Op-Amp: Voltage follower, negative input, positive output?
xanatos
Posts: 1,120
Hi folks,
I need to measure a very high impedance (50k) voltage source that will be in the range of 0 to -5Vdc, using an ADC (probably an MCP3202/4 or similar).
I want to buffer the input since the ADC will significantly load the source, so I want to use an op-amp in a voltage follower configuration to give me - maybe - a few microamps of current drain and not load the source.
Problem is that it's a negative voltage I'm measuring.
So if I use the op-amp in inverting mode with a split supply, will I get 0 to +5V with a 0 to -5V input? Or am I barking up the wrong tree here?
Thanks,
Dave
I need to measure a very high impedance (50k) voltage source that will be in the range of 0 to -5Vdc, using an ADC (probably an MCP3202/4 or similar).
I want to buffer the input since the ADC will significantly load the source, so I want to use an op-amp in a voltage follower configuration to give me - maybe - a few microamps of current drain and not load the source.
Problem is that it's a negative voltage I'm measuring.
So if I use the op-amp in inverting mode with a split supply, will I get 0 to +5V with a 0 to -5V input? Or am I barking up the wrong tree here?
Thanks,
Dave
Comments
Connect your voltage source to the amp's negative input through a resistor of value R. The output is also connected to to the negative input thorugh a resistor of value R. The positive input is grounded. The output voltage will be the negative of the input, andf the input impedance will be R.
If R can't be made high enough to satisfy your impedance requirements, you can precede the inverter with a non-inverting follower.
-Phil
Phil has basically answered this but please note, 50k is NOT "very high impedance"
If you want to have a single supply opamp rather than negative rails you can bias the positive input and with the right resistors it will amplify the difference so you are putting the opamp's differential inputs to good use. It's also unlikely that the source is 50k but more like a few hundred ohms at most but it is specified for 50k loading, this figure being important because you don't want to affect the DC level.
There you received good advice, but .........
If you are not looking for high precision, and are satisfied with a per cent or so, then you could always skip the external A/D and the op amp, and go straight to a Prop SigmaDelta A/D which will handle negative voltages just fine. All you need is a small capacitor and two resistors connected to two Prop pins, and you're done. The software to do this (about 6 or 8 lines of assembler) is probably also simpler than what you would need to interface to an external chip.
Cheers,
Peter (pjv)
As for the output impedance not being 50k... I designed that piece myself. The output impedance is exactly 50k, it's the last leg of a 300.050M voltage divider measuring a -30kv supply. I have 0.1mA available across that last 50k resistor, and any resistance across that under about 10M shows on the output... although I may be wanting more precision than is really necessary, I need to have very precise data collection initially.
I'll post some schematics soon, being called to dinner
The simulations were made with Linear Techs free tool LTSpice:
1) Using VCC & VSS at +5V and -5V. Note the current through R-SRC and the drop caused between R-Sig and VSRC. Also note the VOUT is clamped at +4V and mid is +2V not +2.5V if it was inverting accurately.
2) Added a buffer opamp and the R-Sig & R-SRC track except now the clamp at -4V on VSRC. Notice the R-src current has dropped from 20 micro amps to a few nano amps. Now the clamping of the output at +4V can be seen clearly on the output.
3) VCC and VSS are bumped up to +7V and -7V. This cleans up the output and will present an accurate voltage to the ADC input. If opamp you select won't go rail to rail the overhead will throw off the readings. It also has some to do with load (20k is better than 1K) If the only VCC you have is +5V it could be a problem and you might need to scale your measured voltages.
Dom...
PS: It should work to tie a 30K to a +4V precision reference and monitor the connection at the 30K-300Meg. Then use very low leakage clamp diodes to ground and VCC to protect the opamp input. Just use a single rail opamp as a follower to your ADC and you're done. +1V to +4V equals -30KV to 0V swing and you can use the +5V and no negative supply.
Well in that case you know the characteristics of your circuit then you can either compensate with gain or use a real differential opamp configuration using two opamps. But personally I would just use a single supply rail and bias it so that 0V in was half the ADC reference, that way you measure either polarity as long as you have enough usable resolution.
@Donamik: I did a quick LTSpice as well but got called to "breakfast"
Dinner & breakfast all set
Thanks for all the SPICE work and the ideas.
I've attached a sort-of schematic/block diagram of what I'm doing - the voltage divider shows directly, as well as the moment when - after I had sketched the section - I realized "oh yeah, it's going to be a NEGATIVE voltage", and started thinking about a voltage follower in inverting configuration.
