# BJT amp problem

Discussion in 'Homework Help' started by Petrucciowns, Nov 20, 2009.

1. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
62
0
I am having difficulty find the Voltage Gain, output peak voltage, and the voltage of the capacitor. If this circuit had a load resistor I would be able to find the voltage gain from the formula rC/r'e, but since it doesn't I don't know how I would go about it. The other two values I don't really have a clue. I would not just like to know the answer, but to know why the answer is what it is. I checked out some articles on this site as well as my text, but I haven't run across this particular configuration. I would appreciate anyone's help.

The circuit can be seen by following the link.

http://img149.imageshack.us/img149/8628/bjt.png

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
I guess you want the gain with the output taken directly at the collector....?

If so the gain will be determined by Rc, RE and re (intrinsic emitter resistance - probably insignificant in this case). If there is no load shown then assume it is infinite.

Were you given the source frequency or Beta values?

If not, just assume Beta is large enough to ignore in the calculation of Ie and hence Ic.

If the frequency is mid-band you could also ignore the AC voltage drop across the cap.

There will however be a DC or average voltage across the cap.

3. ### hobbyist Distinguished Member

Aug 10, 2008
802
76
If, you'r asked for V.pk, then you have to know what the quiescent collector current is, and the change in collector current with a incremental change in base current.

So you may need to use extremal analysis, using beta at 20 then beta at 100.

First what is your calculation of the base voltage and emitter voltage, with your schematic shown?

If you calculate 2.94V. for the base, and then subtract 0.7, for Vbe. to get 2.24V. for emitter voltage, it would be wrong.

If you know how to thevenize a network, thevenize the voltage divider ALONE,

then add a diode with a 0.7V drop connected to a resistor of 1K together as one load, then take this load and connect it to your thevenized network, and you'll see that the voltage at the base will be around 773mV. Because the diode and 1K resistor are in parrallel with 39K resistor, so voltage drops considerably.

Now that's not considering the Beta of the transistor, Now take that same load, and multiply it by 20, then recalculate, then again do the same by multiplying the load resistance by 100, and that will give you a extreme coverage of base current with changes in Beta, and then these base currents and assumed beta's are used to calculate collector currents.

And from there you can check the incremental change in collector voltage, with the incremental change in base voltage from 250mV. signal source.

4. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
62
0

Why is this, can't you just use the voltage divider formula source x r2 / (r1+r2)? This works in similar BJT configurations with a voltage divider network.

If you thevenize the voltage divider alone does that make R2 RL? I'm used to thevenizing with more than two resistors. Removing the load, source, putting the resistors in parallel etc.. How would you do this with just a source and two resistors?

The rest of your post is slightly confusing to me, as I am not familiar with this method. Would it be possible for you to draw a schematic or a quick circuit in multisim (or similar), and we can go over it?

5. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
62
0

All you see, is all that I was given. I am guessing the output would be taken directly across the collector as there is no load.

How would rC be calculated in this circuit if there is no load?

6. ### hobbyist Distinguished Member

Aug 10, 2008
802
76

It works as long as the base current isn't loading the voltage divider.

But with the high value of 120K ohms at the top of the divider, and the low value of the base emitter diode, with 1K ohms emitter resistor in series, makes for base current affecting the base voltage.

The 1K and base emitter diode, is in parrallel with the 39K resistor, and now must be included in the equation for finding thevenin voltage.

With the emitter resistor 1K multiplied by Beta:

A beta of 20 would make the VB. = 1.6V.
A Beta 100 would make the VB. = 2.43V.

When considering incremental changes in collector current, this is important,
but if just a quick analysis, then rough calculations would suffice,

Hope that helps.

Last edited: Nov 20, 2009
7. ### Audioguru AAC Fanatic!

Dec 20, 2007
9,462
916
The circuit was designed by one guy who made a voltage divider for the base of the transistor without knowing how much base current is needed.
Then the emitter and collector resistor values were selected by another guy who didn't know how much current there is in the divider and didn't know the hFE of the transistor.

You cannot design a circuit this way.

8. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
62
0
Well, thanks for the help the was given....