Low impedance driver for high voltage (500V) MOSFET  - Page 1
Forum section: electronics / projects

author: Merlyn
author_rank: Contributor
timestamp: December 17, 2016, 10:02:03 pm
content: Hello,To make the long story short what you see on the pic below is a module that controls DC voltage in the range of 25-500V by applying a 0-5V at the opamp input VCS. The high voltage (450-500V) is applied at HT2 and the controlled voltage appears at the MOSFET source (VS).It works as it is but some suggestions were made that it will be better for the MOSFET to be driven from a low impedance source.My questions is what is the easiest way to do that? Will a simple source follower do the job or a totem pole is needed or anything else?My transistor knowledge is basic so I would also appreciate if someone can explain how the whole thing actually works.

author: PartialDischarge
author_rank: Super Contributor
timestamp: December 17, 2016, 10:31:00 pm
content: Your circuit works thanks to a current mirror and a x100 multiplication factor. the current in tr19 is (5-vcs)/3k.That current is mirrored and a voltage of HT2-Itr19x300k appears at the gate of the mosfet. tr15 and 17 are needed to lower the voltage seen by tr13Now, Id use a schematic with feedback, yours is open loop. Get the book the art of electronics, theres a Power Supply similar to this one that uses feedback

author: Merlyn
author_rank: Contributor
timestamp: December 18, 2016, 08:08:37 am
content: Thanks for the replies.evb149: The MOSFET is always ON and it doesn't need to respond to voltage changes fast. Since the voltage that appears on the gate is in the same range as the output voltage I don't think a 40V SMD device would do the job. I was checking for those but I didn't find any that can take 500V or more. Maybe a single ended source follower is the the easiest solution provided speed of response is not important?MasterTech: Thanks for the explanation. From what I understand the voltage that will appear on MOSFET gate of depends on the R72 (3k) resistor value. Currently there is a 10V voltage difference between what is on HT2 and the output. Is there a way to lower that dofference provided that the max current that goes through the MOSFET is around 100mA?This circuit is actually controlled by a DAC and control voltage (VS100) is fed to a uCU so I guess this is the "feedback" here.

author: PartialDischarge
author_rank: Super Contributor
timestamp: December 18, 2016, 08:23:53 am
content: Yes, R72 controls the current of the mirror. The 10V difference probably comes from the Vgs at the output mosfet, but feedback should take care of that.Also transistor TR13 is rated as 1W, and it seems low for the Vce and current it is handling.Search in mouser.com, they have mosfets up to 4500V

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 18, 2016, 10:31:52 am
content: How do you know youneedor evenwantto have high speed?I don't see any current limiting either, which is a red flag.Tim

author: Kleinstein
author_rank: Super Contributor
timestamp: December 18, 2016, 12:39:45 pm
content: There is no problem with the power of TR13: the current is limited to about 1.5 mA (3 K and 5 V). The voltage at TR13 should not be higher than about 250 V due to the zener diodes. So I won't expect more than 400 mW. Still quite some power for such a small case.The circuit is operating in a slow linear mode, not fast switching. So unless really fast reaction is needed, there is no need for an extra low impedance gate driver. The extra capacitance near the gate suggest a rather slow reaction anyway. Having current limiting might be a good idea.The 10 V difference could be the voltage to drive the gate. So there is no easy way to reduce this (using a different circuit with a P MOSFET might do). 100 mA at 500 V is quite a lot of power for the MOSFET. If such a high current is also needed at low output voltages, one should consider even 2 power MOSFETs in series to ease on the SOA requirements. However this would increase the minimum drop a little.

author: Merlyn
author_rank: Contributor
timestamp: December 18, 2016, 01:54:44 pm
content: QuoteYes, no switching speeds are required so besides that a low impedance won't make the gate "feel better" or have any other benefits?If for the sake of the experiment I want to try a source follower how will this go? 100 Ohms from R53/C37 to MOSFET gate, drain to HT2 and 47-100k in the source then R67 to source?QuoteThe current at idle is around 20mA. Max current is 100mA. Lowering the voltage lowers the current consumed as well. The MOSFET is mounted on a heatsink and even at extreme conditions doesn't produce so much heat.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 18, 2016, 04:20:28 pm
content: One could use a second MOSFET as a source follower, however would increase (e.g. double) the minimum voltage drop. The large value of C37 looks like the circuit is intentionally slow, and at higher frequency the impedance at the gate is thus already quite low.Power consumption can be critical because of a limited SOA. The noted MOSFET is a rather modern type so may not be that well suited for power dissipation at higher voltage. The DS I found does not give details. Usually beyond a certain voltage, the permitted power goes down (e.g. proportional to voltage or faster). So at something like 400 V drain to source the safe power limit may be considerably lower than the quoted 160 W - maybe only 40 or 20 W. The limit is not the chip / case temperature, but current localization, similar to second break-down. As this is a rather large type it could well be OK, especially if the current goes down at lower voltage.What is the purpose / context of the circuit.

author: Merlyn
author_rank: Contributor
timestamp: December 18, 2016, 04:48:07 pm
content: The circuit is used for controlling the screen voltage in a vacuum tube guitar amplifier. It's taken from a well known manufacturer's amp schematic (less the bias module). By lowering the screen voltage simultaneously with the negative bias voltage it's possible to overdrive the power amplifier at lower levels. There are different ways of doing that and this is just one of them.Some have suggested that switching MOSFETs are not very suitable for this application and that ones with "larger die" should be used like FQA8N100C for example. It looks like this is confirmed here. That makes me wonder how to discriminate which MOSFETs have "larger die" than others?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 18, 2016, 08:08:27 pm
content: Quote from: Merlyn on December 18, 2016, 04:48:07 pmPower dissipation, roughly.Or any other parameter that scales, given technology type. All the new HV MOSFETs are SuperJunction type, which gives much lower Rds(on) per die area than earlier tech. But within those types, they will all be made for roughly similar Ciss/Rds(on) ratio, so that higher amp ratings have less resistance, more gate charge, and more power dissipation.Crss is characteristic of SuperJunction types, being large at low drain voltages (~nF), dropping sharply around 20Vds to very low levels (~pF), and usually with a very small rebound at high voltages (>200V).Tim