For those who are interested, this is a system to monitor the voltage and current, as well as temperature and vacuum pressure, inside a small deuterium fusion chamber, and display the data on a video overlay on a screen that displays the actual fusion plasma so I don't have to x-ray my eyeballs every time I want to see the plasma configuration
I'll post more when I'm done,
Dave
If you connect one end of your 50K to +5V you can eliminate the opamp. Your 0V to -30KV will have a range of 0V to +5V. This can be fed directly into the ADC. The input leakage of the the MCP3202 is negligible at 1na (0.001% of 100ua). Since the ADC ties the VREF and VDD to one pin the reading will be affected the stability of the power supply. For example if the VDD is 4.5V the readings (0V to 5V) will be off by 10%.
On another note the high side current monitor looks like it's connected to ground. There are some great devices that will measure small high side currents and level shift it down to low voltage ADC inputs.
Looks like an interesting project,
Dom..
Why didn't I think of that?
I read that the MCP3202 actually does have some loading since it has to charge a capacitor to sample, but I guess that only kicks in if the sample rate is high. Is that your understanding? Of course, I'll test all of this out on a breadboard anyway, but I want to be sure I have it all straight in my head first.
Indeed the high side of the current monitor is connected to ground (which will be at full output effectively +30kv above the supply side. ) This allows me to monitor current without the hassles of having anything touching the nasty supply side of this thing (that's going to be about 65mA at 30kv... it would hurt
Thanks for the ideas and feedback,
Dave
When you fire up the PS it will take awhile to stabilize and the ADC input should track nicely. At 30KV and 300meg you have a constant current source of 100ua charging the 10pf cap. To slew from 5V to 0V takes 21us-- it's a linear slope not the typical RC curve you'd see with a resistor cap. It should work just fine for tracking.
The overall inaccuracies will creep in because of: 1) how close to 0V can you measure 2) the 1% resistors 3) the fluctuations in the VREF-VDD. If you want to do better than 2% you'll need to calibrate by measuring the 300meg & 50K resistors to LT 0.1% or measure the 30KV accurately and use math to correct. Large value resistors also have stability problems-- new ones should be "burned in" under load for 60 hours. I haven't had experience with high voltages together with large resistors (just 10G at 10V to get down into picoamp measurements to 1%).
You can Google "high side current monitor". But if one side is "ground" as described you won't need it. Looks like you're measuring current by the voltage drop thru 10ohms at 65ma = 650mv -- so your system ground is referenced to ZERO volts and your power supply is approximately 30,000.65 volts.
You must have a whopper of a power supply. 30,000V X 65ma / 85% eff. = 2,300 watts Wow-- .
Dom..
As for measurement accuracy, I have a good HV probe that "should" be accurate, although it was probably last calibrated shortly before the last ice age (ie., 1989
This thread has been full of good suggestions... I knew this was the place! Thanks all,
Dave
If you had a high voltage DC source (like coming off the input caps of a switching supply) you could:
1) Hook your resistors across it and measure the HV - mine switchers run at approx. 300V
2) Measure the 300V exactly; it might be 289.7V for example
3) Measure the voltage across the 50K. It should be about 50mv.
4) Now you have enough information to find the ratio of your divider and know the real value of the 100Meg resistors
Whatever value you get should correlate with your HV probe calibrated. I've seen those HV probes be off by huge amounts, like 20% and the meter input impedance has to be part of the calculation to use them.
Also a safety concern. Imagine one end of a resistor or wire that is normally close to 0V comes loose and the 300M resistor charges that node up to 30KV. The circuit parasitic capacitance would store the charge at 30KV if it didn't arc. That is until it comes into contact with the circuit or discharges through body contact. The amount of current supplied could be more than the 100ua, it could be more like 3Amps. Two clamp diodes in series would hold the node at -650mv if there's a failure and be out of circuit under normal operation.
Dom...
Dave
Check that, thanks for updating us on this cold case, we know what to do now, puncture the lexan barriers, one more big bang, problem solved, case closed.
Kind Regards,
Your Friendly Cold-case Closer
Just like the thread title, negative input....positive output
Seriously though, stay safe, HV is bad enough without even talking about EHV and serious current (and those FCC types).
You can eliminate your ADC by using the onboard ADC124S021. Here's a link to the hookup and code to make it happen.
http://learn.parallax.com/propeller-c-simple-circuits/measure-volts
The other 3 inputs might be useful for the temp, pressure, etc.
Dom..
Call me paranoid (or why do GE, Siemens, Philips and so many other OEMs go overkill on isolation and grounding in x-ray based system),
Frank.
What a great idea... Isolate the HV by sending a frequency instead of straight voltage. The Analog Devices AD537 is one such device and operates at 10KHz and quiescent current of 1.2ma. You could run it off a 9V battery that is completely isolated from the low voltage P.S. and the prop should easily work as a counter. Don't know what to use as an isolation barrier; perhaps toroidal coil or there might be opto devices with this kind of isolation or a jar capacitor. This project sounds like fun.. Hopefully Dave will share photos and plans for his fusion reactor -- the goal is 1.2 GIGAWATTS!!!!!!!
Dom..