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Good evening. Sorry for my grammatical errors, or any other errors that have connection with English grammar and language in general. I have a little problem to solve. I think that these two BJT's are in Darlington pair. I have to find Zin, Zout, Au and Ai, common BJT characteristics. Problem is I think that this first transistor is in cut off, or am I wrong? If it is in cut off the second transistor isn't connected to the AC source, and the second BJT is in active regime. Then circuit doesnt make any sense. In general:"Can BJT conduct if it doesn't have any DC voltage source connected with it's Base connected through some resistors?

Info: Vcc=12V,Vdd=8V,Rc=1.8k,Re=5.6k,Rb=390k,ro(output resistance for ac)=infinity,betaforvard=120...

I don't expect that someone solve this for me, just answer me if I am on the right path with this conclusion about regimes of this BJT's transistors. Thanx ...:D

enter image description here

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  • \$\begingroup\$ This is not a Darlington. \$\endgroup\$ – Brian Drummond Nov 25 '15 at 22:32
  • \$\begingroup\$ @Brian Drummond you're right, probably because BJT's aren't directly connected through base-emiter, and colector-colector, anyway it does not matter much, for me problems are regimes of this BJT's... \$\endgroup\$ – Ahmo Nov 25 '15 at 22:36
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Problem is I think that this first transistor is in cut off, or am I wrong?

If the first transistor is in cutoff, then the emitter current is zero and, thus

$$V_{E1} = -V_{DD}$$

$$V_{B1} = 0$$

So,

$$V_{BE1} = V_{DD} = 8V$$

But this is absurd (and contradicts the assumption the transistor is cutoff)!

Assuming the transistor is in forward active, KVL 'round the base-emitter loop yields:

$$V_{BE1} + I_{E1}\cdot R_{E 1}- V_{DD} + I_{B1} \cdot R_{B1} = 0$$

Can you take it from here?

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  • \$\begingroup\$ I can. Thank you a lot. I didn't use the fact that emitter current is 0 in cut off. Also i used "Every Circuit" app on my phone to simulate this circuit and it according to it, i don't have any current in transistor which doesn't have any Vsource connected with it's base... \$\endgroup\$ – Ahmo Nov 26 '15 at 12:03
  • \$\begingroup\$ Sorry it is my fault. Now I made simulation in multisim and it works. Thanks a lot, again. \$\endgroup\$ – Ahmo Nov 26 '15 at 12:44
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The first transistor is not cut off. Look at the battery in the emitter circuit. It is taking the emitter negative and biasing the transistor. How much or how little it biases the base emitter junction depends on it's voltage. It's a two stage amplifier and not a Darlington by the way.

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  • \$\begingroup\$ Whether it means that this battery, affects on PN junction of base-emitter by making Ve negative, and this negative voltage draws holes from P junction in base to N junction in emiter and that we have emiter current. But how could i conclude that Vb is positive, or do i just guess it? \$\endgroup\$ – Ahmo Nov 25 '15 at 23:14
  • \$\begingroup\$ Vdd is at -8V feeding the emitter via a 5k6 resistor. The base is loosely tied to 0V and this means there has to be 7.3 volts across the 5k6 which means a current of 1.3mA and this current also flows in the collector (well 99% of it). \$\endgroup\$ – Andy aka Nov 26 '15 at 8:18
  • \$\begingroup\$ You're right. I was confused how can BJT conduct, if it doesn't have any source connected with it's base, but again this battery on the emitter works...Thank you again. \$\endgroup\$ – Ahmo Nov 26 '15 at 12:46

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