# Resistor Divider for Op Amp Supply Rails

#### pdhaslam

Joined Dec 21, 2022
5
Hi kind folks,

I'm trying to built an implementation of the Doug Self Headphone amp from his wonderful book. It's based around an LM4562 Op Amp (Self states TL072) and MPSA42/92 push-pull transistors.

I bought a DC-DC convertor which put out +24v/0v/-24v so I can do a ground and the +/- rails. The op amp needs +/-17v so I thought I could put a resistor divider on each rail with 10k/22k to give 16.5v

Without the amp connected the dividers sits happily at +/-16.5v, once I connect the amplifier circuit it drops to +/-0.9v.

I though it would drop with a load added but does this big drop mean its basically shorting it? I've tested all the components (can't test the op amp but have tried another from the pile) and can't find a fault.

I wondered if my cheap breadboard could be to blame? the transistors are arranged with direct connection to the rail at each end then 2x 10r resistors between the two - at the connection between the two resistors is where the feedback is taken to the op amp inverting input and also goes to the output via a coupling cap.

I feel like I'm missing something very obvious and am just making a newbie mistake. I would be very grateful for some help!

P

#### Ian0

Joined Aug 7, 2020
6,980
The maximum current through your resistive divider will be 24V/(10k+22k) = 750uA, which is clearly not enough for the op-amp.
You would be MUCH better off using a 15V regulator (7815/7915) than attempting to use a resistive divider.

#### pdhaslam

Joined Dec 21, 2022
5
Thanks IanO, that’s sounds sensible.

I have a SMPS which will do 34v so was wondering if it might easier to use that and bias the Ground reference to 17v?

#### WBahn

Joined Mar 31, 2012
28,166
The fault is that you are starving the op amps of the current they need to operate.

Let's see where the current is going when you power the op amp (and assuming that nothing else is connected to your voltage divider).

At 0.9 V, there is 2.31 mA of current in the 10 kΩ resistor and 41 µA in the 22 kΩ resistor, which means that about 2.27 mA is making it into the opamp.

What does the data sheet say that the opamp's needs are when it is just sitting there looking stupid (i.e., the quiescent current)?

So it typically needs 10 mA and as much as 12 mA with ±15 V rails (which is close to your ±17 V rails).

Using a voltage divider to power a device is almost always a nonstarter. Your 10 kΩ resistor would need to be reduced to not only supply the 12 mA that the opamp's internals might need, but also ALL of the current that it might be called upon to deliver to its output, both for the load and for any feedback network.

If you want to allow your circuit to have up to the full 23 mA limit of the part, then your upper resistor has to allow as much as 35 mA of current while dropping no more than 7 V. That alone puts an upper limit on that resistor of 200 Ω.

But it gets much worse, since the opamp's supply voltage will go up considerably as lower currents. If the quiescent current is the 10 mA typical, the voltage would go all the way up to 22 V, which would significantly violate the 36 V absolute power supply rating (it would produce about 44 V across the opamp's supply inputs). Even if this weren't the case, it means that the voltage will vary all over the place under normal operation, which will transfer a lot of noise to your output.

If you wanted to hold the supply voltage to the absolute max rating of ±18 V (and are willing to accept the noise that would still result), then you need a change in current from 10 mA to 35 mA to produce a voltage change of no more than 1 V. That means a value of R1 of only 40 Ω. If you wanted to keep the supply voltage within 0.25 V of the 17 V target, you are now talking about R1 being 10 Ω. With 7 V across it, that's a current of 700 mA.

That means that R2 would need to be sized so that it drops 17 V when 700 mA is flowing through it, making it 24 Ω.

Thus, when nothing is going on, your voltage divider is going to be consuming 16.8 W of power (and the same for the other divider producing the negative rail).

Hopefully this shows why you don't use voltage dividers to provide power to anything.

Another thing that it shows is where the common rule of thumb comes from that the no-load current in a voltage divider needs to be 10x to 20x the current that is going to be pulled from it in order to provide decent voltage regulation. Here we are pulling as much as 35 mA from it and the no-load current needed to regulate the voltage to 0.25 V is 20x that (700 mA).

#### dl324

Joined Mar 30, 2015
15,510
Welcome to AAC!

It's always helpful to include schematics.

How close do the inputs and outputs get to the power rails?

#### WBahn

Joined Mar 31, 2012
28,166
Thanks IanO, that’s sounds sensible.

I have a SMPS which will do 34v so was wondering if it might easier to use that and bias the Ground reference to 17v?
Possible, but you have similar issues in that whatever established the ground reference has to be able to maintain that voltage even under the worst case total ground current in your circuit.

One option is to use a regulator to produce 17 V relative to the negative rail of your SMPS output.

Be sure to provide really good filtering.