Eliminating Optocoupler Turn-On Spike

[PDDD]

Hi Ron,
I think the answer to my question is that pin 5 must be tied to the battery ground you have shown. Any comments would be appreciated.

 

[kal.a]

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.
View attachment 95911
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!
View attachment 95912
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).
View attachment 95913

Hi Ron,

I hope you don't mind me bringing up this old thread. I don't quite understand how, in the bottom circuit, the capacitor gets charged when the FET is turned on.
The battery is at 12+ which turns on the FET and pulls up the source to about 300+ which means that the capacitor will have one side connected to 12+ and the other side to 300+. How does it gets charged to 312+? Doesn't need a negative or ground?
And what happens to the battery when we connect its negative side to 300+?
I'm quite a newbie so excuse me if the questions seem too simple.

Thanks
Kal

 

[ronsimpson]

It has been a long time and I don't remember where we are.
BOTTOM:
This is simple. The driver output sits on the Source of the MOSFET and drives the Gate. The Source is on ground.
TOP:
No so simple. The driver output sits on the Source of the top MOSFET and drives the Gate. The top Source is on the output.
The output voltage is 0 volts or 300 volts. Pin 5 is not ground, it is output. When the bottom MOSFET is on it pulls pin5 to ground.
When the output is at 0 volts current flows from +12V through D1 and charges C2,C3 to about 12 volts.
When the output is high, 300V, pin 8 is at 312 volts. (pin 8 to pin 5 is 12 volts) D1 blocks current flow.
When the top MOSFET is on it pulls pin5 to 300 volts. C2,C3 is disconnected from +12V.

 

[kal.a]

Thanks Ron.
So there's circuitry that controls the sequence of turning on the MOSFETS and D1.

I got lost at the fourth line in your post. "When the output is high, 300V, pin 8 is at 312 volts. (pin 8 to pin 5 is 12 volts) D1 blocks current flow.". Which output?

After Q1 is turned off (i believe it has to be turned off before Q2 is turned on) C2 and C3 are charged to 12+ and D1 is reverse biased which means pin 8 is at 12+ .
Then OK2 is turned on bringing the gate of Q2 to 12+ and turning on the MOSFET ( obviously it happens gradually) which starts to pull its source up to 300+.
Then I get lost.
As the source is now positive and rising how is it going to charge two capacitors one of which is polarized and one is non (am I correct?) while they're connected on the other side to a 12+ coming through D1? Or is D1 now connected to ground?

 

[ronsimpson]

Which output?

The output of Q1, Q2.

To charge up C2,C3 the bottom MOSFET must be on. This pulls pin5 to 0 volts or ground. Current will flow from +12V through D1 and into the caps.

The moment when the output of Q1,Q2 rises above 0 volts, D1 will open up and disconnect C2,C3 from the +12V.

When Q1= open and Q2=closed the bottom of C2,C3 will be almost 300 volts. Because C2,C3 is charged to where there is 12V across them the top of C2,C3 should be 300+12=312 volts.

 

[kal.a]

Thanks again Ron.
I've been trying but I don't get it. I think I understand the circuit as whole but not how the caps are charging to 312+.

"The moment when the output of Q1,Q2 rises above 0 volts, D1 will open up and disconnect C2,C3 from the +12V."
What does that mean, how does D1 open? Is there a switch to disconnect it or do you mean reverse biases?

Accepting that D1 will "open up and disconnect C2 and C3 from the +12V and leave them charged to +12V.
OK, so now I have a capacitor charged to +12V and not connected to anything at all then I connect one plate to +300V and the other plate to nothing , how does it get charged up to +312V? I have one plate at 12+ and the other at +300.

Sorry, I know it's basic but I've spent the day reading about capacitor ad how to charge them and still don't get it.

 

[spec]

Hy Kal,

If Sir Ron does not mind, I will have a go at explaining the operation of his schematic in post #43. It is a strange but clever circuit, and as a result it is tricky to understand the detailed workings. It is an often-used circuit though.

*Work in progress* Issue 09 of 2016_06_21

(1) BASIC CONCEPTS
In order to follow the circuit of post #43, a few basic concepts need to be explained.

(1.1) NMOSFETs
Why use such a complex circuit? Why not use a PMOSFET in the top half of the circuit? The answer is practical. In general, NMOSFETs are cheaper and perform better than PMOSFETs. This was especially the case when MOSFETs first came out, but is not so relevant now that improved PMOSFETs are available. Even so, if you need extreme performance: thousands of amps and thousands of volts, there is only one type available to do the job: NMOSFETs

(1.2) Relative
Electronic components do not care about absolutes. They have no idea where they are in the world. So for example an NMOSFET will always conduct current if its gate is more positive (by a certain amount dependent on specific device) than its source and if its drain is also more positive than its source.

Take an NMOSFET: assume that its source is at zero volts, its drain (VDS) is at 10V and its gate (VGS) is at 7V. Also assume that, under theses conditions, the NMOSFET is conducting 2 amps.

