Search OpAmp chip as a voltage follower

Hello, I am now looking for an Operation Amplifer chip able to be a voltage follower, which I need to use as a buffer between my sensor circuit and the digitizer. Can anybody help me?

Since the output impedance of my sensor circuit is about 0.2 Gohm, the follower needs to have a input impedance much higher than that (Maybe a JFET one could work).

The input signal is always sinusoidal, with an amplitude = 1~2V, and a variable frequency = 10KHz~1MHz, and the noise at the output is required to be

Reply to
lvwener
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I suggest OPA2350. But your requirements just can't be met by any technology.

Vladimir Vassilevsky

DSP and Mixed Signal Design Consultant

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Reply to
Vladimir Vassilevsky

We'll ignore the difficulty of reconciling a 200M source with a 1MHz bandwidth (what about capacitance?). Let me suggest you evaluate the TI Burr-Brown OPA657 family. These are JFET opamps with 1.6GHz bandwidth and 700V/us slew rate -- that should be fast enough!

Reply to
Winfield

On a sunny day (Sun, 28 Oct 2007 21:28:42 GMT) it happened Vladimir Vassilevsky wrote in :

Perhaps, if the source allows it, he can drive the sensor directly into a low impedance source, like a BJT. Then it behaves as current source, and capacitance becomes not important.

Reply to
Jan Panteltje

Input impedance using Tera Ohm IC's or the equivalent transistor circuits are fairly easy. (It's in the book) Much lower than that and you'll be counting electrons :-). Ken

Reply to
Ken S. Tucker

Thanks so much for the kindly replies!

Well, I am sorry about the typo in my post: the OUT IMPEDANCE of my sensor circuit is typically "20 MOhm" (but NOT 20G).

The > snipped-for-privacy@gmail.com wrote:

Reply to
Wener

"Wener"

** Even with 20 Mohms, noise a 1 MHz band is 570 uV.

( En = sqrt 4kTBR )

You have not got the tiniest clue what are crapping on about.

........ Phil

Reply to
Phil Allison

A 1 MHz, 0.2 Gohm signal basically can't exist in any form that can be connected to an opamp. And at that resistance, the noise density is almost 2 uV per root Hertz.

John

Reply to
John Larkin

I thought perhaps something like the LH0063 is what the fella might want...

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Ken

Reply to
Ken S. Tucker

Wener schrieb:

Yes! - as others have remarked, you do not seem to know what you are talking about. By now, you have mentioned three different values for your sensor source impedance: 0.2 Gohm (that's 200 Mega-Ohm), 20 MOhm (I take this to be Mega-Ohm), and 20 G (Giga-Ohm?). Before somebody succeeds in selling you a $50 opamp, I suggest you try this:

Cut off most of the signal pin of your sensor, leaving the smallest amount of metal that you can solder to. Make sure no other metal parts are close to the pin. Obtain a FET (field-effect transistor) with a small reverse capacitance - the 2SK161 comes to mind. That should cost you less than $1. Cut off as much of the gate pin of the FET as you dare and solder it directly to the sensor pin. Pick off the signal at the source pin of the FET, using a resistor from source to ground to feed the FET a current of a few milliamps (chose the resistor such that the current stays below Idrain-source at zero Vgate-source, so measure your FET first). Connect the drain pin of the FET to a positive voltage supply (not exceeding 18V with the 2SK161).

Note that even with a FET capacitance of 0.1 pF, 100 MHz signals will be damped very significantly: the R-C time constant of the setup would be 20 Mohm * 0.1 pF =3D 2 =B5s (micro-seconds). And, as others have remarked, your noise requirements were nonsense anyway: if kept at room temperature, the 20 Gohm sensor itself will be the dominant noise source (look up the Nyquist formula) - you can forget the FET noise. Once it works, you should measure the frequency response of your sensor-plus-preamp setup and correct all later measurements for it.

That's it.

Martin.

