3
\$\begingroup\$

I would like to build an amplifier can serve as an electrometer (measuring voltage with very low input current/high input impedance) at high bandwidth, ideally up to 10MHz. Think of it as a hybrid between an electrometer and LNA :) Desired specs: input current as low as possible (femtoamps), noise as low as possible, bandwidth >10MHz. This is the front end of a custom instrument in which I am trying to measure microvolt signals from a source with impedance >100MOhm at 5-10MHz. The source is physically small, fits in a 1mm cube.

There seem to be at least two big problems with this:

  1. The standard electrometer-grade op-amps have very low bandwidth, for example OPA129 has offset/bias current of 30fA but unity gain bandwidth of 1MHz. (Another device, the LMP7721 looks nice with 3fA offset and 17MHz bandwidth)

  2. I assume the source would have to be very close (essentially located between the input pins of the op-amp), but it seems even the tiny parasitic capacitance would be enough to short-circuit a lot of the signal at higher frequencies.

Any suggestions for how to approach this problem?

Update: Thank you for the helpful discussion. I am not completely sure of the source properties, I made an order of magnitude guess from a rough physical model. What I do know is that most simple amplifier configurations hooked up to that source show no signal, so the signal has got to be very weak. It appears that accurately modelling the source is key for making any progress.

I am aware of thermal noise, and indeed the signal I am looking for may be quite weak compared to thermal noise. It may be possible to see it by integrating over time (after digitizing) despite that.

\$\endgroup\$
  • 1
    \$\begingroup\$ With a source resistance of 100Mohms and a bandwidth of 10MHz, you'll be getting an equivalent noise voltage signal at ambient temperature of 4mV RMS. How do you plan on measuring a "microvolt" signal? \$\endgroup\$ – Andy aka Jul 15 '13 at 13:45
  • 1
    \$\begingroup\$ Are you sure that 100M output impedance is legitimate? Or is the question indicative of an attempt to over-spec the design? \$\endgroup\$ – Anindo Ghosh Jul 15 '13 at 14:23
  • 2
    \$\begingroup\$ Yeah, 100 MOhm alone (let alone the filter due to parasitic capacitance) is not trivial to handle. At least that is doable in isolation, but you have to be extra careful. Don't touch the insulating material with your fingers. The dirt of your fingerprint and from other sources plus atmospheric moisture can be significant relative to 100 MOhm. \$\endgroup\$ – Olin Lathrop Jul 15 '13 at 14:29
  • \$\begingroup\$ @Andyaka: That at least is possible, collect a lot of digitized data and do an FFT :) \$\endgroup\$ – Alex I Jul 16 '13 at 5:08
  • \$\begingroup\$ @AnindoGhosh: Not a deliberate attempt to over-spec, although perhaps my source model was too simple/not accurate. I need to go back to that and come up with an accurate source (signal level, equivalent resistance/capacitance, ...) \$\endgroup\$ – Alex I Jul 16 '13 at 5:11
6
\$\begingroup\$

Fahgeddaboudit.

Do the math. The capacitance to roll off at 10 MHz with a 100 MΩ impedance is only 160 fF. There is basically no chance you can keep the parasitic capacitance that low with any off the shelf chip, and then there is whatever parasitic capacitance in the thing that is originating the signal, and in the connection between the thing and the chip intended to receive this signal.

The usual way to receive high impedance signals is to run them directly into the gate of a FET. However, whether MOSFET or JFET, those will have substantially more gate capacitance than 160 fF. The last resort is usually a vacuum tube, but even those have more grid capacitance than that. I didn't look for low grid capacitance, but even a medium mu amplifier like a 12AU7-A has 1.5 pF grid to anode and 1.6 pF grid to cathode. That's still 20x more than you can tolerate, even with a circuit that holds the plate voltage fixed. Perhaps there are some specialty tubes that can do ultra-low grid capacitance, but 160 fF is a stretch.

