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Problem with a Rin measuer

F

Francesco

Jan 1, 1970
0
Hi,I have a problem with this circuit
http://digilander.libero.it/checco23/Schemi/Per_sito1.GIF , I try first
to calculate and after to measure in a simulation the value of Rin, but
the two results are much different!

Where is the mistake?


Data:

Vi= 100mV;

Ii( AC current at the transistor's base ) = 794nA;

hfe = 83.6;

re = 42 Ohm;

Ri ( calculate ) = (hfe+1)*( re + RE ) = 84.6*1042 = 88153.2 Ohm;

Ri ( measured ) = Vi/Ii = 100m/794n = 125944.6 Ohm;

The difference between the two result is too much!

Help me to understand where is my error!

Thanks !

Francesco
 
M

Mantra

Jan 1, 1970
0
Francesco said:
Hi,I have a problem with this circuit
http://digilander.libero.it/checco23/Schemi/Per_sito1.GIF , I try first
to calculate and after to measure in a simulation the value of Rin, but
the two results are much different!

Where is the mistake?


Data:

Vi= 100mV;

Ii( AC current at the transistor's base ) = 794nA;

hfe = 83.6;

re = 42 Ohm;

Ri ( calculate ) = (hfe+1)*( re + RE ) = 84.6*1042 = 88153.2 Ohm;

Ri ( measured ) = Vi/Ii = 100m/794n = 125944.6 Ohm;

The difference between the two result is too much!

Help me to understand where is my error!

Based on the diagram, I'm assuming R5 is intended to be the output
load. For an emitter follower, it's on the wrong transistor terminal.
It should be on emitter for an emitter follower - note this would
change the RE value to R3 || R5 ( '||' = parallel reduction ).
Strictly, as drawn, C2/R5 do nothing right now - for small signal they
are not even part of the circuit since all DC sources are effectively
grounded for analysis.

Perhaps you are measuing the correctly wired circuit rather than the
displayed circuit diagram: the change in RE is ~0.1% but 0.1% *
(Hfe+1) is 8.4% difference in the formula. Not enough to explain the
whole difference (formula deviates 29% from measured).

More likely: remember that the Ri/Zi formula you are using is only
valid for "small signal".

Try measuring with an even lower input signal level, e.g. 1 mV or 100
uV. 0.1 V is probably not "small signal" enough. Roughly speaking
with an 0.1V input and hfe of 83.6, your collector current is going to
move significantly so your Q-point/bias point is not going to be where
you thought it was.

For a 100 mV DC level you've shifted the Q-point, while for 100 mV AC
level you're seeing an integral input impedance of all the traversed
Q-points. Considering that, the difference of 88K vs 125K seems like
a reasonable result.

Another source of error: is the Hfe measured or from the datasheet?
If the latter, there is statistical variance between any actual
transistior and what the data sheet says (which is only the average
value of Hfe). If measured, how? In reality Hfe = Hfe(Ic) so it's
Q-point dependent also. Again gets back to not being small signal.
BTW the datasheet value of Hfe is always Hfe(Icmax) where Icmax is the
Ic that gives maximum(Hfe) - obvious marketing prefers listing *that*
value.

A remote possibility is Rb: the "full form" is Zin(CE/CC|s.s.) = (rb +
RB) + (Hfe+1)*(re + RE). Usually Rb is ignorable but not always.

Small signal can be a tricky concept: small signal is small only in
the sense of bias point shifting (as in Taylor expansions and the
resulting shift of y (or in this case Ic) ), but *not ever* in the
human intuition of when signal level seems "small" sense. Thus
small-signal really means "*small enough* that I *feel* comfortable
ignoring the *actual deviations* from my *assumption* of linear
operation". It's an important subtlety.

MM
 
F

Francesco

Jan 1, 1970
0
Perhaps I have find the error: in my preceding calculation I used, to
calculate beta, the DC Ic and Ib, instead of the AC values! Now infact
using the new beta ( AC values ) all is OK!
 
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