Using a low voltage opamp for feedback with high voltage mosfet? - Page 1
Forum section: electronics / beginners

author: new299
author_rank: Regular Contributor
timestamp: June 17, 2015, 10:56:14 pm
content: I've been looking at Piezo actuator drivers recently. They amplify a 10V input signal to a 100v output, and generally source ~60mA, with bandwidth up to a few KHz.The drivers I've seen mostly use high voltage opamps like the Apex PA41 opamps which are relatively expensive parts.I was wondering if it's possible to use a power MOSFET with a lower voltage opamp for feedbackThe drivers I've seen mostly use high voltage opamps on the output (Apex PA41s for example). However those opamps are very expensive. Is it possible to use a low voltage opamp to provide the feedback to a high voltage MOSFET something like the following schematic?Perhaps using something like a IRFI640G on the output? Or is there a reason to go with an integrated solution?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 18, 2015, 06:43:13 am
content: Exactly as shown, you have a source follower (made ideal by the op-amp). The voltage gain is exactly 1, so the op-amp needs to support an equal voltage range, either by design or by providing floating supplies.It is better to use common-source amplifiers so the op-amps can be supply referenced, and only need to provide gate voltage. Multiple stages of level translation may be necessary to go from a ground-referenced signal input to positive and negative rails.Tim

author: Zero999
author_rank: Super Contributor
timestamp: June 18, 2015, 07:39:19 am
content: The trouble is because the gain is 1, then the op-amp would also need to be powered from 100V to work.It's possible to add a buffer to the output stage of an op-amp and include it in the feedback loop but it needs to have some gain to increase the voltage, although it can't have too much gain, otherwise it will oscillate.There are several ways to do this, see the articles linked below:http://www.ti.com/lit/an/snoa600b/snoa600b.pdfhttp://cds.linear.com/docs/en/application-note/an18f.pdfhttp://www.ka-electronics.com/images/pdf/Op_Amp_Booster_Stages_pt1.pdfhttp://www.analog.com/media/en/technical-documentation/application-notes/28080533AN106.pdf

author: new299
author_rank: Regular Contributor
timestamp: June 18, 2015, 09:45:53 am
content: Thanks! The links are very interesting.I guess I'm confused as to why the gain is 1. If the feedback resistor is sized to give a x100 gain what prevents this? (to clarify my understanding).

author: mikerj
author_rank: Super Contributor
timestamp: June 18, 2015, 10:43:42 am
content: The voltage gain of the MOSFET is 1, because it's used as a source follower. The gain of the op-amp is irrelevant, it still needs to provide 100v+Vgs to the MOSFET gate to get 100V output of the MOSFET source.

author: new299
author_rank: Regular Contributor
timestamp: June 18, 2015, 01:49:23 pm
content: Quote from: mikerj on June 18, 2015, 10:43:42 amAh right I see, because the current gain is high, but the voltage gain of a MOSFET is 1?Perhaps I need to review my understanding on MOSFETs. If I put 1V in on the gate, does this mean the MOSFET will only ever pass 1V?

author: PSR B1257
author_rank: Frequent Contributor
timestamp: June 18, 2015, 03:14:17 pm
content: It has nothing to do with the MOSFET.Your so called feedback resistor is simply useless (since there is no current through it). Vout is therefore basicly directly tied to the opamp, therefore the Vout=VinThe suggested common surce amplifier however is not that trivial hence you have to sence a voltagedifference referenced to the positive supply.U1, U3 and U4 can also be replaced by an integrated instrumentation amplifier like the INA128.

author: mikerj
author_rank: Super Contributor
timestamp: June 18, 2015, 03:41:58 pm
content: Quote from: PSR B1257 on June 18, 2015, 03:14:17 pmYes the feedback resistor is incorrect, but this doesn't fix the fundamental problem. The op-amp output has to rise above 100v to give a 100v output because the MOSFET is configured as a source follower (which has a gain of 1). Fixing the feedback resistors simply means Vin could be lower for a given output voltage.If the op wants want to use a low voltage op-amp then his choices are:1) Use an N channel FET on the low side of the load2) Use P channel FET on the high side, with an additional transistor to ground reference the drive voltage.3) Use P channel FET on the high side, and float the op-amp to the supply voltage, requiring additional components to ground reference the input voltage.

author: PSR B1257
author_rank: Frequent Contributor
timestamp: June 18, 2015, 04:40:36 pm
content: QuoteN-channel high side works also, with an additional transistor as a level shifter.This circuit looks basicly like this.How good it (as well as my previous circuit) will perform atseveralkHz remains to be seen. This one however is a bit simpler.

author: Zero999
author_rank: Super Contributor
timestamp: June 18, 2015, 08:46:36 pm
content: These are all good suggestions but aren't piezo actuators suppose to be driven from a push-pull amplifier?The original schematic which uses a resistor to represent the load is incorrect. A piezo actuator is capacitive.

author: PSR B1257
author_rank: Frequent Contributor
timestamp: June 20, 2015, 02:53:16 pm
content: You're right, I didn't thought of that. In this case it requires in deed a push pull output stage in order to quickly discharge the capacitance. Well, that doesn't makes it easier, does itBlessed is he who has a Keithley 2400