Speeding up High side switch turn-off time - Page 1
Forum section: electronics / beginners

author: mrkev
author_rank: Regular Contributor
timestamp: March 27, 2017, 07:50:07 am
content: Hi,I am working on device that thas at one of it's outputs high side mosfet switch with SI2307 switching about 20V. I had to speed up the turn off time as I wanted to use PWM modulation. Gate capacitance of this mosfet is about 500pF which, together with 4k7 resistor, gives time constant about 2,4us. However, turn off is significantly longer since you have to discharge the gate cap from about 10V to 1,5V (and that is multiple of time constant).In my effort, I came up with this (second picture), this actually works great, the turn off time dropped to about 100ns,BUT when I wanted to use this with unfiltered and not stabilised VCC1, I realise that when you have >3V fast voltage peak on VCC1, it can (and I tested that it will) turn on T9 momenteraly. Does anyone have any ideas on how to supress this behavior or how to remake this?

author: langwadt
author_rank: Super Contributor
timestamp: March 27, 2017, 08:28:13 am
content: without thinking too much.remove r20, replace c14 with resistor, resistor from gate of T9 to vvc1

author: bktemp
author_rank: Super Contributor
timestamp: March 27, 2017, 08:37:02 am
content: Quote from: langwadt on March 27, 2017, 08:28:13 amThis.And using a NPN transistor for T9 is probably a good idea, otherwise T9 will discharge the gate only to ~3V because of its gate-source voltage requirement. A BJT it will stay on until the gate has been discharged to around 0.7V.By the way: The mosfet symbol looks wrong:https://en.wikipedia.org/wiki/MOSFET#Circuit_symbolsBetter put the body diode symbol into the middle of the symbol and add a seperate line for source.

author: Zero999
author_rank: Super Contributor
timestamp: March 27, 2017, 09:36:43 am
content: A PNP transistor to turn off the MOSFET would be ideal.

author: mrkev
author_rank: Regular Contributor
timestamp: March 27, 2017, 09:59:39 am
content: Quote from: langwadt on March 27, 2017, 08:28:13 amT9 and T10 can't be both opened at the same time.Quote from: bktemp on March 27, 2017, 08:37:02 amHi, thx. It doesn't really matter if I use NPN or MOSFET (of same polarity) drop voltage on FET is probably even lower. Mosfet symbol is correct, it even is in your link (4th column) and I am not the one who made it...Quote from: Hero999 on March 27, 2017, 09:36:43 amYeah, I tried this, but it has the very same properties as my design. When you aply some voltage peak on V1, the Q1 will be momenteraly opened (throught C1) and that is gonna quickly close M1. Even worse, this is susceptible to peaks above 0,5V; mine will resist up to 2V...

author: Zero999
author_rank: Super Contributor
timestamp: March 27, 2017, 10:27:58 am
content: Quote from: mrkev on March 27, 2017, 09:59:39 amNo, if anything, voltage pulses on the +V line will cause M1 to open (turn off).

author: Ian.M
author_rank: Super Contributor
timestamp: March 27, 2017, 11:06:26 am
content: You need a cap-free speedup circuit. The best option is a complimentary BJT emitter follower buffer stage that can slam lots of current into the MOSFET gate.LTSPICE sim attached, including original (slow) circuit for comparison.Torture test it by making the Vcc sourcePULSE(22 28 0 100n 100n 5u 50u)

author: langwadt
author_rank: Super Contributor
timestamp: March 27, 2017, 11:10:42 am
content: Quote from: mrkev on March 27, 2017, 09:59:39 amit doesn't seem like you actually read any of what was replied...

author: snarkysparky
author_rank: Frequent Contributor
timestamp: March 27, 2017, 11:37:04 am
content: also a Schottky diode between base and collector works wonders to speed up BJT turnoff by preventing saturation. I just did this to my mosfet drive circuit and now I can get reasonable square 1us pulses on the gate.Google "Baker Clamp"

author: mrkev
author_rank: Regular Contributor
timestamp: March 27, 2017, 11:42:51 am
content: Quote from: langwadt on March 27, 2017, 11:10:42 amHow so? Did YOU read those replies?Quote from: Ian.M on March 27, 2017, 11:06:26 amThank you very much, I was looking for something like that.I don't know why, but I just couldn't figure out how to build that driver.

author: Ian.M
author_rank: Super Contributor
timestamp: March 27, 2017, 12:22:20 pm
content: Quote from: mrkev on March 27, 2017, 11:42:51 amWell the first thing to do is to discard any prejudice you have against BJTs . . .Next, ask yourself why are you using discretes instead of a dedicated gate driver IC!

author: Zero999
author_rank: Super Contributor
timestamp: March 27, 2017, 01:00:10 pm
content: Quote from: Ian.M on March 27, 2017, 11:06:26 amI notice you have a 24V supply on that circuit which is above the maximum rating for the MOSFET. As drawn the circuit will apply the full 24V to the gate which is only rated to 20V.

author: Ian.M
author_rank: Super Contributor
timestamp: March 27, 2017, 01:12:14 pm
content: No it wont. R1 and R3 form a potential divider, limiting the max Vgs to a bit under 12V

author: Zero999
author_rank: Super Contributor
timestamp: March 27, 2017, 01:13:31 pm
content: Quote from: Ian.M on March 27, 2017, 01:12:14 pmOh yes, sorry missed that.Any way, here's another way of doing it.

