HP E3610A Power Supply Design Logic

Thread Starter

JimmyTriceps

Joined May 11, 2022
3
I have a few grey areas when it comes to the design of this power supply.
1) Why have they referenced the op amp supply ground to the positive rail of the power supply output?
2) How does the current sense section work and why haven't they used the traditional differential amplifier configuration to sense the current?
3) Why are the opamps used to sink current rather than directly drive the pass transistors using a darlington or sziklai pair?
4)What is the purpose of anti parallel diodes at the input of the current error amplifier also diode CR6?HP Power Supply .jpghttps://neurophysics.ucsd.edu/Manua...P361XA 30W Bench Series DC Power Supplies.pdf

I have attached an image of the schematic and a link to the full datasheet.

Thanks,
Jimmy.
 

RPLaJeunesse

Joined Jul 29, 2018
213
1) Because it makes it easier when the output current sense resistor (R2) is in the + line.
2) Hard to answer not knowing the values of R27 and R34, but I'll try. Let's assume CREF is some voltage below the output terminal, and pulls U4B-6 below U4B-5 and U4B-7 Is high. As load current rises the high side of R27 rises, which pulls up R23. With enough current the input to U4B-6 goes above U4B-5 and U4B-7 swings downward. It will eventually go below U1-6 and pull Q2 base down, reducing the output.
3) Designer's choice. Makes it easier to use the CR4-CR5 steering technique to achieve CV-CC crossover.
4) Some devices get permanently unhappy with a large differential input. The clamp diodes prevent that.
5) I'm not certain but they probably help in protecting the series pass transistors. They will limit the maximum gate-drain voltage, and might be sinking some current from the gate which helps speed turnoff in a suddenly-shorted-load situation.
 

Wizard56

Joined Jun 11, 2020
1
I can add a few additional comments:
1) Why have they referenced the op amp supply ground to the positive rail of the power supply output?
This arrangement changes the control input to the voltage error amp to a current source due to Op Amp action. I.e., the Op Amp tries to keep the difference between it's + and - inputs at 0 volts forcing a constant current through R37. Why is this done? Because it keeps the noise gain of the Op Amp constant throughout the output voltage range. This makes the stabilization of the circuitry over the output voltage adjustment range easier than other topologies. It also means that the power supply circuity is essentially "floating", so that means that HP/Agilent/Keysight can use the same circuitry on the entire E361x range.
2) How does the current sense section work and why haven't they used the traditional differential amplifier configuration to sense the current?
A differential amplifier is not needed in this circuit topology because the (+) input of the op amp is at circuit ground and the voltage drop across R2 (<1v ) is well within the common mode range of the op amp.
3) Why are the opamps used to sink current rather than directly drive the pass transistors using a darlington or sziklai pair?
The op amps aren't sinking much current - the diodes are used to create an "or" gate with R5. Q2 sinks the current provided by R3. This creates a voltage drop across Q2 that is used to drive the MOSFET pass transistors. MOSFETs need very little gate current in linear mode so a darlington driver would be "overkill".
5) What is the purpose of Q4 and Q5
It is a way to keep the current balanced between the two MOSFETs. Without current balancing one of the MOSFETs would take over due to minor variations between devices at manufacture. A description of this very arrangement can be found in "The Art of Electronics" on Page 213. If you aren't aware of this book, it is a must have.

Here is a link to the datasheet for an IC that was popular in the 1970's that has a similar topology as the HP/Agilent/Keysight. You might get a better idea of what is going on and why from this simplified circuit:
https://www.mikrocontroller.net/attachment/160010/mc1466l_tiff.pdf

A more modern IC is the LT3080 from Analog Devices. The datasheet speaks to the use of a current source for the control circuitry and what problem that solves.
 
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