Added:

You had originally asked for a very high impedance amplifier, but now you say it is permissible to short the signal. That makes a huge difference. Now you can measure the short circuit current instead of the open circuit voltage. As others have said, you want a transimpedance amplifier, which means it takes current in and produces a proportional voltage out.

Your specs are still demanding because the signal is so tiny. There will be lots of inherent sources of noise at that level. You may need to actively cool the amplifier front end.

You said the impedance of this source is 100 MΩ and the open circuit voltage of the desired signal would be microvolts. 1 µV / 100 MΩ = 1 fA. That's tiny.

Rethink

You need to step back and look at the overall problem, and not pre-suppose a particular solution. What are you really trying to measure. Or perhaps pop up another level and explain what you are trying to control, prove, or whatever, that you think making this measurement is part of.

\$\endgroup\$
  • 2
    \$\begingroup\$ Don't forget about noise. For \$100M\Omega\$ at room temperature the noise is around \$1.3\mu V/\sqrt{Hz}\$. At a bandwidth of only 1Hz this is already on the same order of magnitude as the "microvolt signal" being measured. Cryogenic cooling, perhaps? \$\endgroup\$ – Phil Frost Jul 15 '13 at 15:09
  • \$\begingroup\$ That's why I suggested magnetic or capacitive receptor to a narrow band radio with RSSI. Then 1uV sensitivity is easy. dBµV = 106.98 + dBm50Ω \$\endgroup\$ – user26366 Jul 15 '13 at 16:07
  • 1
    \$\begingroup\$ @Olin - I think the main problem is noise - using a transimpedance amplifier can largely overcome the capacitance issue seen on conventional amp configurations. \$\endgroup\$ – Andy aka Jul 15 '13 at 16:27
  • \$\begingroup\$ @Andy: Transimpedance (basically measuring the short-circuit current) is a good strategy if it's OK to short the signal. Maybe it is, but the OP specifically asked for a high impedance amp, which just isn't going to happen with these specs. \$\endgroup\$ – Olin Lathrop Jul 15 '13 at 17:25
  • \$\begingroup\$ @OlinLathrop: Thank you. A few comments... "roll off at 10 MHz with a 100 MΩ impedance is only 160 fF" but that assumes this is the -3dB. I don't need a flat frequency response, I just need to get something at 10MHz, even if it is rolled off a lot more. "the connection between the thing and the chip" - I was assuming that the source would be literally placed between the input pins. If that is not enough, I might look to more exotic solutions, basically remove the op amp packaging and bring the source closer, possibly directly on chip. \$\endgroup\$ – Alex I Jul 16 '13 at 5:24
1
\$\begingroup\$

Ignoring the noise issue and bandwidth I commented about, you will (I'm sure) have to use a trans-impedance amplifier.

In effect the 100Mohms source is shorted to a virtual ground (hence input capacitance issues are largely removed) and the amplifier amplifies the current from the source via the 100Mohm impedance it has.

They are used to good effect in photo-diode amplifiers but your yield (volts out per micro-amp put in) is going to be very low and you will struggle with 10MHz bandwidth.

For instance, an OP380 with 50kohm feedback resistor and maybe 0.4pF of anti-peaking capacitor (to keep frequency response flat) will have a bandwidth of barely 10MHz and for every micro amp from your source, the output voltage will be only 50mV.

enter image description here

Maybe there are some better devices so keep searching but a trans-impedance amp is your only hope I think.

\$\endgroup\$
  • 1
    \$\begingroup\$ What does the response look like with inherent capacitance across the 100MOhm resistance? +1 slope? \$\endgroup\$ – user26366 Jul 15 '13 at 16:04
  • 1
    \$\begingroup\$ @JoeSCADA how much inherent capacitance do you want to guess at? It could be fempto farads if the source is an ion beam as would be found in a gas-mass-spectrometer. It may be a lot higher. Noise is the real problem I suspect. \$\endgroup\$ – Andy aka Jul 15 '13 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.