Download Dear Gerd - USC Physics

Dear Gerd,
Please see the added notes directly under the schematic, and I hope
the schematic survives the transmission!
Typically you can expect gains of 100,000 to 3 E06 for voltage gain
from common opamps. A few types give more and very fast ones
often give less gain, but I doubt either of these cases is your case.
My first choice would be LF356, which I believe you have and which
is very common and relatively cheap and rugged. One important
question when you want to work with only one nA is the input current
of the opamp, and the LF356 needs bias current of only about 25 pA
at room temperature (It could be as high as 200 pA in the worst
case, but then you just have to make a slight correction, or look for a
CMOS amp, but I do not think you have any of those. Better is to just
take a different LF356 until you find one good enough or use a
correction circuit. Drawing will follow.) I looked up the gain of the
LF356; it is 50,000 minimum, 200,000 typical, and unspecifi ed
maximum. These are enough for your job in any case.
Don't be shocked by what you see! All is very simple except that you
have selected to work with a quite tiny current which does not make
much voltage across a 100k resistor. So part of what you see is to
make the 1nA, and other parts are messy to measure the 1/10 mV
the small current produces in the test object.
#########################
Added after phone call:
You can put the "unknown" 100 k resistor in place of the 1M which is
fed from the junction of R3 and R4. Then the voltage applied to "-10
volts" will be divided by 1000 and applied to the "ubnknown" resistor.
You can then put a 1 M or so resistor in place of "Resistor being
meaured". This resistor will then develop 1 mV per nA measured at
its junction with R1 and R2, while the opamp will then have 1000x as
much at its own output. You should retain the "Current offset
compensation".
In the lab I would select one of the chopper stabilized opamps with
low input bias current. Most of the major makers of linear ICs offer
something which is good. These will have offset voltages in the
range of a few microvolts and drift around 50 nV / K. Bias currents
should be just a few pA (I remember one from Linear Technology
with input current around 3 to 6 pA.) If you want to go to still smaller
currents life is somewhat harder because the range of devices
decreases, but as you know, the amplifiers I planned for Irvine could
work with 20 fA input and the drifts were a couple microvolts per K.
Beyond this you should consider finished, commercial products if
possible.
Resistor being
measured (100k)
" -10 volts "
R3
R4
+9 to +15 volt
10k
10
Voltage offset
Adjust
1M
20k
Try + Power but maybe
- Power
R1
100 k
10 ohm ????
1k
Current offset
compensation
-9 to -15 volt
R2
10k
10
1 Meg (10 meg if possible)
Thanks for the call: It definitely added some reduction in pressure!
########################
I have tried to select colors which you can handle but they are not
essential. The green resistors (R1 and R2) allow the output of the
opamp to be 1000 x as large as the voltage on your sample. (I am
cheating by letting 1001 and 1000 be equal.) Your DVM can easily
measure 100 mV, but it probably barely responds to 0.1 mV. 10
ohms is so small that loading by 100 k can be totally forgotten unless
you make truly precision measurements.
R3 and R4 perform the same function for the applied 1 nA current
except here you have the trouble that a 1 meg resistor needs 1 mV to
yield 1nA, so we divide the applied voltage by 1000 (a 1 volt supply at
"10 volts") to 10,000(a 10 volt supply at "10 volts"). You should
adjust something here to fit your choice of circumstances. You can
apply 10 volts as I marked and use a 0.0001 divider (100k and 1
ohm) or use a 9 volt battery and adjust R3 to match, or whatever you
want. The point here is that the simple solution would be to install a 9
Gohm resistor and run directly from your 9 volt battery. But you will
not like the effort of trying to buy a 9 Gohm resistor. 10 M ohm is
roughly the border between easy and painful, though 100M is still
easy at certain suppliers, but not all. Already 10M is somewhat
uncommon. One other issue is the sign of the voltage.&nb sp; This is
a leakage current and a bias current and I expect the negative supply
to be the correct source but there is a small chance a positive supply
will be needed. Another motivation for the " ".
Last, we have the cyan parts for current offset compensation. If you
are lucky and your LF356 does not use more than about 25 pA bias
current you could omit the these parts, including the 1M resistor
which limits the current. If you do not believe in depending on luck
and do not want to accept 20 % error such as you would get with a
worst allowed LF356, the extra parts allow you to compensate the
bias current. The 100k resistor allows the possibly required range of
currents to be spread over a larger portion of the potentiometer shaft
angles. The same goes for the 10 ohm resistor. This is marked with
both italics and question marks as the value is only a first guess. If
you can find a 10 M resistor of decent quality and you also need the
whole 200 pA bias current you might need to change the 10 ohm to a
larger value. Conversely, for a typical LF356 and / or typical bi as
currents 10 ohm is probably good enough. Finally, your power supply
voltage goes into the calculation of the value as well. (There are 10
ohm potentiometers but they are usually unpleasant things better
suited to rather coarse applications. On the other hand, the value of
the potentiometer is not at all critical.)
The 20 k ohm pot is to compensate the input voltage offset of the
LF356. You can use 22 K or 25 k as well, depending on what is
easier to use. The data sheet descibes this part. (
http://cache.national.com/ds/LF/LF155.pdf )
To set this up, apply no power to either the current compensation
input or the 1 nA supply. You might put a 10 k resistor in the place of
the unknown later to be measured and 100 ohm shorting between the
two opamp inputs (pins 2 and 3). Try adjusting the offset voltage pot
(20k) to zero the output. You can leave out the 10 k resistor and use
only the 100 ohm or even a wire, but the output will jump from full
scale with one sign to full scale with the other sign. You will get a
setting but you will hate the procedure. Now remove the 100 ohm or
short. Set the current offset compensation to again give zero out,
with maybe 1 M at the place of the unknown. Finally, insert the
inknown itself and measure the output voltage, which is directly
proportional to the 1 nA current, with a factor depending on R1 and
R2.
Perhaps I have explained it in excessive detail, but better too much.
Bob