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In the circuit given below we have to find output voltage. What is Vo No specifications about any of the transistors are given just that all the components are "discrete". Can we find out how much current is passing through which BJT? Also how to tell how much voltage is dropped across MOSFET i.e. VDS ? What is the procedure to do analysis of such circuits?

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  • \$\begingroup\$ Step 1, assume all components ideal. Assume diff pair are balanced. Vout = 3x base voltage of 2nd transistor = 7.5v. Does that get you started enough? Once you've got node voltages for the ideal case, you might add small changes due to finite BJT beta (100 is a reasonable assumption) for extra marks. \$\endgroup\$ – Neil_UK Apr 5 '18 at 5:00
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Current thru each transistor depends on load impedance and matching Vbe but it can range form 0 to 4mA with the sum of both not to exceed 4mA.

{ assuming real-world question, change Nfet to Pfet}
Otherwise Nfet is OFF and body diode drops 0.7 not shown to Vout with Q1=4mA Vg1=2V.

Since Vg is limited to between 10V and 2V with an 8V difference , for best regulator the Vgs(th) must be around 1/3 of 8V or 2.7 max so use Vt or Vgs(th) ~ 1V. Then RdsOn will determine what load you can handle.

Since there is no short circuit protection, you may want to add current sense to limit Vgs drive.

Voltage regulation is not as good as LDO's here due to BJT gain of single stage diff Amp. and higher Iq due to bias R's

Also since the R ratio is 2:1 the Vout is (2+1)*Vref(2.5)=7.5V out ideally but with bias error closer to 7.7V

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  • \$\begingroup\$ Can't Vo be zero? See if full 4mA current passes through transistor with 2.5V base supply, then 2V will be the voltage on the GATE of MOSFET. Its body is at 10V potential so channel between source and drain wont form. thus no current will pass through it. Is this explanation incorrect? I think the statement in bold isn't right. Please clarify. @TonyStewart \$\endgroup\$ – user175306 Apr 5 '18 at 5:22
  • \$\begingroup\$ It's incorrect Pch FET will conduct heaviliy with 2.5x Vgs(th) in sub-threshold types and 3x Vgs(th) for older types. 2~4V. When Vg=2V, Vgs=-8V thus 3*Vt will conduct near rated RdsOn \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 5 '18 at 5:31
  • \$\begingroup\$ Sir, this is a Nch FET. Please see link \$\endgroup\$ – user175306 Apr 5 '18 at 5:38
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    \$\begingroup\$ I assumed that was newbie error.... then it wont work and internal diode will drop 0.7 to output, but I guess this is an academic trick question \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 5 '18 at 5:40
  • \$\begingroup\$ Okay. Will that means output will be 9.3V? If so, then 2K and 3K transistor have to conduct currents. And as BJT can't supply any current so that has to come from MOSFET. But it isn't conducting as we saw earlier. \$\endgroup\$ – user175306 Apr 5 '18 at 5:44

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