Input impedance of a BJT in CE config

Discussion in 'Electronic Basics' started by riccardo manfrin, Oct 2, 2013.

1. riccardo manfrinGuest

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

2. riccardo manfrinGuest

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?

R

3. George HeroldGuest

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...

4. riccardo manfrinGuest

This was definitely something I was missing and I realized only after posting. My fault for not investigating more.

Now that the I realize that we are talking of DC bias current I_B, I still miss the motivation behind the use of V_t instead of using the DC voltage between base and emitter (which would be V_{BE} ~= 0.7V >> V_T ~= 26mV).

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

5. Julian GrodzickyGuest

The book is available in its entirety online.

6. riccardo manfrinGuest

The book is available in its entirety online.

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..

7. riccardo manfrinGuest

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.

8. riccardo manfrinGuest

How much formal education do you have in circuit design?

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..