0
\$\begingroup\$

enter image description here

Hi

Can someone please help me out. I've been battling this problem for ages and I don't know what I'm doing wrong

My AC input is 50mV and the output is measured as -190mV giving a voltage gain of -3.8. If I were to calculate the voltage gain without the graph I would do -RC/RE = -3900/1000 which gives me -3.9.

Why are the 2 results not equal? am I missing something?

Thanks

\$\endgroup\$
3
  • 5
    \$\begingroup\$ 3.8 does equal 3.9 to many engineers. What is the hfe of the transistor, what is its hoe, what is its hie, what voltage is dropped on the input C, is there a finite resistance to the output meter, is there a finite resistance to the voltage source? Try replacing the transistor with an ideal voltage controlled current source and simulate again. \$\endgroup\$
    – Neil_UK
    Nov 30 '20 at 13:29
  • 5
    \$\begingroup\$ Your calculation of A = -\$R_C/R_E\$ is an approximation which for example ignores the influence of the \$g_m\$ of the NPN transistor. With such approximations, don't expect "spot on" results, actually calculating -3.9 and getting -3.8 in the simulator sounds pretty good to me, more than close enough. \$\endgroup\$ Nov 30 '20 at 13:30
  • \$\begingroup\$ In addition to the finite transistor gain, 1 uF and (2.2k in parallel with 9.8k) forms a potential divider. Part of your loss will be there. 3.8 is actually pretty close as these things go : as transistor gain can vary by 4:1 or more there's not much point aiming for more precision (and better ways of doing it) \$\endgroup\$ Nov 30 '20 at 14:57
3
\$\begingroup\$

Inside the emitter is \$r_{E}\$ and that adds onto the 1 kohm external emitter resistor: -

$$r_{E}=\dfrac{\text{26 mV}}{I_C}\hspace{2cm}\text{at an ambient temperature of about 27 degC}$$

And \$I_C\$ will tend to equal half the voltage rail divided by 3900 ohms = 1.53846 mA hence \$r_{E}\$ = 17 ohms. Now, the gain is 3900/1017 = 3.835.

Add a little \$h_{OE}\$ (maybe 200 or 300 kohm in parallel with the external 3900 Ω collector resistor) and your done. Add a little bit of Early effect and you're also done.

\$\endgroup\$
3
\$\begingroup\$

In small signal modeling of a bjt transistor, there is an intrinsic emitter "resistance", \$r_e = \frac{25mV}{I_E}\$. You need to add this to your external resistance \$R_E\$. \$r_e + R_E\$ is the total emitter resistance to use in your gain equation. When \$R_E\$ is small, or 0, the effect of \$r_e\$ on gain becomes significant.

(A completely unrelated side note: Although the phase of your output is inverted in relationship to your input, I think it is bad practice to carry the negative sign into your gain. When gain is described in decibels, negative gain is loss. Just a suggestion. Your teachers may strongly disagree!)

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Please, make a distinction between STATIC (R) and DYNAMIC (r) resistances. Otherwise, one could think that the quantity 25mV/IE would allow a DC drop. (By the way: In fact, it is not a resistive quantity although it has the unit V/I; it is simply the inverse of the transconductance gm) \$\endgroup\$
    – LvW
    Nov 30 '20 at 15:04
  • \$\begingroup\$ I added "In small signal modeling of a bjt transistor, there is an intrinsic emitter "resistance",". Does that do the trick? \$\endgroup\$ Nov 30 '20 at 15:09
  • \$\begingroup\$ No, it needs to be little r as in \$r_E\$. \$R_E\$ is the external emitter resistance. \$\endgroup\$
    – Andy aka
    Nov 30 '20 at 15:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.