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The schematic above is a very good voltage regulator, excepting the high power dissipation for Q1 (also I'd use an NPN instead of a PNP). Can I replace Q1 and Q2 with a high current n-mosfet and expect the same behavior? A mosfet has lower power dissipation than a bjt.

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    \$\begingroup\$ It's also not a very good voltage regulator unless Vin is already regulated : the voltage across zener D1 is somewhat dependant on the current, which is (Vin-V(D1))/ R. That dependence, or "slope resistance" of the Zener differs for different voltages of zener diode. \$\endgroup\$ – user_1818839 Apr 25 '13 at 21:31
  • \$\begingroup\$ I'd also say it's not guaranteed to be a good regulator in general applications because there is a good chance it will oscillate unless A1 is a very poor op-amp. You can't say it's a good regulator without knowing what the component values are too. Ditto Brian's comment about the zener and I'd add that if there was significant impedance in the line feeding it it'd be even worse!! \$\endgroup\$ – Andy aka Apr 25 '13 at 22:05

To make the new circuit work the same way as the existing one, you'd need to replace Q1 with a p-channel MOSFET, not an n-channel one.

However, you would not save any power dissipation by doing this. Because you wouldn't operate the MOSFET fully switched, you'd be operating it as a variable resistor.

You can see that there's no way to significantly improve the power dissipation of this circuit by the equation for device power: P = I * V. In this circuit, the current through Q1 is equal to the load current, so you can't change that. And the voltage across Q1 (collector to emitter) is equal to Vin - Vout, and you can't change that. So no matter what you replace Q1 with (however good a MOSFET you find), the power dissipation will still be Iload * (Vin - Vout).

If you want to reduce the power dissipation from your regulator, you need to look to a totally different type of circuit, like a buck switching regulator.

  • \$\begingroup\$ Exactly, these regulators work by converting the difference in voltage to heat. The one thing that a MOS based design might help with is in the design of a LDO regulator, but this design may not lend itself to this. LDO = Low Drop Out \$\endgroup\$ – placeholder Apr 25 '13 at 21:08

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