# just 5 hours remaining : analyze this circuit ..please

Discussion in 'Homework Help' started by tallo, Mar 4, 2011.

1. ### tallo Thread Starter New Member

Mar 4, 2011
12
0
hi everybody

please i need it as fast as possible

thank you so much

Last edited: Mar 4, 2011
2. ### blueroomelectronics AAC Fanatic!

Jul 22, 2007
1,758
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We're not a homework service.

3. ### mjhilger Member

Feb 28, 2011
119
16
R1 & R2 form a voltage divider. You should be able to determine the voltage and thus the currents through each. Since the transistor has no identification I must assume that hfe is high enough for the base current to be negligible; if not then you must add it in. So once you determine the voltage divider point voltage at the base, the emitter will be 0.7 less or so. That sets the voltage present at the top of the 2 emitter resistors, so you should be able to determine the current at the emitter. The collector current is the emitter current minus the base current; but as stated above, if we ignore that saying it is too small, them Ic & Ie are the same for the DC bias point.

From there you should be able to determine the impedance of the caps at the frequencies of interest and determine the voltage gains from there.

That should get you started

4. ### PRS Well-Known Member

Aug 24, 2008
989
36
Without actually doing your work, I can give you procedure to follow, not only for this circuit but for others as well.

1. R1 and R2 form a voltage divider. As such they, together with Vcc of 12 volts determine the voltage at the base of the resistor as

Vb = [R2/(R1+R2)] * Vcc

2. Having the base voltage, use it to find the emitter voltage. The base-emitter junction of a BJT is merely a diode drop of 0.6 volts. So that

Ve = Vb - 0.6

3. Determine the emitter current using ohms law

Ie = Ve/(Re1+Re2)

4. The collector current is apporoximately the same as the emitter current, so

Ic = Ie

5. Determine the voltage at the collector

Vc = Vcc -Ic*Rc

6. This give your bias point, Vce = Vc - Ve

7. The gain is the resistance at the collector divided by the resistance at the emitter. At the collector, we have Rc in parallel with RL and Ro, the latter being about 100kohms. Since RL is so high, I'd just ignore it in an approximate analysis such as this, so Rc is our resistance at the collector.

Looking into the emitter, we have re = VT/Ic where Vt is a constant 26 mV at room temperature. Thus re is several ohms and for the signal, which has a path to ground through Ce, this re is in series with only Re1.

The maximum mid-band gain is given by

Av = Rc/(re + Re1) , which by inspection is a little less than 10V/V

When Ce is removed, the path to ground is through both emitter resistors and our midband gain becomes

Av = Rc/(re+RE1+RE2), which is smaller than the above.

But your teacher wants the gain at specific frequencies. This involves making a bodie plot. More on this if you want.