If you then increased these three voltages equally by 1000V (1KV) the relationship between the three NMOSFET terminals will be exactly the same and the PMOSFET will continue to conduct 2A. The same would be true if you added -1KV to all three terminals.

(1.3) Bootstraping
Bootstaping is a pig to understand. The name describes the function well. It is akin to pulling yourself up by your own braces (suspenders). But once you get the hang of the principle it is dead simple, like most things in electronics.

Take a normal domestic mains supply of 110V 60Hz. The actual voltage of the sine wave will be 2 * 1.414 * 110V = 311.08 V peak to peak (P/P), call it 310V P/P. This will be a sine wave swinging between -155V and 155V. Also take an ordinary 3V lithium button cell battery.

Connect the negative terminal of the battery to the earth pin of the mains supply. The negative terminal of the battery will be zero volts and the positive terminal will be 3V; no big mystery there. If you connected the battery negative terminal to the live of the mains supply the negative terminal of the battery will be a sine wave of 310V P/P, but the positive terminal of the battery will be a sine wave of -152V to 158V. Again, no big mystery: just adding two voltages in series.

Now take the 3V battery and make a sine wave oscillator with a transformer that has an isolated secondary winding of 310V P/P. Connect one end of the transformer secondary to earth. Connect the battery negative terminal to earth. The battery negative terminal will be earth and its positive terminal will be 3V, obviously.

But next, instead of earth, connect the negative of the battery to the live output of the transformer. The battery negative terminal will now be a 310V sine wave, but the positive terminal will be a sine wave swinging between -152V and 158V. In other words the battery will have been bootstrapped.

(1.4) Capacitor as Battery
Capacitors are like buckets. They accept electrons and stores them. The more electrons that they store the higher the voltage across their terminals. This voltage can be used as a power supply for circuits. In general, a capacitor used in this way, is akin to a rechargeable battery.

(1.5) MOSFET Gate Drive
Why is there all this complexity with driving the gates of MOSFETs (and IGBTs)? Why are there special gate driver chips that can source voltages of plus and minus 12V and currents up to 4A? After all, the input resistance of a gate is very high, in the order of 10M Ohms and, taking NMOSFETs, all you need to do to turn them on is put around 12V on their gates and to turn them off all you need to do is put 0V on the gate. 12V/10m Ohm = 1.2uA so what is with the 4A drivers and why do the drivers bother with -12V?

The answer is speed. It is very important to turn MOSFETs on and off as fast as possible for two reasons:
(1.5.1) To protect the MOSFET
(1.5.2) To reduce wasted power and thus increase efficiency

MOSFETs have unbelievably large capacitance between their gate and drain. This capacitance is made up of real capacitance and virtual capacitance reflected to the gate from the drain. Believe it or not, this capacitance can be as high as 50nF (not an error) and this is the problem. MOSFETS are voltage driven but to get that voltage, the massive gate capacitor must be charged up to turn the MOSFET on and discharged to turn it off.

To change the voltage across a capacitor fast requires a large amount of current, hence the 4A capability typical of gate drivers.

But why take the gate negative to turn MOSFETS off. Once again speed. =12V simply discharges the gate capacitance faster than 0V. Note that it is generally harder to turn MOSFETs off fast than on.

 

[crutschow]

One way to prevent a power-on pulse affecting the bridge is to delay the power to the bridge until after the optos have been powered for a few ms and turn the bridge off a few ms before removing the opto power.

 

[kubeek]

... Accepting that D1 will "open up and disconnect C2 and C3 from the +12V and leave them charged to +12V.
OK, so now I have a capacitor charged to +12V and not connected to anything at all then I connect one plate to +300V and the other plate to nothing , how does it get charged up to +312V? I have one plate at 12+ and the other at +300...

The capacitor is charged to 12V, so one plate has 12V more than the other plate with respect to ground. So whatever voltage you have on the first plate, the other will be that plus 12V. So if the voltage is pulled to 300V above ground the other will be at 312V above ground, but the cap is still charged to the 12V it has between its plates.

 

[Misterbenn]

HCPL3120 opto has under-voltage lockout to ensure the active opto circuitry is in a defined state prior the the output becoming active. this would remove the turn-on spike.

Teh circuit in post #43 is good although i'd use an isoltaed DC-DC converter in order to prevent possible connection of the high voltage circuit to the low voltage power lines.

 

[kal.a]

Quote spec: "
Connect the negative terminal of the battery to the earth pin of the mains supply. The negative terminal of the battery will be zero volts and the positive terminal will be 3V; no big mystery there. If you connected the battery negative terminal to the live of the mains supply the negative terminal of the battery will be a sine wave of 310V P/P, but the positive terminal of the battery will be a sine wave of -152V to 158V. Again, no big mystery: just adding two voltages in series."

The capacitor is charged to 12V, so one plate has 12V more than the other plate with respect to ground. So whatever voltage you have on the first plate, the other will be that plus 12V. So if the voltage is pulled to 300V above ground the other will be at 312V above ground, but the cap is still charged to the 12V it has between its plates.

Oh. So the idea wasn't to top up the cap to 312 but to use the 12v already in the cap to maintain the potential difference. I think I finally got it!