1) The article says T5 serves as a surge protector. But if the gate is connected to the bottom of RS (RS-SEL in this schematic), wouldn't the
Vgs of the JFET be 1V? Therefore, the FET would never turn on under
normal conditions, which kind of defeats the current source? I'm
obviously missing something.
They're pulling a bit of a trick here — a JFET can also function as an ordinary diode. For example, look at T9 down below. If a positive surge is applied to K2, the gate-channel junction of T5 will be forward biased, which connects it directly to the output of the opamp, a low-impedance point.
The advantage of doing this is that in normal operation, T6 has very little leakage and very little capacitance, so that it doesn't disrupt the low-current settings.
Remember that a JFET is a depletion-mode device. It will conduct current unless the gate is driven more negative than the channel (relative to either the source or drain terminal) by the specified threshold voltage. With a VGS of -1V, the FET still conducts. The drain-source resistance doesn't upset normal operation because of negative feedback, IC6 will raise its output voltage to the level required to force the desired current through RS.
2) According to the article, all of the source current from IC6 travels through R59 as long as the user-selected current is 100uA or
less. I believe I understand that because at 100uA, Vbe of T6 would be
360mV, which is less than Vbe(on) for T6, so T6 would be off. But
wouldn't R59 contribute a very large error in series with R28? The
combined resistance of R59 and R28 is 13.6k, which would result in a
current of 1V/(10k+3.6k) = 73.5uA. That's pretty far from 100uA.
No, because R59 is inside the feedback loop for IC6, the opamp automatically compensates for its effects.
3) For user-selected currents above 100uA, how does R59 and T6 affect the voltage going into the resistors downstream? Wouldn't they
contribute significant voltage drops that mess up the 1V reference
calculations? I can't figure out an intuitive feeling for how the
resistor and transistor work together here.
Again, because of negative feedback, when T6 conducts, IC6 reduces its output voltage to maintain the correct voltage across RS.
The general principle is that components between the output of the opamp and the feedback point don't matter (within certain limits), because the opamp will act to reverse their effects and maintain the desired voltage at the feedback point. It can be tricky sometimes to understand exactly where the "feedback point" is in some circuits. In this case, it is the node labeled TP6.
1 - The first thing you need to do is rethink your choice of amps. An OP37 simply will not respond in nanoseconds. Neither will an AD625. And that 100 pF feedback cap is great for stability, but it also reduces bandwidth. If you want low-voltage, high-speed op amps, it's time to move away from DIPs. And you do want high-speed amps. Your current source needs nsec response times if it is going to respond to nsec transients in load impedance.
2 - For better results, put the 100 ohm resistor to ground, and float the sensor. That way, you won't have to worry about common mode response to your sensor fluctuations.
3 - I take it as given that your control voltage is not actually +/- 15 volts.
4 - If your sensor is ESD sensitive, the last thing you want to do is to connect/disconnect the sensor while the circuit is hot. Just don't.
5 - If you wish to drive 1 mA through 650 ohms, you are going to get 0.65 volts - and that is incompatible with your 400 mV number. Even so, you should consider diodes that are not optimized for relatively large currents, such as the (1 A) diodes you're using.
EDIT - Furthermore, I'm curious about how you're measuring circuit performance. On the one hand, if you're getting a lot of noise, what bandwidth is it? Is it AC line noise? And just as importantly, what sort of scope do you have that will let you detect 1 nsec, 0.1 volt spikes, which you say can kill your detector? This is pretty sophisticated work.
Best Answer
There appears to be a couple of problems with your design:
D5 needs to be a Schottky diode (and as noted in another answer needs to be able to withstand the string voltage). I doubt that D1 is reverse avalanching since you don't report excessive heat, but you should select a more appropriate diode.
The chip draws pulses of current up to 750mA, but with L1 being only 10uH you may well be putting the chip into overcurrent shutdown. When this happens the soft start is initiated so it is likely this what is preventing you achieving the right string voltage.
Why did you select an inductor half of the value recommended in the datasheet? See page 8.
Are you aware that this changes the peak current radically?