PWM boostrap issue - Page 1
Forum section: electronics / beginners

author: step_s
author_rank: Regular Contributor
timestamp: March 28, 2016, 12:22:17 am
content: Hi EEVblog.I'm current playing around with some PWM circuits and wanted to try out something for driving an N channel mosfet on the high-side.After playing around a while I came up with a simple design, and in theory it should work, but it doesn't.As you can see from the circuit, the two complimentary mosfets drive the gate of the last N channel sitting on the high side.What is strange is that the gate of the high side N channel is oscillating from +7.5V to 0V (GATE on the circuit), and the mosfet turns on fully, but when driven low to 0V, it doesn't turn off fully, even though the Source-Gate voltage is lower than 0V and the N channel should be shut-off.The source-gnd voltage seems to be sitting at 2.5V.I have included the oscilloscope readings from the setup. The yellow is from the marker "GATE" and the green is from the marker "SW".Anyone know what is going on?

author: station240
author_rank: Unknown
timestamp: March 28, 2016, 12:40:56 am
content: Tried using a higher value for R1 ?Seems logical that could be holding the voltage high.I think you should have a low value resistor on the gate of Q4 eg 1-10ohm.

author: step_s
author_rank: Regular Contributor
timestamp: March 28, 2016, 12:47:57 am
content: The 300ohms resistor is giving a decently square wave, and as you can see the gate wave is behaving like it should, so that's why I'm confused.The whole idea of this circuit was to prevent the use of low value pull-up and pull-down resistors to prevent powerlosses.Will driving the gate of the last N channel mosfet below it's source, make it behave strangely?

author: amspire
author_rank: Super Contributor
timestamp: March 28, 2016, 01:00:06 am
content: The traces do not make sense if they are connected the way you described. It would make complete sense if the yellow trace was the gates of Q1 and Q2 instead of the gate of Q4.A bit of an odd circuit, but excluding the ringing (that may be just a ground loop in the oscilloscope probes), it looks like it is behaving the way the design is expected to behave.The reason that the output voltage only goes to 2.5V when Q4 is off, is that C1 has about 2.5V DC across it and current can flow through D1 and C1 to R2. When Q4 is off, the voltage on R2 is (5V - D1-voltage-drop) - C1 voltage. That means the R2 voltage is about 5V-2.5V = 2.5V if D1 drops are ignored.Q4 is definitely completely off when the output is 2.5V.Richard

author: amspire
author_rank: Super Contributor
timestamp: March 28, 2016, 01:25:50 am
content: Quote from: station240 on March 28, 2016, 12:40:56 amDefinitely if R1 can be much bigger (like several kOhms) or R2 can be much smaller, you will get closer to 5V across C1, closer to 0V on the SW output when Q4 is off and the yellow trace will peak closer to 10V.You didn't mention the value of C1, but from the droop in the green trace, I am going to guess at 22uF.

author: step_s
author_rank: Regular Contributor
timestamp: March 28, 2016, 02:14:29 am
content: Thank you for your insight Richard.I just looked at it myself and realized that the cap is being discharged to 2.5V and is trying to charge itself when Q4 is off. Since this is through a resistor, the voltage will be 2.5V across the cap and 2.5V across the resistor in series.I figure this was what you were saying as well?Also there isn't a lot of shoot throuhh in Q 1 and Q2 but the little there is will discharge C1 as well.The frequency is 500kHz so making the R1 bigger, will cause the wave to slope, since the pull-up of Q1 needs to happen quite fast.C1 is 0.1uFI guess I will need to find a way to prevent the cap from discharging through R1 and Q3. . .

author: amspire
author_rank: Super Contributor
timestamp: March 28, 2016, 02:29:49 am
content: Quote from: step_s on March 28, 2016, 02:14:29 amNot 22uF... That is what happens when you try and do quick mental arithmetic.Just pulled out the calculator - 0.1uF gives a charging dv/dt of about 0.5V/uS. So that means you should get about a 0.5V droop - sounds right.

author: step_s
author_rank: Regular Contributor
timestamp: March 28, 2016, 02:37:12 am
content: Under what circumstance does it discharge 0.5V?Through the 300ohm resistor?

author: amspire
author_rank: Super Contributor
timestamp: March 28, 2016, 03:00:32 am
content: Yes, it is discharging through the 300 ohm resistor, but it does have 7.5 volts across it, where as the 50 ohms only has 2.5 volts when charging.I am sure if you properly analysed all the voltage drops around the circuit and the real ON-OFF ratio (the D4 mosfet turn off is much slower then the turn on), it would all make sense.This is one of those clever minimalistic circuits that can end up taking ages to tweak till it is just right. Sometimes it is never "just right". The simplistic way is to use a MAX660 to turn the 5V to 10V.

author: step_s
author_rank: Regular Contributor
timestamp: March 28, 2016, 03:21:42 am
content: Ye true. But there's a chip for everything these daysSomestimes it's just good fun to play around with some things.I have tried this with a mosfet driver, but that's just too easy!The mosfets are quite fast, and can make very nice square waves, when connected to a mosfet driver, but just wanted the challenge this timeIf what you have calculated with the 0.5V drop, then that would make it 9.5V, and the last 2V must be lost to shoot through?

author: amspire
author_rank: Super Contributor
timestamp: March 28, 2016, 03:51:15 am
content: I will leave it to you.I still have nightmares from circuits like this in production. After a week on the bench, the designers tweak it with the tongue at just the right angle and it works perfectly in the prototypes. Then it goes to production and with the real world parts variation, it is inadequate. Some poor bugger then has to find a solution within hours and there is no room to move. Lower the OFF voltage but slow down the rise times. Or replace the 50 ohm resistor with a 5 ohm resistor and waste 2.5W of power on every board just to get the production line going again.