# Capacitor that connects the base and the collector of BJT makes it hard to study multistage amplifier

I'm studying ECE and at this point in microelectronics they've taught us that when performing DC analysis on low frequencies we must short circuit the capacitors of the circuit.

In the circuit below, I've been asked to find the value of $$\R_B\$$ so that the dc voltage at the collector of $$\T_1\$$ is equal to $$\5V\$$ ($$\V_{C1}=5V\$$) and then calculate the voltage gain $$\\left(\frac{V_{out}}{V_{in}}\right)\$$. The capacitor $$\C_{BC}\$$ makes everything difficult! I've short-circuited the capacitors and have calculated the following.

Assuming that for both $$\T_1\$$ and $$\T_2\$$ it is $$\\beta=246\$$, $$\V_{BE}=0.78V\$$ and they operate in the active region:

$$\V_{B1}=V_{C1}=V_{B2}\$$

To calculate $$\I_{B1}\$$:

$$\V_{B1}=V_{BE}+I_{E1}\cdot 0.1k\Omega\Rightarrow\$$ $$\5V=0.78V+(\beta+1)\cdot I_{B1}\cdot 100\Omega\Rightarrow\$$ $$\I_{B1}=170.85\mu A\$$

Thus, $$\I_{C1}=\beta\cdot I_{B1}=42.0291mA\$$

To calculate $$\I_{B2}\$$:

$$\V_{B2}=V_{BE}+I_{E2}\cdot (R_{E21}+R_{E22})\Rightarrow\$$ $$\5V=0.78V+(\beta+1)\cdot I_{B2}\cdot (0.33+1.8)k\Omega\Rightarrow\$$ $$\I_{B2}=8.02136\mu A\$$

Now, to calculate $$\R_B\$$:

$$\V_{C1}=15V-(I_{B1}+I_{C1}+I_{B2})\cdot\left(220k\Omega+R_B+1.8k\Omega\right)\Rightarrow\$$

$$\5V=15V-(0.17085mA+42.0291mA+0.00802136mA)\cdot\left(220k\Omega+R_B+1.8k\Omega\right)\$$

and I get $$\R_B=-219.72716...k\Omega\$$.

Note that $$\R_B\$$ is an ohmic resistor. I've found many similar examples/problems in textbooks but none of them had a capacitor like $$\C_{BC}\$$. Using the low-frequency hybrid-$$\\pi\$$ model of the BJT proves to be a bit difficult, once again because of $$\C_{BC}\$$.

I'm not quite sure what it is that I'm not doing right. Any sort of hint, help or even a reference to a similar example in some textbook is much appreciated!

• "performing DC analysis on low frequencies we must short circuit the capacitors of the circuit" Strange. At low frequencies (DC) capacitors behave as open circuit. At high frequencies they behave as short circuit. Commented May 31, 2022 at 11:47

they've taught us that when performing $$\\color{red}{\text{dc analysis}}\$$ on low frequencies we must short circuit the capacitors of the circuit.

That is incorrect.

When performing AC analysis it's usual to short capacitors. When performing DC analysis it's usual to open-circuit capacitors.

I've short-circuited the capacitors

That's incorrect for DC analysis; you need to open-circuit the capacitors. So, when you initially start your solution and say this: -

$$V_{B1}=V_{C1}$$

That is incorrect for DC analysis.

Rb is incorrect.

For Vc=5V and Vcc=15 with Rc=1.8k it will have a 10V drop thus Ic=10/1.8k= 5.56 mA and Ib=Ic/hFE=22.5 uA yet hFE has a tolerance >>+/-50% at best if binned parts with special suffix.

Let Ie = Ic and Ve=IeRe=10VRe/Rc = 0.55V

Assumptions are incorrect for Vbe. Use Vbe=0.6 for 1mA and 0.65V for 5mA

1. Thus now you get Rb= 614k minus the fixed 220k.
2. For gain it is usually Rc/Re or slightly less for DC gain for hFE >>100 within 1%. This is attenuated by series rBE when shunt Re is small (<=100). You may compute that and result is around 17 gain instead of 18.
• but this will be reduced by the negative feedback of Ccb driving both Rb and Cin for attenuation of the open loop gain. When impedance of C matches Rb you have -3dB. But there is also an attenuation of AC from Ccb to Cin attenuating the input with Rc as a source impedance, so the voltage net at Vb has a low pass effect. The high pass effects are from CinRin and Cout to Rc&load as the current loop.
1. the 2nd stage will have unity gain with no load as Rc/Re=1 but attenuated as expected with the load.