Hi all, I had a bit of time this afternoon and I tried this,
Where R1 was 100 ohms and the Potentiometer was also 100 ohms. I looked at the step response and was able to 'tune out' the over shoot by pulling the pot wiper off the top point.
Thanks for all your help and interest,
"mischief managed" George H.
I forgot to note that the U1 gain is -100 not +100 Does that make more sense?
The drain of the FET goes down, the output of U1 goes up, the source of FET goes up making it negative feedback.
[...]When you get up in frequency, the phase of what you feedback starts to matter hugely. Just a little error makes for a lot of reduction of the impedance of the thing.
George Herold writes
I am looking at a biased photodiode which is, I suppose, very high impedance indeed. It strikes me that my topology is fundamentaly different to yours - I am using two different coaxes, and driving the shields of both with images of the signals on their cores (which is why I use a dual op amp). That's why I assumed there is no capacitance from inner core to outer shield, because I have two independent shields. I then wrap a metal mesh round the pair of them to act as an RFI screen, I'll connect that to 0V. This is all try-it-and-see stuff, I can't say it works yet! Hope to improve leakage (mainly), noise (hopefully) and see if I can push bandwidth up to 1MHz.
The points about phase inversion etc is very interesting. Eeek, hadn't thought of that! This is going to be fun 8)
-- Nemo
t
Hmm, I'm not quite sure. I put thin in my notebook and if I ever need it I hope I can find it. I re-drew your circuit to look like this,
--------------- ! ! [R] |\ V] +------- \ ! !---+ GND+ >----+--+---- in ---->! | / ! !---+--[R]------------- ! R2 [R] ! R1
(Sorry a bit hard to squeeze the ground in there.)
But I'm not sure that's what you intended.
Yeah that's what I found, If I tried to 'tune out' too much capacitance I got gain peaking at high frequency. My solution is a pot on the output to adjust the 'feedback'. I still have to see how it works on a noise signal.
George H.
A photodiode is a whole different ball of wax. You can look at the current and run it into the inverting 'virtual ground' input of an opamp. The impdedance of a photodiode changes with the amount of current its generating. It's kT/e (25mV @RT) divided by the current. (I think? I'm sure someone will correct me if I'm wrong.) Since it's a current source you have to put the impedance in parallel with it. So at higher currents the impedance goes down.... which is not what you want for a current source.
I was thinking I could also look at the current noise from a resistor (put it into the inverting input.) But then how to do the driven shield? Can I put TIA's in series? like this,
|\ i--->--- \ | >--- to shield +--+ / | |/ | | |\ +--- \ | >--- to signal GND--+ / |/
The current noises may add, but with fets and 1Meg ohm you hardly care.
George H.
n
Opps, I forgot the feedback resistors there....
Hmm that doesn't look like it will work... Sorry I should think more before posting.
George H.
Big sinps...
That is total Crap! I forgot how to make a current source.
OK that is better.
(Sorry) George H.
The circuit above is the right one. Try spicing it up and you will see why. Use an E for the amplifier.
The drain of the FET sees a low impedance because of the inverting op-amp. This improves the bandwidth just like a cascode circuit does.
R1 and R2 provide the negative feedback to the FET's source The FET is running at IDSS where the gm is big and the noise low.
Darn, the 'blind pig' finds an acron again. (I'm the blind pig)
Thanks MooseFET I'll try spicing it at home.
What's an "E"? "Use an E for the amplifier."
George H.
Note the added capacitor. It isn't a change to the topology. Ideally it would be infinite or nearly so
[C]
If you are using LTSpice, go into the place where you get the Voltages and currents etc and look for the "E" part.
An E is a voltage controlled voltage source. It is like a perfect amplifier.
The E in this case can be set to have a gain of 1 for your first run and then 10 and then 100. As you crank the gain up, you should see the operation depend less and less on the details of the JFET.
In real life, the gain of -100 is made with an op-amp. The feedback local to the op-amp is a series / parallel.
---+-----[R]---[C]----+--- ! ! -------[C]--------
At DC the gain is near infinite
At some place where the JFET starts to roll things off, the zero from the resistor kicks in and keeps the phase reasonable.
Near the 0dB gain point of the op-amp, the extra capacitor and the drain node of the JFET form a divider that makes the op-amp happy about the frequency of gain crossover.
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