Eliminating Optocoupler Turn-On Spike

="PDDD, post: 1241526, member: 259786"]

Thank you, Ron! Now I see why you have so many likes! Actually, I'm glad you mentioned 300 to 400 Volt H-bridges. Are you using enhancement mode MOSFETS or something else? I wanted to run higher VDD, but I can never get more than the Gate voltage through the MOSFET!

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The circuit Ron posted is called a bootstrap circuit - you can google it. What happens is the cap gets charged up to about 11.4 volts either thru your load or the bottom FET when it is on and the diode. That voltage gets "added" to the voltage on the source so the gate will be 11.4 volts higher than the source at all times. The cap must be large enough that it does not get discharged very much when operating at the lowest frequency - in your case 8 ms. If you need the top FET to be able to stay on forever the circuit won't work. I suspect that is why Ron ask you if your load was a motor.

The opto gate driver spikes to VDD for 5 to 15uS at power up and power down, in the same way a monostable multivibrator does if the reset is not held low during power up. So turning off the opto causes the same problem as turning it on.

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The opto won't create a pulse on power up. It has a circuit that keeps it's outputs safe until the voltage is 13.5 volts. Since we don't know what the whole circuit looks like it is possible that the 30 volts is on and the opto is being told to turn on by the circuit that drives it. Running the opto and the FET at their maximum rating is not a real good idea anyway, so maybe if you have 15 volts somewhere you can use that and power your other circuit from that as well.
Having said all that if there is no voltage on the drain unless the 30 volts is blowing them something else is wrong.

Bottom half:
This MOSFET pulls down very well. Does not pull up at all. The G-S voltage should be in the 10 to 15V range. 20 absolute max.
The Drain voltage can be very high (if the MOSFET is rated for it). The voltage can be almost anything and is not effected by what the +12V is.

The MOSFET turns on when the G to S voltage is more than 3V. (depends on the part).
In this case the gate voltage (compared to ground) goes from 0 to 30V.
So the MOSFET-S will go up to about 27 volts and no more. If the S gets any higher the transistor turns off.
The D could be at 300V but the S is at 27.
This is not the way!

This is the way to drive the top side.
Note the 12V battery.
The isolator/driver is connected G-S and applies 0 or 12V on the G-S. It is not referenced or connected to ground in any why.
When on the driver applies 12V from G to S. The D=300, The S=about 300,
The gate is at 312 volts. The battery is connected to S (output).

Beautiful!! That opens up a whole new world of possibilities.! Thanks, Ron!

What is making your digital signals? micro processor?
What supply voltage? 30V?

PDDD said:

since mosfet gates do not pass current.

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The gate capacitance charging/discharging can involve considerable current.

PDDD said:

How do I go about passing 300V at the drain with a 20Vgs FET?

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You use an appropriate high-side driver with a bootstrap capacitor. Or you could use a transformer driver for the gate.

alec_t said:

The gate capacitance charging/discharging can involve considerable current.

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Because I am switching at 200khz or faster, I drive the gate at amps. The current is only when the voltage change. Just like charging a capacitor.
Because this H bridge only switches at 100hz it is not necessary to drive with much current.

alec_t said:

You use an appropriate high-side driver with a bootstrap capacitor. Or you could use a transformer driver for the gate.

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Because of the very slow speed; the booststrap capacitor must be LARGE. also a transformer must pass the 100hz so its size is also big.
Yes you need a good high side driver. (or use a P-MOSFET on the top side)

Remember that, if you're using a bootstrap driver, you'll have to get rid of the 1k G-S resistor shown in the schematic you posted in #10. That resistor will discharge the bootstrap capacitor all the while the high side MOSFET is on, limiting its maximum on time significantly.

Ron, thanks again for all your help.
I see how by raising the gate voltage to 312 you are able to pass 300 through the upper mosfet of the H-bridge, but I'm confused as to how the lower mosfet will pass the 300 V to ground when its gate is at 12V. My limited mind says it would limit the voltage passing through to about 12V. I am anxious to know how this really works!!!

PDDD said:

I am anxious to know how this really works!!!

