I frequently find myself doing the following circuit (for low frequency).
A problem I have recently encountered is high voltage switching (circa 100v) and the cost and availability of such p-fets. They also tend to be large, for example I've just built a circuit with 10 SOT-223's (DMP10H400SE) merely to get a 100v rating. Normally I don't switch the p-fets on at this higher voltage but they have to withstand it (such as an over-voltage lockout). For very low voltages the circuit above works down to about 2V. The floorspace needed for all these p-fets is a pain.
I'm aware of the reason for n fet drivers for high-side but I'm having difficulty finding any drivers that makes the N+driver greater value than an expensive p-fet. One popular one is the LM9061 yet it doesn't work until 7V. Another, the MAX1614 works only to 26V.
I typically bootstrap a PIC using a sideband shunt which controls this high-voltage circuitry but believe I'd need to include a p-fet to switch off the N-Driver (pin 5 for the above) which powers the n-fet (down the rabbit hole!), and the only benefit that brings is the high frequency and current possibilities - neither of which I require.
Assuming I can find a driver that works at low voltage up to about 20V, perhaps the equivalent of pin 3 can be low sided?
Dude, you gotta tell us what you're trying to do with your circuits. You're clearly leaving a lot unsaid because you seem to think we know it somehow.
For example:
1. I don't see how using 10 DMP10H400SE is supposed to get you a rating of 100V over just using one.
2. You imply quite a few times that you have a lower and upper operating voltage requirement, but you never say what they actually are.
P-MOSFET high side: Use a high voltage low current transistor to drive the gate. A digital signal is connected to the Base of the transistor. This places the emitter at 3 to 4 volts. Set the emitter resistor for 100uA to 1mA. The G-S mosfet resistor should be 3 or 4 times higher in value to get good drive voltage on the Gate. A 12 to 20V Zener is added to protect the Gate.
N-MOSFET high side: There are many different ways to do this. A Photo Voltaic OptoIsolator is a good way to drive a gate of the N-MOSFET. (assuming the turn on/off time can be a little slow) VOM1271 is an example.
Ronsimpson, Driving an N fet high side requires voltage at gate to be source + VGS, so unless you're measuring it that VOM1271 is either going to be too low to fully turn the fet on or too high (which would be strange) leading to the fet being damaged.
dknguyen, I'm primarily switching AC which is where the high-side requirement comes from. I'm using 2x DMP10H400SE with one in reverse to block both directions.
One possibility is using a bridge like the wonderful DMHC10H170SFJ and not using the N fets. It's less than 50% the footprint of 2 x DMP10H400SE with the same rating. Yet, something feels wrong with using parts of an IC to do a job!
It really depends on the current you are trying to switch. The ON resistance of PFETs at high voltage can be up in the Ohms region, so switching more than a few 100 mA makes this solution impractical. The LTC7000 generates a high side VGS so you can use NFETs which have a lower ON resistance and are generally smaller in package
Hi Simon, the LTC7000 has it nailed for DC, but for AC is where the issues start. For DC I'd mostly just use the N-FET low side anyway.
They have the LT4320 which is doing something similar.
My hypothesis is for 0-100V AC the design would be:
1. A traditional bridge rectifier fed into a transistor shunt "LOW-V" fed into a pump generates voltage "FETMAX" equal to the intending working voltage of the circuit (say 10V) plus VGS, say 10V.
2. An op-amp measures the voltage at the drain of the N fet's (which would be connected together for reverse switching) and feeds FETMAX - FETNOW into the gates. Thus the gate always receives 10V over VGS.
Ronsimpson, Driving an N fet high side requires voltage at gate to be source + VGS, so unless you're measuring it that VOM1271 is either going to be too low to fully turn the fet on or too high (which would be strange) leading to the fet being damaged.
The type of isolator I chose makes a voltage. It works like a light bulb or LED and a solar panel. It does not make much power but just enough to turn on the gate of a MOSFET. For Gate turn off, inside the isolator there is a J-FET that pulls the output low when there is no light.
I agree.
The isolator is not making a voltage starting at ground. It is making a voltage from Source to Gate. The Source can be at -100 or +100 it does not matter.
I agree.
The isolator is not making a voltage starting at ground. It is making a voltage from Source to Gate. The Source can be at -100 or +100 it does not matter.
To re-iterate, the opto-isolator's output pins are floating relative to the output. Very similar to a transformer or an isolated power supply.
I do the same thing for back-to-back MOSFET AC switches except I've not found an opto-isolator fast enough for my needs so I use a regular gate driver that is powered by small isolated converter, and the "ground" output terminal for that power supply is connected to the MOSFET source. Whatever voltage is produced is always (source voltage + converter output voltage) because the output is floating.
As for value, I would think that the opto and NMOS would be cheaper than your original circuit. The isolated supply + gate driver + NMOS is probably going to cost a little more mainly because the isolated supply costs ~$4USD. Both these solutions are more rugged and flexible than you original circuit. Safer too because they provide isolation.
That makes sense, and there I was always thinking opto's were for voltage/safety isolation!
I've gone with the MIC5019 for high-side which is only $0.50 however this is only because I'm currently working with DC and at 5V. For AC I shall surely keep the opto method in mind.
It was designed for a solid state relay.
With a 200pF 10meg ohm load its on/off time is 53/24uS. (LED=20mA)
Your Gate has capacitance so speed is a little complicated. 10mA LED current = 15uA output current You can drive the LED harder. (40mA = 47uA)
If speed is important I wold drive the LED with a R parallel with a RC so the current is 40mA for 100uS and then drops back to 10mA.
It was never designed for 100khz PWM applications. Try some thing different.
Suppose I want a 100V rectifier such as this with the P channel fet's swapped out. How might the opto work with reverse voltages? It looks like I'd need a 100v diode to protect against reversal therefore making the turnon about 2.4v?
I'm having difficulty finding 100v P-channel fets in a small QFN style package so worth looking into if it's possible.
Suppose I want a 100V rectifier such as this with the P channel fet's swapped out. How might the opto work with reverse voltages? It looks like I'd need a 100v diode to protect against reversal therefore making the turnon about 2.4v?
I'm having difficulty finding 100v P-channel fets in a small QFN style package so worth looking into if it's possible.
Why do you need an opto for that at all? Or MOSETs? Or resistors? I dont think that circuit does what you want think it does. Just looks like a diode rectifier to me with a bunch of components that do less that nothing. 4 regular diodes does the exact same thing.
Almost seems like you forgot MOSFETs have a parasitic body diode and those diodes are the only things in your circuit doing anything.