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What we know about the circuit is that in DC-analysis (when \$V_{in}=0\$), \$V_{out}=10V\$.

We also know that the circuit (even outside of DC analysis) will have \$I_{C}=2mA\$ and \$V_{CE}=5V\$. \$R_{1}\$ and \$R_{2}\$ should be so that \$I_{B(max)}=\frac{1}{10}I_{2}\$ where \$I_{2}\$ is the current over \$R_{2}\$.

I'm assuming

\$V_{out}-V_{B}=0.7V\$

and

\$V_{B}=V_{CC}(\frac{R_2}{R_2+R_1})\$

And in DC-analysis if \$V_{out}-V_{B}=0.7V\$ holds, I can conclude that

\$V_{B}=V_{out}-0.7V\$, so that \$V_{B}=10V-0.7V=9.3V\$

But I'm not 100% certain on what I should do to figure out the formulas to calculate the resistors to match the assumptions and how to proceed overall.

Edit 1

If \$V_B=9.3V\$, \$I_2*R_2=9.3V\$ and \$I_1*R_1=20V-9.3V=10.7V\$

\$I_1=I_2+I_B\$

assuming \$I_B=max\$

\$I_1=I_2+\frac{1}{10}I_2=1.1I_2\$

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Vout is on the collector. That 0.7v figure you have in your head is for the base-emitter drop.

If you want 10v out, and 5v VCE, then the emitter will be at 5v, and the base at 5.7v. With Re and Rc drops of 5v and 10v respectively, then at 2mA you can calculate their values as 5/2m = 2.5k and 10/2m = 5k respectively.

You mention Ibmax, but you don't give any figure for beta. 100 minimum is a reasonable value. So if you set the base bias chain current at Ic/10 = 200uA, then 5.7/200u = 28.5k for R2, and 14.3/200u = 71.5k for R1.

The actual bias point will be slightly different to the voltages assumed, due to ignoring base current. However, the point of this sort of bias arrangement with a large Re is that it's very tolerant to transistor variations.

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  • \$\begingroup\$ If I'm not completely mistaken, was it not apparently possible to find out the required parameters purely by DC-analysis instead of having to worry about AC at all? Also, thanks for the answer \$\endgroup\$
    – Grak
    Jan 29, 2019 at 10:45

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