The main function of R2 is to limit current thru the opto's transistor, but it also provides a pull-down for the gate.

This brings other questions to my feeble mind. If using a resistor in this fashion to limit current is possible, why isn't it done with a led?

Since the mosfet gate is in reality a capacitor, and a lower resistance than R2, wouldn't the current/voltage first "fill" the capacitance before shunting to ground through R2? And before someone says mosfet gates are voltage, not current driven, it takes current to turn on a mosfet gate fast, but voltage to keep it on. If the last statement isn't true, then why do gate drivers supply so much current?

Not trying to be a troll, just trying to better understand.

 

Sounds like a good question to me. I guess if it acts like a capacitor it simply fills up with a charge based on the voltage and stops. In the mosfet case the charge pushes ions into different parts of the channel causing conduction or not. I think the current is only involved in charging the capacitor which stores a potential that reacts with the substrate. you helped me out as I was looking at it as requiring high resistance to prevent damage. Hope I have this right.
Much appreciated.

 

If using a resistor in this fashion to limit current is possible, why isn't it done with a led?

I'm missing your question. It's common to use a current-limiting resistor with an LED - they're rarely used without one. I know that you know that, so again I do not understand what you are asking.

 

If using a resistor in this fashion to limit current is possible, why isn't it done with a led?

Since the mosfet gate is in reality a capacitor, and a lower resistance than R2, wouldn't the current/voltage first "fill" the capacitance before shunting to ground through R2?

1) because an LED is a diode. It doesn't respond to voltage or current in a linear fashion.

2) It's simultaneous.

Here is a (bad) drawing that pretends to represent a MOSFET gate.
There is a capacitance which needs to be "filled".
There is a pair of zeners that represent the maximum voltage that can be applied between the gate and the source.
There is a fuse that tells you what will happen if you exceed the maximum gate to source voltage.

When you apply voltage to the gate, the capacitor gets charged, but there is also current through R1. It's simultaneous.
When you stop the current through the optoisolator, R1 empties the capacitance. You can't use an LED for R1 because it still wouldn't have any current limiting. If the opto did not supply enough current to smoke the LED, the voltage on the LED wouldn't be enough to turn on the MOSFET. When you turn off the opto, the LED would not discharge the input capacitance to zero volts.

Is this helping?

 

...wouldn't the current/voltage first "fill" the capacitance before shunting to ground through R2?...

Yes, the voltage develops across the capacitor and thus R2 over a brief period. By brief I mean the RC time constant is very small. The R comes from the internal impedance of the power supply and the transistor's "on" resistance, and the C comes from the MOSFET's gate. This is two small numbers multiplied together. Unless you are switching at high frequency (>100kHZ is "high" to me), the time is virtually instant. The current might be high for part of that instant, but the time is so short that the transistor has time to dissipate the heat and we don't usually worry much about limiting the gate current with a resistor.

For big MOSFETs (higher capacitance gates) switched at high frequency, things change and we need to use gate drivers to move that charge in and out.

 

I'm missing your question. It's common to use a current-limiting resistor with an LED - they're rarely used without one. I know that you know that, so again I do not understand what you are asking.

I was meaning the OPs circuit. You said R2(between opto and ground) is limiting the current to the mosfet gate. I thought to limit current R2 had to be between opto and gate, in series. Like a resistor is in series with a led.

Just never seen a resistor connected like that to limit current. Always have seen them in series- base resistor for transistor, limiting resistor for led, etc, etc.

Now as a pull down for shutting off the gate, when opto shuts off, I've seen that. In fact driving a gate with an opto it's necessary. Because there is no path to ground to drain the gate capacitance.

 

Yes, the voltage develops across the capacitor and thus R2 over a brief period. By brief I mean the RC time constant is very small. The R comes from the internal impedance of the power supply and the transistor's "on" resistance, and the C comes from the MOSFET's gate. This is two small numbers multiplied together. Unless you are switching at high frequency (>100kHZ is "high" to me), the time is virtually instant. The current might be high for part of that instant, but the time is so short that the transistor has time to dissipate the heat and we don't usually worry much about limiting the gate current with a resistor.

For big MOSFETs (higher capacitance gates) switched at high frequency, things change and we need to use gate drivers to move that charge in and out.

I never knew a "shunt" resistor, like in the OPs circuit, worked as a RC circuit. The resistor, R2, in series with the gate, I see a RC circuit. But not R2 shunting to ground like it is.

Again, I'm trying to learn, not be a smart a$$. Every time lately I think I'm getting a leg up on this stuff, things seem to change.