author: mrkev
author_rank: Regular Contributor
timestamp: March 27, 2017, 01:51:14 pm
content: Hey, thanks to everybody. I am not really a beginner, it is just so basic stuff, that I didn't want to post elsewhere. And I honestly needed just a good kick... The circuit I came up with after Hero999's post and that I am gonna use is this:The edges look pretty nice (delay for falling edge is because I am feeding signal from function gen. TTL output and it has problems with discharging M2 at those low voltages):Quote from: Ian.M on March 27, 2017, 12:22:20 pmI dont have any prejudice against BJTs, that T9 in the first schematic should really be FETAnd using of dedicated gate drivers? I would love to, but I can't, and hey, it's not that hard.

author: bktemp
author_rank: Super Contributor
timestamp: March 27, 2017, 01:56:39 pm
content: Quote from: mrkev on March 27, 2017, 01:51:14 pmThat is exactly the circuit langwadt descriped in the first answer...

author: mrkev
author_rank: Regular Contributor
timestamp: March 27, 2017, 02:26:17 pm
content: Quote from: bktemp on March 27, 2017, 01:56:39 pmYup, I see that now. It didn't make sense with that MOSFET so I somehow disregarded whole answer... Sorry.

author: Ian.M
author_rank: Super Contributor
timestamp: March 28, 2017, 02:03:37 pm
content: A combination of Hero999's circuit and my push-pull driver. It can theoretically do 50V/us slew rate at the MOSFET gate.Turnoff is speeded up by a bootstrap cap to the top end of the pullup resistor, and turnon is speeded up by a RC network in the emitter circuit of Q3, the inverting level shifter. Both have been tuned for optimum response, with minimal overshoot. Q3 acts as a switched current source, and R1 converts that current into the desired gate drive voltage. I don't think you can do much better with 'jellybean' transistors without a far more complex circuit.

author: JanP
author_rank: Newbie
timestamp: August 03, 2023, 10:38:41 am
content: I built this circuit to speed up high side switch turn-off time, based on the post of Ian M on March 27, 2017:It works fine with a switch-off delay of 1 us (which is ok-ish), but only after I lowered R3 and R4 from 10K to 1K:(Yellow: input 0 - 5V pulse on J1-1, Green: Collector of Q1, Magenta: Drain of Q4)With 10K values the switch-off delay measured was a whopping 10us!Can someone explain why R3 and R4 need to be so low? Thanks.

author: mikerj
author_rank: Super Contributor
timestamp: August 03, 2023, 11:45:29 am
content: You chosen power MOSFET has many times higher gate charge than the OP's MOSFET, so the BC33x driver transistors are having to carry higher peak currents and therefore more base drive current is needed.

author: Ian.M
author_rank: Super Contributor
timestamp: August 03, 2023, 03:33:12 pm
content: Decrease R4 even further - your MOSFET has characteristic curves for -10V Vgs, and Abs. Max. Vgs of +5V/-16V. As-is you cant get more than -7V Vgs. Try 470 ohms which should give 10V swing at the complimentary emitter follower bases and >9V gate drive

author: Zero999
author_rank: Super Contributor
timestamp: August 03, 2023, 07:17:56 pm
content: Perhaps an extra transistor and some speed-up capacitors?

author: JanP
author_rank: Newbie
timestamp: August 04, 2023, 11:22:51 am
content: Thank you all for your help, much appreciated!I brought down R4 to 470E, which reduced the switch-off delay to about 700 ns:Ch 1 yellow = input on J1 -1, Ch 3 magenta = Drain Q4, Ch 4 green = bases of Q3 & Q4I coincidentally found that disconnecting probe 4 (green) reduced the switch-off even further to about 550 ns, but which an undershoot of -12V:Apparently the probe capacitance has a big influence!Adding the C2 of 100 pF across R1 as in schematics of Zero999 caused by far the largest improvement; delay of just 200 ns but with an undershoot of -15V!I don't understand where the undershoot is coming from; the inductance of the wirewound load, resistor R5?To reduce the undershoot I added a 100 pF from the bases of Q3 & Q4 to ground, now 'just' 5V undershoot for a slightly increased delay of 350 ns.But this is somewhat silly I think, adding a capacitor slows things down again ...So I removed it and added a flyback diode 1N4007 across the load instead. This removed the ringing all together, but still an undershoot of -7V!How can this be?What is the best way to deal with this undershoot?@ Zero999 : As the switch-on and switch-off delay of about 200 ns is now ok for my purpose, I will probably not build your more elaborate solution.But for my understanding, is the following reasoning correct:the Schottky diode was removed because of R1 = 100K instead of my 1K8 doesn't drive Q1 in saturation and C2 speeds up switching times considerably.The 10K from base Q1 to ground was also removed. I always thought it is good practice to add a resistor from base to ground in order not to have a floating base (when no signal is applied), not?The sole purpose of Q4 is to speed up switch-on time, as it bypasses R3 to provide a larger base current for Q2 to supply charge to the gate.D1 is needed to discharge C1.PS: I don't understand why my pictures don't show up inline as yours do, I did read the instructions but it doesn't seem to work.

author: ledtester
author_rank: Super Contributor
timestamp: August 04, 2023, 03:31:18 pm
content: Quote from: JanP on August 04, 2023, 11:22:51 amThe inline option doesn't work on this site. To get inline images you need to edit your message after posting it and insert img tags to the images - use the URL from the download link.