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Hey. I am a biomedical engineering student focusing on the biomaterials side. However, my university makes me take a few electrical engineering classes as well. So I am not very good at this. I was hoping if anyone could help me out.

As of rn, I have tried to use DC bias to find Ie. But I feel that Ie should be just 1mA due to the current source and Ic should be 0.5mA. But if I use Ic/alpha I get 505 microAmps. Could anyone point me in the right direction please?

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  • \$\begingroup\$ How do you calculate \$I_C\$ to be 0.5 mA if \$\beta=100\$? \$\endgroup\$
    – The Photon
    Apr 8, 2020 at 0:06
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    \$\begingroup\$ The collector cannot go below the emitter voltage. The problem specifies that voltage. At most the collector current is \$\frac{5\:\text{V}-\left(-700\:\text{mV}\right)}{10\:\text{k}\Omega}=570\:\mu\text{A}\$. Given the emitter is having \$1\:\text{mA}\$ pulled out of it by force, this means the base current is at least \$430\:\mu\text{A}\$. So I don't know what the \$\beta=100\$ means in this case. The \$-5\:\text{V}\$ voltage reference is irrelevant, too. As the BJT is heavily saturated, so collector looks like a voltage source, the voltage gain is very close to 1 (slightly more, perhaps.) \$\endgroup\$
    – jonk
    Apr 8, 2020 at 6:23
  • \$\begingroup\$ It's a trick question : as an amplifier, it is broken. \$\endgroup\$
    – user16324
    Apr 8, 2020 at 14:24
  • \$\begingroup\$ @ThePhoton 5V/10K should get me Ic right? \$\endgroup\$ Apr 8, 2020 at 18:28
  • \$\begingroup\$ Not if the BJT is biased into forward active mode. You can't assume the collector is at 0 V. \$\endgroup\$
    – The Photon
    Apr 8, 2020 at 19:50

1 Answer 1

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Your analysis is correct.

This circuit as biased is not an amplifier.

1/4 mA would be appropriate.

In that case, zin(base) would be 26 * 4 = 104 ohms

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  • \$\begingroup\$ Where are you getting the 1/4 mA from? \$\endgroup\$ Apr 8, 2020 at 18:30

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