Limit output voltage - Page 1
Forum section: electronics / projects

author: delphes123
author_rank: Contributor
timestamp: August 21, 2023, 03:21:05 pm
content: Dear all,I have a mosfet with VDS = 48V, this mosfet is controled by PWM from 0 to 50% duty cycle in order to modulate voltage output up to 24 volts. Needed current is 1 Amps maximum on the output.I was wondering how to protect the ouput to stay below 24 Volts, in case of PWM DC goes from 50 to 100 %.Zener Diode ? TVS ?any idea ?Thanks

author: magic
author_rank: Super Contributor
timestamp: August 21, 2023, 09:29:14 pm
content: What kind of MOSFET and how is it connected?TVS is a brute force solution which may endanger the MOSFET when it triggers. However, it will also protect from short-circuit failure of the MOSFET if a fuse is added too.TVS is only effective if the output voltage is averaged somehow, not pulsed.Either the output voltage or the PWM signal could be monitored for abnormalities and then the MOSFET turned off, overriding PWM signal.Some latching mechanism may be desired, so that once it turns off it stays turned off even when output voltage drops below 24V.

author: langwadt
author_rank: Super Contributor
timestamp: August 21, 2023, 09:55:58 pm
content: 48V and 50% pwm does not give you 24V, it gives you 48V half the time, which depending on what you are driving may or may not be equivalent

author: delphes123
author_rank: Contributor
timestamp: August 22, 2023, 01:08:44 pm
content: Charge is a solenoid runing at 24 volts maximum, schematics is on the picture

author: magic
author_rank: Super Contributor
timestamp: August 22, 2023, 01:44:14 pm
content: You clearly haven't built this circuit yet because it wouldn't work at all the way you want.When PWM is low, the FET is off. So far so good.But when PWM is high, this gate driver outputs 9V or so.The FET acts as source follower and applies maybe 5V to the load (which is connected between the source and ground, I guess).All current drawn by the load from the source passes through the FET, which has over 40V across it, generating plenty of heat - another problem.Do you know the inductance and coil resistance of the solenoid?If you are lucky and the parameters are right, it may be possible to fix the circuit with little change:- ground the source- connect the solenoid between 48V and the drain- add a diode rated for full solenoid current (whatever it is) in the direction from the drain to 48VThis is essentially a buck converter, using the solenoid itself as the inductor.

author: delphes123
author_rank: Contributor
timestamp: August 22, 2023, 01:52:25 pm
content: Thanks for answerI forgot 2 points :I need to switch high side , and by connecting as you wrote this will be not the case and I also need to use N Mosfet.

author: langwadt
author_rank: Super Contributor
timestamp: August 22, 2023, 01:56:27 pm
content: Quote from: delphes123 on August 22, 2023, 01:52:25 pmthe gate driver will not work for a highside switch on 48V

author: magic
author_rank: Super Contributor
timestamp: August 22, 2023, 02:08:49 pm
content: There are high side drivers which can drive N-ch FET with source voltage far higher than VCC of the driver.Most of them only work with PWM (can't keep the FET permanently on), but that's OK.P-ch could also switch the high side easily and another option are integrated load switches, which are gate driver and either N or P channel FET in one chip.What's your reason for wanting N-ch? At any rate, I don't think there is any way to switch load on high side with N-ch and your gate driver, so the circuit must be modified one way or another.What's the current of this solenoid? If it's a low power thing, maybe a linera solution could be viable too...

author: delphes123
author_rank: Contributor
timestamp: August 22, 2023, 02:18:26 pm
content: The main reason of using N Mosfet is the place on the PCB with 100 V VDS and low RDSon but I am open to use P Channel if I can find one.Current is 1 Amps on 24 volts maximum

author: Andreas
author_rank: Super Contributor
timestamp: August 22, 2023, 02:23:22 pm
content: Quote from: delphes123 on August 21, 2023, 03:21:05 pmHello,usually the voltage is not interesting.Self heating of the solenoid easily changes current by 40%.Valves are usually controlled by current.So the typical power stage would be a PWM controlled current source.with best regardsAndreas

author: ledtester
author_rank: Super Contributor
timestamp: August 22, 2023, 02:25:31 pm
content: Quote from: delphes123 on August 22, 2023, 01:08:44 pmSome datasheets:https://ww1.microchip.com/downloads/en/DeviceDoc/MIC5019.pdfhttps://www.onsemi.com/pdf/datasheet/fdma037n08lc-d.pdfWith a 3.3V supply the MIC5019 can only achieve around a 10V gate drive. I don't see how you can use it in a high-side switch with even a 24V load supply.How about using it as a low-side switch? The FDMA037N can tolerate a V_DS of 80V.