Reply to
clicliclic

On a sunny day (Mon, 29 Oct 2007 04:32:08 -0700) it happened " snipped-for-privacy@freenet.de" wrote in :

Very nice solution, but let's look again: +12 | 2k2 R1 |------------------------------- out c ------------------------ b T1 NPN low noise | Z1 e beta ~500

20MOhm | | 2Vpp | ~ | | | /// ///

So what happens? Delta IbT1 is 2 / 20.000.000 = .1uA Delta IcT1 is 500x.1 uA = 50uA Delta Ur1 is 50uA x 2200 = 110mV. in 2k2! Any cable and component capacitance at the input sees the base emitter capacitance of T1 in parallel, changing the time constant to much shorter (will be a few kOhm).

Although the output voltage swing is lower, you have a much wider freq response in a nice low impedance. I have only showed the AC path, one should bias the emitter of T1 for the right Ic (or the generator). Simpler even if AC coupling is allowed. The non-linearity cause by the be junction of T1 is insignificant when you see that IbT1 is mainly set by Ugenerator / Zgenerator.

In some cases this circuit may be a very good solution,. I have used to as preamp in a vidicon based camera, where there is also high impedances like this, and lot of noise from defection coils etc in the immediate vincinity. Worked very well.

Transistors are good current amplifiers, why not use it?

Reply to
Jan Panteltje

Reply to
Winfield

Jan Panteltje schrieb:

pacitance of T1

hm).

sponse in

right Ic

u see that IbT1

igh impedances

cinity.

Three remarks:

- Loading the 20 Mohm sensor with (say) 2kohm will divide the signal by 10^4 and the noise by 10^2. So the signal-to-noise ratio will decrease to 1/100 - which is bad.

- I fear that at frequencies around 100 MHz the Miller capacitance of real beta=3D500 transistors will bring down your output signal to around

1 mV - as bad as with the FET circuit.

- My phrase "using a resistor from source to ground" should have been "using a resistor from source a negative voltage supply" - after all, we have to accomodate a large voltage swing. The total voltage across the 2SK161 must not exceed 18 V.

Two $1 solutions - it's up to the original poster.

Martin.

Reply to
clicliclic

On a sunny day (Mon, 29 Oct 2007 09:34:35 -0700) it happened " snipped-for-privacy@freenet.de" wrote in :

I think you see this the wrong way. You are using the current in the circuit, and the current is in no way attenuated.

The OP mentioned *1MHz* IIRC.

Yep.

The circuit I mentioned worked for me though. I had this camera preamp, with nice FET input stage, and no way could I get it 'noise' free, and here it was not so much thermal noise, but the fact that these extremely high impedances pick up _any_ noise from nearby (and probably far away too) sources. So I tried about 3 or 4 different high impedance FET circuits. None if these worked like I hoped, and then accidently I touched the target wire (from the vidicon tube) on the input of the NEXT stage (a normal NPN common emitter amplifier). Great picture, no noise, flat frequency response!! It was one of those moments where you sit for some minutes, stunned, wondering 'how can that be?'. I figured it out, removed the FET stage, changed a few things to optimise it, and it worked OK ever after. In those days (sixties) you had no spice...You had to build things to try it. As to the Miller cap, you can drive the collector into the emitter of a next NPN, that has the base at say +5, so Uce is constant (cascode).

Reply to
Jan Panteltje

Jan Panteltje schrieb:

I must disagree: the noise injected by a base-emitter junction that constitutes an input resistance of 2 kohm roughly corresponds to the voltage (and current) noise of a 2 kohm resistor. But we should let others comment on this point.

You're right! That's so much easier for both circuits. (And we may also hope for the sensor impedance to be corrected to 2 Mohm!)

(Of course my correction should have read "using a resistor from source _to_ a negative voltage supply".)

Martin.

Reply to
clicliclic

On a sunny day (Mon, 29 Oct 2007 11:53:16 -0700) it happened " snipped-for-privacy@freenet.de" wrote in :

OK, but for that to be amplified, there must be Ib flowing, that requires an external path, and in this case that path is via the 20 MOhm generator Ri? That would result in a really really low contribution to Ib... But indeed, others may be better at theoretical noise calculations then me. I only know for sure it worked for my application :-) And then mainly to reduce noise pickup from external sources.

Reply to
Jan Panteltje

You aren't looking for an op amp. What you want is an electrometer amplifier.

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So, does that mean we're coming to the end of anything new and exciting?

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