 

You said R2(between opto and ground) is limiting the current to the mosfet gate.

No, I said it limits current thru the transistor, which would be shorted to ground if R2 were not there. It is true that the portion of current thru the transistor that goes to the gate is not limited by R2, but this is relatively much less current over a period of time.

I thought to limit current R2 had to be between opto and gate, in series. Like a resistor is in series with a led.

This circuit has no limiting resistor on the gate. It probably should, for good form, but the transistor can likely survive without it. The downside of protecting the transistor - by limiting the gate current - risks a slower turn-on, turn-off of the MOSFET, and that leads to heating. Trading one risk for another.

 

I am operating this is the range of 1 - 100 Hz (very slow). Can you recommend a minimal safe amount of resistance to limit the inrush current on the transistor from charging the capacitor on the gate? I suppose R2 acts to limit he current on the transistor after the gate capacitor charges. then it is acting as a series resistor.
Do I have this right?
Thanks.

 

No, I said it limits current thru the transistor, which would be shorted to ground if R2 were not there. It is true that the portion of current thru the transistor that goes to the gate is not limited by R2, but this is relatively much less current over a period of time.

Thank you for the answer. But wouldn't the current just stop when the capacitance of the gate is fully charged? Since without R2, the only path to common/ground would be the gate.

I know this may be every day/no brainer stuff to you, but not to me. I can build stuff from a kit or schematic, but I'm trying to learn how to try to come up with my own schematics for my ideas. So asking about new things I see being done.

 

I'm not the expert here but I believe your correct. the only thing to remember is that you have to have a resistor that will drain the charge on the capacitor when you shut it off, otherwise it won't shut off. I proved that theory the hard way.

 

...But wouldn't the current just stop when the capacitance of the gate is fully charged? Since without R2, the only path to common/ground would be the gate...

Yes! That's one of the great features of a MOSFET, that no current is required to hold it on.

A downside is the current required to switch the MOSFET at very high frequencies. (They have an inductance as well as capacitance.) I don't know where the crossover occurs, but at a high enough frequency the current needed to switch a MOSFET will exceed the amount needed to switch a BJT.

 

Thank you for all the help. It has been very helpful to say the least. It seems that whatever resistance I use to limit the current on the opto would also protect the gate especially at the low frequencies I am working at. Author #12 made a good point about the current through the opto-isolator. Will have to check out the requirements for a 4n24. It seems like the resistor for the opto and R2 could limit the turn on time, while R2 would limit the turn off time.

 

Why? You want at least 12V on the gate to be sure the MOSFET is operating at its minimum Rdson. Otherwise, they get hot.

Probably good with 8 - 10 volts, but 12 is a nice round figure.

Anything over 20 can degrade the oxide layer over the long term.

 

Thank you for all the help. It has been very helpful to say the least. It seems that whatever resistance I use to limit the current on the opto would also protect the gate especially at the low frequencies I am working at. Author #12 made a good point about the current through the opto-isolator. Will have to check out the requirements for a 4n24. It seems like the resistor for the opto and R2 could limit the turn on time, while R2 would limit the turn off time.

I do know that you need the resistor R1 in the circuit going to the led side of the opto, that limits the current on the led in the opto.

And the resistor R2 is needed to give a path to ground for the gate. Both to shut it off fast and make sure it stays off.

It was the 'current limiting' to protect the opto, that I was questioning.

And Laurence, I apologize for hijacking your thread with my questions, but if you don't ask the question, at the time, it's harder to get it answered later. So please stick around and don't be mad. This is the best place to learn electronics on the web.

 

@wayneh, thanks for taking the time to explain this.

 

It was the 'current limiting' to protect the opto, that I was questioning.

Yes, my earlier comment on that may have been misleading. It only limits the opto transistor current when compared to an incorrect hookup, where R2=0 and the transistor shorts to ground if turned on. Hooked up correctly, R2 is chosen to turn off the gate at the speed desired without drawing too much thru the transistor.

 

Shortbus,
I didn't think you hijacked anything. I totally appreciate this highly educational discussion.
Many thanks to all.
I learned a lot.

 

One other point of clarification, I might have been misleading when I referred to the resistance on the "opto", that might have implied R1, I meant the Potentiometer on the transistor side in combination with R2 would limit current through the transistor of the opto.

 

Rather than a pot, a resistor divider would be a more 'professional' way to do it. This would also keep anyone in the future from changing the setting(gate voltage). But in reality you don't need a divider or a pot, your 12V power rail is just right to switch the gate using the opto.