Measuring load voltage

[chrome]

I'm trying to measure the voltage across a load, the max voltage can be +60VDC so I figured a resistance divider would work here.



Now I need to keep the current trough the resistance divider below 2.5µA so I chose a high value (41M75 total) but now i ran into problems because the ADC typical Differential Input Impedance is 2M4 so that's obviously gonna skew the value/resistance divider.

Anyone know of any better methods of getting the reading? Maybe an opamp between the resistance divider and the ADC with a high (>G) input impedance?
Or another method altogether.

[PA0PBZ]

60 Volts @ 250mA, so your load resistance is like 240 Ohms?
And why do you want to limit the current trough your divider so much? that's like 10^^-5 of the current trough your load, are the load and ADC that accurate?
Just wondering, what are you making?

[chrome]

It's a current source from 0-250mA which can be 0-60V and I want to keep the current trough the measurement low as to not offset the other current reading I have already in place.

[PA0PBZ]

What is the other current reading you already have in place? I'm still thinking that you are aiming at a lot more precision than you can ever read.
Things might be a lot simpler if you go for 10^^-4, so 25uA max which translates to 2.4M. 2M and 180K would give you 0-5 volt, the 180K being ok for your 2M4 ADC input. (This assuming your ADC is 0-5 volts)

[chrome]

The reference is 2.5V.

The ADC can do (theoretically) 28.6µV (over the full 60V range) where I only need 1mV.

The current reading is a measurement over a very accurate resistor.

[digsys]

There's definitely no technical reason NOT to use a buffer, and you can easily find one with that sort of input impedance.
The tricky part is - Input offset and temperature drift !! (which will also include the resistors)
With a 2.5V Ref and 2.5V max Input, If you wanted 1.0% precision, then you need 1 order of magnitude below 25mV,
accuracy and stability. For 0.1% precision, then it's 2.5mV / 1 order .. all quite doable.

[chrome]

Well I'm stuck again, I could go for a Instrumentation amplifier or a differential amplifier but those also have downsides:
Instrumentation amplifier : I don't think I can put 60V on the inputs.
Differential amplifier : Because of the high resistance divider before it and having to use a sort of resistance divider for the gain (1) it needs even higher resistors which is pretty hard to do cheaply (right?).

[Berni]

You could put two opamp buffers on to it. You can make a 1x gain buffer if you tie the negative inputs to its output and feed a voltage in to the positive one. I am assuming the ADC is pretty slow so you can use one of those chopper opamps that automatically compensate there offset voltage to near 0. So put one of these buffers on the + and - input of your ADC and you have the input impedance you want.

[chrome]

I don't quite understand what you mean, could you draw me a little schematic?

[Berni]

Just built two of these circuits http://en.wikipedia.org/wiki/File:Op-Amp_Unity-Gain_Buffer.svg and put them right before the + and - input of your ADC. The input of this thing is near infinite resistance and the opamp will do its magic to keep the output voltage identical to the input voltage. And you can feed that output voltage directly in to your ADC.

Just be sure to use a opamp with a very low offset voltage. The offset voltage of a non ideal opamp will add or subtract a few mV from the output of the circuit and so affect your accuracy.

[chrome]

Just built two of these circuits http://en.wikipedia.org/wiki/File:Op-Amp_Unity-Gain_Buffer.svg and put them right before the + and - input of your ADC. The input of this thing is near infinite resistance and the opamp will do its magic to keep the output voltage identical to the input voltage. And you can feed that output voltage directly in to your ADC.

Just be sure to use a opamp with a very low offset voltage. The offset voltage of a non ideal opamp will add or subtract a few mV from the output of the circuit and so affect your accuracy.

I don't think that will work since the voltage can be between 0 and 60V so I would need an opamp that can handle +60V on it's rails.

[Berni]

No i meant to put it after the voltage divider and before the ADC to buffer the dividers voltage and turn it to near 0Ohm impedance to drive the ADCs input. If you wanted to buffer before it then yeah you would need a opamp that can handle that. I know TI makes one that accepts an input 100V above its supply rail so that could be used but its expensive so just put it behind the voltage divider

[chrome]

Even after the voltage divider the voltage can be 60V (Short or very low resistance on output)

[chrome]

Oh man, I think I solved it, using two separate voltage dividers I ensure I always go below 2.5V or lower.

[Crumble]

Make sure to clean your pcb. High-impedance circuits are sensitive to resin or finger prints left over from building the circuit. The late Robert Pease recommended cleaning a pcb in a dish washer. I never tried, but most electronic components are usually quite water-proof (exception include connectors, relays, most inductors etc).

[Berni]

That circuit should work yes, but as Crumble said its very important to keep the board clean of flux and keep large clearances because that 40M resistor can quickly get changed to 35M by some dirt. Its also important to use a high input impedance op amp like a JFET input one, but then also add some clamping diodes since this is used around power stuff and JFET inputs quite are sensitive.

[TerminalJack505]

If your 40M voltage dividers don't work then you might try something like the following.  It is based on the JFET source follower in the Art of Electronics book (figure 3.29.)  By using a FET gate for the input stage the input current will be on the order of 10 nA.

The problem with the Art of Electronics circuit for you is that 1) high voltage JFETs are hard to find, and, 2) you can't put a voltage divider directly on the output without affecting accuracy.

This circuit uses depletion mode FETs rather than JFETs so that higher voltages can be supported.  It uses a third FET which you can place a voltage divider on with a more sensible impedance (470k in the simulation)  than your 40M.  If precision isn't important then the third FET can be replaced with a BJT.  (The value of R1 would need to be changed.)

Note that your attenuator could 'fix-up' some of the output error with resistors of suitable value.

The FETs used for the simulation were BSS229s but you should be able to use pretty much any so long as they support the system voltage.  The closer the FETs match, the better, obviously.

The simulation shows that you should be able to get pretty accurate results.  I wouldn't expect +/- 0.2% like the simulation shows but +/- 2% should be easy to achieve.  Assuming, that is, that I haven't missed something.  I have only simulated the circuit.  I haven't actually built it.