looking into the base is solely given by the internal impedance of the BJT from the base to the emitter, which is V_T/I_B. For V_T ~=26mV and I_B in ranges from 0.1 to 1mA isn't the input resistance going to be low enough to be the dominant component in the parallel with the base biasing voltage divider resistors?

I'm reading further to see if something comes out..

• posted

How much formal education do you have in circuit design?

The AC impedance of the emitter terminal comes directly from the equation for emitter current, which, if my memory serves, is

i_e = Iss * (e^(v_be/V_t) - 1),

where Iss is the saturation current of the BE junction. If you solve for Iss at any given combination of v_be and i_e, then take the derivative

1 / R_e = di_e / dv_be

you'll get the whole R_e = V_t / I_e thing.

When I took circuits they practically drilled a hole in your head and shoved in a note card with this analysis written on it, so they must have thought it was important.

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Tim Wescott
Wescott Design Services 
• posted

If you're looking at the same picture I'm looking at, the emitter resistor is not bypassed. Generally if you choose a set of bias resistors and emitter resistor that give good performance over a wide range of part variations and temperature, then his statement will be true.

If you bypass the emitter resistor then yes, the base resistance may start to matter.

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Tim Wescott
Wescott Design Services 
• posted

In the image, the 0.2KOhm is not bypassed, the 1K is, so you are right to say that R_E is not bypassed. Nevertheless, the 200 Ohm R_E is not that much and I assumed that the resistance seen through base to emitter started to matter.

• posted

Definitely not enough

That is what I was looking for (and actually also found in "the art of electronics" meanwhile). I was considering resistance as the ratio between voltage and current whereas for a non linear device it is rather the derivative of these two guys around the point of operation. I was missing something so simple as linearization..

• posted

Well, do the math. At low frequencies the base resistance due to emitter resistance is beta * Re, with Re being the resistance both internal and external.

In that particular schematic, because the base bias network is fairly high resistance, the base resistance will, indeed, have an effect.

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Tim Wescott
Wescott Design Services 
• posted

This may be overly pedantic, but the _AC_ resistance comes from dv/di. "DC resistance", and by extension "resistance" is fairly meaningless. It's probably more accurate to talk about "small signal resistance", or even "differential resistance".

Be careful with this AC resistance stuff, by the way -- I went into college an electronics hobbyist, so when I hit this small signal analysis stuff I was overjoyed. I immediately went home and designed myself some circuits that worked like absolute crap, because my "small" was about 20 times bigger than the book's "small". The distortion was bad enough that the bias got all messed up, and -- well, it was a disaster.

Doing some checking with circuit simulations or just plain math to see how much distortion you'll be getting with seemingly-tiny signals is a Good Thing if you're at the level that I think you are.

--
Tim Wescott
Wescott Design Services 
• posted

Here's a simple example...

...Jim Thompson

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| James E.Thompson                                 |    mens     |
| Analog Innovations                               |     et      | 

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