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I have a question regarding the circuit below.

I am unsure how to calculate the current flowing through the 1k resistor.

I think it is reasonable assumption that the collector of the PNP BJT must be sitting around (5-0.7) = 4.3V as it's base is tied back round to the collector (and assuming the PNP BJT is fully on).

But what assumption should we make about the voltage on the collector on the NPN BJT? I know it is the same as the voltage dropped across the 1k resistor, but I don't know the current flowing through it to work out the voltage dropped.

Is the current through the 1k resistor simply controlled by the base current on the NPN BJT, where the (base current x beta) gives the current through the 1k resistor?

Thanks.

enter image description here

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    \$\begingroup\$ Do some circuit analysis on it. What do you know about the relationship of \$\beta\$ (and perhaps \$\alpha\$) and the currents of a BJT? This is really just Kirchhoff's Current Law, assuming of course each transistor is in active region. \$\endgroup\$ – KingDuken Oct 11 at 18:50
  • \$\begingroup\$ You certainly cannot assume that both BJTs are "fully on" (saturated?). In fact, with the component parameters shown, neither one is fully on. \$\endgroup\$ – Dave Tweed Oct 11 at 18:50
  • \$\begingroup\$ As written, your question doesn't make sense. Your stated assumptions are not consistent with the schematic and the marked values. \$\endgroup\$ – Elliot Alderson Oct 11 at 18:57
  • \$\begingroup\$ @KingDuken are you talking about circuit analysis with the pnp or the npn transistor first? \$\endgroup\$ – David777 Oct 11 at 18:59
  • \$\begingroup\$ More like this electronics.stackexchange.com/questions/462469/… \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 11 at 19:12
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But what assumption should we make about the voltage on the collector on the NPN BJT?

You don't need to make any assumption about the collector voltage of the NPN.

You have enough information to find its base current, and you know its \$\beta\$. From this you can answer the question without even calculating the collector voltage. (Although you should calculate it to verify your assumptions about the operating state)

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  • \$\begingroup\$ Thanks, that answers my question. I thought there was another way, but I am familiar with this method. \$\endgroup\$ – David777 Oct 11 at 20:04

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