I´m having trouble with this circuit. Is it possible to get Q2 --> saturation using the VI voltage?

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When VI = 0; the VCE2 = 14.457 V. By calculations I know that Q1 and Q2 cannot be on saturation at the same time.

Is possible to get Q2 saturated with the VI voltage?

  • \$\begingroup\$ You'd need about 50 A coming out of the Q2 collector to get to saturation. What problem are you actually trying to solve? \$\endgroup\$ – The Photon Oct 4 '17 at 18:06
  • \$\begingroup\$ What calculations have you performed? I don't see any, at all, beyond statements. How did you get the value of 14.457, for example? How would you approach answering your question? Show some work. Any work. (This isn't the first time I've seen this specific schematic tossed in here with this particular question. Must be on someone's course work.) \$\endgroup\$ – jonk Oct 4 '17 at 18:07
  • \$\begingroup\$ @ThePhoton I don't think the OP is trying to solve a problem, I'm pretty sure, since this exact problem has been posted here before. I remember something very close. \$\endgroup\$ – jonk Oct 4 '17 at 18:08
  • \$\begingroup\$ EEK upside down circuits..... \$\endgroup\$ – Trevor_G Oct 4 '17 at 20:02

This circuit won't saturate.

The Q2 base will not be pulled lower than ground by Q1.

To reach satureation, then the collector needs to be pulled up to around -0.2 V. But because of the low value of R1, this would require about 50 A out of the collector.

However to get 50 A out of the collector would drop 50 kV across R2, which would reverse bias the b-e junction and put the device in cut-off rather than saturation.

If you want to get Q2 into saturation make R2 small and R1 large instead of vice versa. Whether that will actually make the circuit do what you actually need it to do, I can't say since you haven't explained what the actual purpose of the circuit is.


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