# Op amp voltage regulator using N-Mosfet

I am trying to make a voltage regulator using an op-amp and a MOSFET:

schematic http://no4nwo.com/images/724278393.png

I'm really new to electronics. Why is $V_{DS}$ always so low ?

Connect the inverting input to the MOSFET source /output - that's the first thing. Also If your zener voltage is too high you may get additional problems if: -

• The op-amp isn't a rail-to-rail output type
• The MOSFET Vgs(threshold) is too high be able to sustain the output under load conditions

Here's a similar circuit: -

This uses an NPN BJT - to use a FET connect an N channel device with source to the circuit output and drain to positive supply.

• Thank you, that's exactly what I needed. Now I understand what I've done wrong. I think it was just too late and I was tired, which lead to mistake. – Bartek Szablowski Jun 4 '13 at 10:37

Here's a rule that's true for most op-amp circuits with negative feedback:

An op-amp will attempt to make both its inputs the same voltage through the negative feedback path.

Here, the feedback path goes from the output, through the gate-source capacitor of the MOSFET, through the $10 \mu F$ capacitor, and back to the inverting input. No DC current can flow through the capacitor, so there is no stable DC operating point for this circuit.

Put as a rhetorical question: what's the voltage at the inverting input? Without considering leakage currents and random noise, there is no answer.

This will be more of an extended comment rather than an answer, probably.

• Consider skimming through: this, this, this, which describe different LDO topologies.
• In the original drawing, why is the feedback AC-coupled through a capacitor? That's incorrect. The OpAmp doesn't see the DC component, which it's supposed to regulate.
• Where does the OpAmp's supply voltage come from (or do you ignore that for the purposes of this simulation)? In the N-channel LDO, the OpAmp has to drive the gate of the MOSFET several volts higher than the source. That implies that there should be a sufficient supply voltage for the OpAmp. If the input voltage is not sufficiently high for this, then the OpAmp supply can come from an auxiliary voltage source (usually it's a charge pump). Don't know if the O.P. needs to consider this in his simulation.