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I apologize in advance for posting a homework problem. I am given this simple BJT circuit as shown below. I have to find the base-emitter voltage with reverse beta equal to 0.5 and forward beta 50. Also a saturation current of 10^-16 A is given, which doesn't sound right to me, but is written anyway in the problem.

I haven't encountered this biasing configuration before and so I have no idea how to proceed except that the collector current can be found to be 12.5 mA using the provided current gain. Aside from that I don't see any reason why the base-emitter voltage would not be 0.7 volts as usual, but the problem states that the correct value is 0.84 volts. I also want to know what this kind of bias setup is called.

schematic

simulate this circuit – Schematic created using CircuitLab

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The clue is in the reverse saturation current of 0.1 fempto amps. It is (I suspect) leading you to use the ebers-moll model of a BJT: -

enter image description here

I'm not going to reverse engineer this equation to solve for Vbe (given a collector current of 12.5 mA) but if I plug in the answer of 0.84 volts, I get a collector current of 26.23 mA.

If I plugged in 0.83 volts I get 17.7 mA so maybe the author of the question has either done an over-simplification or you haven't provided some information that is vital like ambient temperature.

So, if Ic = 0.0125 amps I get Vbe to be 0.82 volts.

Maybe (?) also the internal emitter degeneration resistor (\$r_e\$) has been factored into the answer: -

enter image description here

With a collector current of 0.0125 amps \$r_e\$ is about 2 ohms and assuming emitter current = collector current, the base will need to be 25 mV higher bringing Vbe up to 0.845 volts.

I also want to know what this kind of bias setup is called.

It isn't a real bias setup used in practice - the current injected into the base (as shown) could come from a voltage supply in series with a resistor but this would create an inaccuracy in how you treat the problem.

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  • \$\begingroup\$ This is very handy. I did not know this. Btw room temp de facto around here is 300 K. Thanks Andy. \$\endgroup\$ – DorkOrc Aug 10 '17 at 3:04
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Assume the transistor behaves as a diode

Ie Vbe 1nA 0.7 - 6 * 0.06 = 0.7 - 0.36 = 0.34 volts

1uA 0.7 - 3 * 0.06 = 0.7 - 0.18 = 0.52 volts

1mA 0.7volts (assumed)

10mA 0.7 + 0.06 = 0.76 volts

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Get the collector current from Beta is 12.5mA. Ignore the base current or not. This equation solution is (ln(0.0125A/1e-16A))*0.026V = 0.84V You can ignore subtracting the one when you solve the diode equation. I'm assuming Vt is 0.026V.

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