author: delphes123
author_rank: Contributor
timestamp: August 22, 2023, 04:56:47 pm
content: New schematics with P Channel on high side, if this is ok, how I can insure by HW that I will never have more than 24 volts on the output ?

author: langwadt
author_rank: Super Contributor
timestamp: August 22, 2023, 07:04:55 pm
content: Quote from: delphes123 on August 22, 2023, 04:56:47 pmwith those resistors you are exceeding the maximum allowed gate-source voltagelimiting the duty cycle to less than 50% is not so simple, you'd have to average the pwm, compare to a maximum and you that to disableI'd expect that with out a flyback diode that mosfet is quickly going to pop

author: delphes123
author_rank: Contributor
timestamp: August 22, 2023, 07:53:56 pm
content: Hi,Vg = (7 / 17) * 48 = 19,76 Volts And Vgs of this mosfet is +- 25 V could you explain why this is wrong please ?Thanks

author: langwadt
author_rank: Super Contributor
timestamp: August 23, 2023, 06:17:03 am
content: Quote from: delphes123 on August 22, 2023, 07:53:56 pmgate-source is the over the 10k resistorVg = (10 / 17) * 48 = 28.24V

author: PCB.Wiz
author_rank: Super Contributor
timestamp: August 23, 2023, 07:51:24 am
content: Quote from: delphes123 on August 21, 2023, 03:21:05 pmHow likely is that ?There are many ways :Simple brute-force* An active clamp, tho that generate many watts and you need to protect the P-FET* A crowbar, that triggers on overvoltage, you also need to protect the P-FETmore finesse, takes the feedback upstream to the driver* You could use a Voltage Monitor device that turns off the driver, which avoids P-FET stressIn most cases this would hiccup on overdrive, as they have small hysteresis.It would need to be fed from a low pass filter, to give a low ripple average of the PWM drive.* You could use a monostable and gate, so that no matter what the PWM > 50%, it cannot output more than a defined ON time.Quote from: delphes123 on August 22, 2023, 01:08:44 pmIf you have a solenoid load, you will need flyback diodes.

author: Terry Bites
author_rank: Super Contributor
timestamp: August 23, 2023, 09:58:30 am
content: A PWM symbol is not a circuit description. Just as the word dog is not a dog and designing a practical proper PWM controller requires a fair bit of ee knowhow.You need to demodulate the pwm waveform applied to the load to obtain a useful control signal.A simple way would be to use an RC filter to create a running average.Then take that average and compare it to a reference value.If the average is above the reference, use the result to disable or reduce the PWM Drive. Easy....Here’s an illustrative circuit providing "bang-bang" control. You have to tweak the filter time constants and possibly the comparator properties to suit your exact needs.It can't respond to an over voltage instananeously becase the of the finite averaging time, here a few mS.You'd likely want to be careful with the control logic to prevent starting the load on 100%.There is probably a way of achieving all these functions with a dedicated PWM controller. TL494 etc.

author: delphes123
author_rank: Contributor
timestamp: August 23, 2023, 10:40:24 am
content: Many thanks I will try to implement this solution.

author: Terry Bites
author_rank: Super Contributor
timestamp: August 23, 2023, 04:22:09 pm
content: Thats untested of course. I can offer you some rough and ready simulation of what to expect.You have to compromise between ripple amplitude and the response time here.The comparator needs to have a wider hysteresis band than the maximum ripple.As you can see, the enable line is set to go low at a duty cycle of 51%, 24.8V.It will stay low if the switch fails short.For simplcity's sake

author: langwadt
author_rank: Super Contributor
timestamp: August 23, 2023, 07:15:12 pm
content: something maybe a bit too clever

author: delphes123
author_rank: Contributor
timestamp: August 23, 2023, 07:23:47 pm
content: @Terry Bites : Thanks for the simulation@langwadt : Thank you . Very interesting approach without monitoring output, could you explain briefly ?Rgds

author: langwadt
author_rank: Super Contributor
timestamp: August 23, 2023, 07:47:43 pm
content: Quote from: delphes123 on August 23, 2023, 07:23:47 pmit monitors the pwm instead, when it averages more than ~50% the comparator output goes high and M1 can't turn on