Is there a linear regulator that can follow a reference?

The broader problem I have is that I have an ADC with reference that can only source 50mA (I have no control over the ADC or reference circuit) but I have some sensors that are low impedance and measured with a Wheatstone bridge so I need the same reference voltage but more current.

I thought of doing this (below) but there must be an IC that does the job.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ How much current is needed? \$\endgroup\$ May 29, 2013 at 8:23
  • \$\begingroup\$ 1A to be safe (I'll be running multiple sensors) \$\endgroup\$ May 29, 2013 at 8:37

2 Answers 2


The requirement basically translates to the ability to source high current at a specified reference voltage.

A low component count way to implement this is to use a high power high current op-amp as a voltage follower, with the reference voltage at the non-inverting input.

For instance, the Burr Brown (now TI) OPA549 can deliver 8 Amperes continuous current with a suitable heat sink. With its 300,000 open loop gain, this should be sufficient for using the available low current reference to generate a stable high current output.

There are many other such high current op-amps, which dispense with the need for the external MOSFET.

From this page:


In other words, a voltage regulator IC is not required.


Let's start with a circuit that works: - enter image description here

The output equals the input because the op-amp has to have both inverting and non-inverting inputs at the same voltage. OK choose an op-amp that has low Vos and low Vos drift. Choose an op-amp that has a very low noise figure if the sensor circuits you are feeding are used for very small measurements.

Your circuit - It won't work - you have placed gain into the feedback loop of an op-amp and it will sing and scream with oscillations because op-amps aren't designed to have extra gain like this. Where is the extra gain coming from? The MOSFET is common source into a load resistor. Its gain is drain resistance divided by source resistance. You have zero source resistance and your drain is 100 ohm. Need I explain more?

Emitter follower - You need to supply 1 amp to your sensors and using an emitter follower with a gain/Hfe of say 50 means your op-amp is having to feed 20mA into the base - this should not be a problem given that your supply voltage is 12V and you only need to produce 3.14V. However, you will have a power dissipation problem. 1A through the transistor with about 9V across it means 9W of power given off as heat.

Lower the power rail - I would strongly urge you to run the circuit from a lower power rail like 5V. Then you'll find that the power is only about 2W and a moderate heat-sink will do the job.

But beware - there are now extra problems in doing this. You'll find that the BJT is working closer to saturation and maybe the gain will drop to 20 meaning the drive from the op-amp will need to be as high as 50mA. Most op-amps won't do this so you'll have to find one that does. I'd start by looking at the AD8605 - it can source +/-80mA and will be nearly rail-to-rail with more normal loads like 10mA. It has fairly low noise too. Vos might be a little high though for your application.

MOSFET source follower - Alternatively use a N channel mosfet instead of the BJT - this requires virtually no drive current from the op-amp - downside - pick one with fairly low Vgs(threshold) or you won't get the output from the op-amp high enough to properly turn it on under heavy load conditions.

  • \$\begingroup\$ The OP's circuit will likely oscillate as drawn. But it can probably be compensated with capacitance at Vout, just like what's recommended/required for just about every linear regulator in existence. \$\endgroup\$
    – The Photon
    May 29, 2013 at 16:20
  • \$\begingroup\$ @ThePhoton this could possibly work but it will need to be slugged with a big cap. The FET might as well be anywhere in the loop - it could, for arguments sake be an extra transistor stge in the op-amp BUT, if the op-amp is anything decent it will be running close to it's instability limit (with unity gain feedback) and adding a gain of between 10 and 100 will make it sing. It'll be stabler on heavy loads for sure but i don't see much of an advantage trying to get this working. I'd bet voltage regs have spare capacity in their error amp compared with off-shelf op-amps \$\endgroup\$
    – Andy aka
    May 29, 2013 at 16:32
  • \$\begingroup\$ Yes you need to explain more. Why is high gain with negative feedback bad? Also isn't the transistor "gain" here the slope of the Vgs-Isd times the load resistance? I am fine with using a BJT though. \$\endgroup\$ May 29, 2013 at 16:58
  • 1
    \$\begingroup\$ An op-amp has a DC gain of maybe 1e6 and by only 10Hz, the slew rate limit is already eating into it. From here until about 1MHz when the op-amp has fallen G=1, it is 1/2 way (90º shift) to instability and usually, by the time unity gain is reached, there'll be some more parasitic capacitance around that nearly turns the corner again and brings about full instability on the op-amp. So if you add a bit of gain to the the negative feedback - it sings. See this electronics.stackexchange.com/questions/69506/… \$\endgroup\$
    – Andy aka
    May 29, 2013 at 20:30
  • \$\begingroup\$ It's not that I'm suggesting using a BJT - i'm suggesting using a source or emitter follower and these both have unity gain \$\endgroup\$
    – Andy aka
    May 29, 2013 at 20:32

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