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Think of it as turning on a tap. Only a small hand pressure is needed to control a high-pressure water supply.

Hey Alec, thanks for the anology. I see how it applies to the top mosfet. With only a 12V Gate-Source differential (ie: 312V at gate) I can control 300V at the drain. But if the top mosfet only had 12V on the gate and 300V on the drain, I could only pass 12V through it. So how is it that I can pass the same 300V through the lower mosfet with only 12V on the gate?

PDDD said:

Hey Alec, thanks for the anology. I see how it applies to the top mosfet. With only a 12V Gate-Source differential (ie: 312V at gate) I can control 300V at the drain. But if the top mosfet only had 12V on the gate and 300V on the drain, I could only pass 12V through it. So how is it that I can pass the same 300V through the lower mosfet with only 12V on the gate?

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The gate only needs to be about 10 volts higher than the source for the FET to "turn on". So in the case of the bottom one the source is ground.

As Ron says. Note, it is the gate voltage relative to the source which controls the FET switching.

Alec and RoV, Thanks for trying to help, but respectfully, my question was NOT how many volts does it take to turn on a FET. My question to RonSimpson was how do you get more than 300 volts through the H-Bridge with only 12V on the gate. His answer was you don't. You must put 312V on the gate. That's cool, but if it takes 312V on the upper FET-G to get 30oV through, then why doesn't it take 312V on the lower FET-G to get 300V through it to ground. Is it simply because the lower FET-S is tied to ground? Thanks, guys, for taking the time to answer this.

PDDD said:

Is it simply because the lower FET-S is tied to ground?

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Yes.

Thank you so much!

The MOSFET does not really know what voltages are on its pins. It only knows what voltage is from G to S.
Here is circuit. V1 and V2 drive voltage from G to S of the MOSFETs. 100 volts is the power supply.

Here is the G1 signal and G2-S2. (G2-out) Notice I do not have both transistors on at the same time. There is 1uS of time when no transistor is on.

Here we see G1 which is 0 to 15V, G2-out which is 0 to 15V, and G2 which is 0 to 115 volts. (because the Gate and V2 ride on "out".

Here I added the red trace which is "out".


Hope this helps! Picture = 10,000 words +/- 9,999.

PDDD,
If you had posted your schematic at the start (Which you still have not done.) then all of the members that had tried to help would have seen that the problem was in the way you are trying to drive the high side mosfets in the H bridge. NOTE that in Ron's post #36 that V2 MUST BE A FLOATING SUPPLY. Note you cannot use the same floating supply for both high side drivers - they must be isolated from each other. This could be from a DC to DC converter, from another secondary winding on the power supply transformer or by using bootstrapping. I think your negative coments about the members that have tried to help are totaly unfounded. It is you that is being unresonable by not providing the requested information.

Les.

Les Jones said:

If you had posted your schematic at the start (Which you still have not done.

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See post #10.

Hi Ron,
I was hoping you would answer one more question about the circuit you posted on Dec 7 with the upper gate drive optocoupler attached to the upper Mosfet Source. I'm using an Avago HCPL-J312-000E it goes high when power is applied to pin 8 because pin 5 is attached to the Mosfet -S instead of ground. Of course the lower opto's remain low until signalled because their pin 8's are attached to ground. I know your circuit works great for you, so I assume I'm going to need a different optocoupler that is compatable with the circuit, but I don't know what to look for in the spec sheets.

I really appreciate your help. Thanks Ron.

sorry, typo in the previous post.

Hi Ron,
I was hoping you would answer one more question about the circuit you posted on Dec 7 with the upper gate drive optocoupler attached to the upper Mosfet Source. I'm using an Avago HCPL-J312-000E it goes high when power is applied to pin 8 because pin 5 is attached to the Mosfet -S instead of ground. Of course the lower opto's remain low until signalled because their pin 5's are attached to ground. I know your circuit works great for you, so I assume I'm going to need a different optocoupler that is compatable with the circuit, but I don't know what to look for in the spec sheets.

I really appreciate your help. Thanks Ron.