2
\$\begingroup\$

Suppose we have this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The BJT transistor is in saturation mode.Which are the equations which help us find the different currents(Emitter , Base and Collector current)?

I only know how to solve BJT transistors in forward active mode and I couldn't find anything online.Help really appreciated

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Actually an extra interesting question since this is different than how you usually see BJTs in saturation! I didn't notice this at first, but VBB is greater than VCC. In saturation, Vbc is forward biased, but you usually still have the voltage source at the collector greater than the base. I think this may actually not be saturation, but some kind of negative saturation since IC is negative. I don't know why people are down voting this question, it's quite interesting, IMO. \$\endgroup\$ – KD9PDP Apr 2 at 15:00
  • \$\begingroup\$ @KD9PDP I think the values of VBB and VCC are pretty much irrelevant in this case. What matters are the voltages VBE and VCE. \$\endgroup\$ – Elliot Alderson Apr 2 at 17:00
  • \$\begingroup\$ @ElliotAlderson yes, Vbe and Vce is what matters for biasing the transistor - but the consequence of VBB and VCC being inverted like this is that Ic flows out of the NPN instead of in - that's the weird thing I'm point out, and it happens because the source voltages are chosen such that the base source is much larger than the collector's source. \$\endgroup\$ – KD9PDP Apr 2 at 17:51
3
\$\begingroup\$

Help really appreciated

I'm going to trust you on that point Miss Mulan.

Your circuit won't be behaving like an active (and saturated) BJT any more - there is too much forward bias between base and collector for this to happen and, all rules about transistor amplification are laid-to-rest and, the BJT behaves like two forward biased diodes: -

enter image description here

Compare voltages and currents side-by-side and you'll see that they are virtually identical. The main point to note is this: -

the collector current is flowing from the collector to the 1 volt power supply rail via the 100 Ω collector resistor.

That means all bets are off when trying to analyse this circuit as a bona fide BJT problem because collector current is flowing the wrong way.

BTW the diodes simulated above are 1N4148 devices. So analyse it as two diodes instead of a regular transistor.

\$\endgroup\$
10
  • 1
    \$\begingroup\$ I believe there's a typo in the first sentence: it won't behave like an active BJT (not "saturated" BJT). \$\endgroup\$ – KD9PDP Apr 2 at 11:48
  • 1
    \$\begingroup\$ @KD9PDP yeah good point. I changed it to active (and saturated). \$\endgroup\$ – Andy aka Apr 2 at 11:50
  • \$\begingroup\$ Can anyone tell me why this answer has been downvoted? \$\endgroup\$ – Andy aka Apr 2 at 13:51
  • 1
    \$\begingroup\$ @KD9PDP I'm not suspicious of your activity. The simulation on the left shows that the current flows from the collector and the simulation uses a much more sophisticated model of the 2N3904 than ebers moll. However, you don't need to use ebers moll to see that the base-collector is forward biased. Simple math tells you that and, once that is established, no amount of ebers moll is going to make that any different. \$\endgroup\$ – Andy aka Apr 2 at 14:55
  • 2
    \$\begingroup\$ No, because normal transistor action is now suspended due to the biasing being so off-beat. \$\endgroup\$ – Andy aka Apr 2 at 16:50
2
\$\begingroup\$

You kind of don't need too many equations. Look at the datasheet to get Vce,sat and Vbe,sat - then make those the junction voltages. Then solve for currents and voltages. So maybe Vbe is 0.7 V and Vce is 0.2 V. Then just do nodal analysis and verify that current is flowing the correct way for a BJT in saturation.

EDIT: This answer is for a saturated BJT. But the circuit shows an unusually biased BJT that is is actually not saturated, and probably more accurately described by two BJTs from base to collector and emitter, respectively.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.