Need help with the process for finding input resistance of atransistor ??

[[email protected]]

I have been trying to follow along with a books calculation for the
input resistance of a transistor which they define as Rin = delta
(Vbe) / delta (Ib) or Rin would be 1 / (slope of the diode curve) or
1 / (derivitive of the diode curve) . Doing the math and assuming
that KT/Q = 26 mv and assuming that (Is) the reverse saturation
current is negligible, I get the desired Rin = 26 mv / Ib.

My question is why is Rin = delta (Vbe) / delta (Ib) and not just
Vbe / Ib . My some how flawed reasoning is that if you look at ohms
law E = IR, then R = I/E and therfore Rin should be Vbe / Ib . Why
is Rin equat to the change in current divided by the change in
voltage?

Any help would be greatly appreciated. Thanks

 

[Kevin Aylward]

I have been trying to follow along with a books calculation for the
input resistance of a transistor which they define as Rin = delta
(Vbe) / delta (Ib) or Rin would be 1 / (slope of the diode curve) or
1 / (derivitive of the diode curve) . Doing the math and assuming
that KT/Q = 26 mv and assuming that (Is) the reverse saturation
current is negligible, I get the desired Rin = 26 mv / Ib.

It may be better to think of this as:

rin = hie = rbb' + (1 + hfe).re

where rbb' is the base resistace, usually of the order 10-500 ohms, and
re=1/(40.IC) = 1/gm

My question is why is Rin = delta (Vbe) / delta (Ib) and not just
Vbe / Ib . My some how flawed reasoning is that if you look at ohms
law E = IR, then R = I/E and therfore Rin should be Vbe / Ib . Why
is Rin equat to the change in current divided by the change in
voltage?

Any help would be greatly appreciated. Thanks

The fundamental reason is:

del_y = f'(x).del_x

if del_y, and del_x are small, as a general math result, from differential
calculus.

For small signals, i.e., when an input signal only changes a small amout
about a fixed bias point, what is the resulting change in the output, about
a fixed bias.

rin, is a small signal resistance. It is not equal to VINDC/INDC. Since the
inputs and outputs that are wanted are the small signal changes, than one
needs the small signal resistance, not the large signal resistance. If one
actually wanted large signal values, than one would use such values
throughout.

Kevin Aylward
[email protected]
www.kevinaylward.co.uk

 

[Kevin Aylward]

Because the resistance is not ohmic.

Interesting point. The resistance does not follow ohm's law, so in that
sense is not ohmic. However, it does have characteristic similar to ohmic
resistors but not similar inductors and capacitors. That is, the current and
voltage are independent of time, i.e not +/-j

For DC conditions, the actual value of V/I may be important, especially for
values of Vdc of around 0.5 to 1V.

Ohmic resistances hold
a constant ratio of voltage to current, regardless of the
magnitude of voltage or current. Diode junctions do not.
So resistance can be defined for such devices only
incrementally (over tiny ranges of voltage and current).

Well... technically one can actually define the large signal resistance as
V/I.

So that:

R = Vd/I = (Vd/Io).exp(-Vd/Vt)

And using the results of
http://www.kevinaylward.co.uk/ee/widlarlambert/widlarlambert.html, for the
Lambert W() function

One can obtain the inverse relation;

Vd = -Vt.W(-R.Io/Vt)

I will leave it as an exercise for the reader to deduce formulas expressing
R in terms of I, and I in terms of R

Kevin Aylward
www.blonddee.co.uk
www.anasoft.co.uk

 

[neon]

NO such a thing as input resistance. but there is inpedance. and using boltzman constant you can find it . it is a function of current