Power Switching with mosfet

It's all about how Vgs is developed for saturation.

High Side: (Vgs >Vdd )

Low Side: (Vgs < Vdd )

D from BC

It is completely symmetric, or shall we say identical. But this means you have to take the high-side gate as far above its source, e.g. 3.5 volts over the drain, as you took the low-side gate above its source=drain. This can mean taking the high-side gate higher than the supply voltage, which requires a special kind of driver circuit, called a high-side driver. These have a charged capacitor to act as the flying gate-driver power supply. That works for periodic switching, but for DC switching you need an isolated gate supply. I just finished a fast DC-capable 600V high-side pulser and powered the driver with its own 60Hz transformer.

It is completely symmetric, or shall we say identical. But this means you have to take the high-side gate as far above its source, e.g. 3.5 volts over the drain, as you took the low-side gate above its source=drain. This can mean taking the high-side gate higher than the supply voltage, which requires a special kind of driver circuit, called a high-side driver. These have a charged capacitor to act as the flying gate-driver power supply. That works for periodic switching, but for DC switching you need an isolated gate supply. I just finished a fast DC-capable 600V high-side pulser and powered the driver with its own 60Hz transformer.

You kind of answered your own question here. The 2N7002 won't even start to turn on before the Vgs reaches 2V. You are saying you can only supply 1.7.

The recommended Gate to Source ON voltage is 5V. When the source is connected to ground that means a gate to ground voltage of 5V. For high side switching you will bring the source voltage up to near what you call V, and the gate to ground voltage now has to be V + 5. If you can't supply enough gate voltage, you need to use a P channel FET.

Tam

What sort of IC are you switching? In the case of low side switching, any input midway between V and G will end up below the load's ground rail. Likewise, for high side switching, an input may end up above the ICs supply rail. Some ICs don't like either (or both) of these situations.

Others have pointed out the gate drive level issues.

Ooops ...let me fix that..

High Side: (Vg > Vdd )

Low Side: (Vg < Vdd)

D from BC

Ok guys, I think I got it. I was sorta thinking about it that way but couldn't put the pieces together cause I thought the gate voltage had only to be greater than V_DS to turn it on.

So basically the gate voltage has to be so many volts above the source voltage? When the source is grounded, as in a low-side, then its very simple. If it's high side then you have increased the source voltage and have to increase the gate voltage to compensate.

Ok, I think I got it for the most part(somewhat makes sense but not sure exactly why it works that way).

I can't see what the difference between them are. Like why is one used over another. (I understand why one uses the mosfets but not when I should know to use one but not the other).

This site talks about a high side using an n-channel,

But it seems I can use a p-channel without all that mess?

There are 4 combinations, n-channel high side, n-channel low side, p-channel high side, p-channel low side.

The n-channel low side and p-channel high side seems to work the same? Is there a time when one is better to use than the other?

What about a p-channel high side vs an n-channel high side? Why go through what that site says when a p-channel will simplify it?

Thanks, Jon

For a given price, you can get a N-channel with much better specifications than P-channel (on resistance, current and voltage handling). Especially at the high voltages and powers, where N-channel becomes the only option.

I guess this might be where my confusion came from?

"If the gate voltage is larger than the threshold voltage, the transistor is turned on"?

But shouldn't that be the gate to source voltage?

So its just a practical issue rather than a theoretical one...

Thanks, Jon

Its a Pic24.

How can it end up above the rail? You mean input to the IC? Since the mosfet might not pull the down completely or pull up completely?

If thats the case then how does one go about solving that problem? In my application my inputs the the IC will be at the rail since I'm using a pullup and very little current will flow. It will only be a few mV I imagine but maybe that will make a difference? R_DS is about 10 ohms which may or may not be significant? What kinda wiggle room do I have to work here?

Oh, I think I can just use the mosfet to power up the pullups too though... so when the mosfet is triggered it will power all pullups used for inputs/outputs and so should reduce there voltages be the same amount... But assume I just didn't think about that ;)

Thanks, Jon

Paul, You hit the nail on the head. From personal (unintentional) experience, at least for digital ICs, opening up either the ground or the Vdd lead can lead to unexpected and bizarre operation.

Tam

By definition gate voltage means gate to source voltage when talking about the device characteristics. Also, see Paul's comments from a few posts back. It is not obvious that turning off power to the IC will do what you want it to. What do you expect the IC outputs to do when it is unpowered?

Tam

Sure. If you are only switching a small, low voltage load, by all means use a p-fet. (And it is usually a better idea to switch the +ve rail than the 0V).

Why is that? I was thinking it might be better since it cuts the power(so the device isn't "floating" in some sense) but not sure...

Huh? I want to cut the power to the IC because it requires a specific power on sequence to enter programming mode. Its not entirely clear that have to cut the power but I don't see the big deal in having that ability and I might need it so I can automatically reset the device if an error occurs.

You are either going to leave VDD floating, or GND floating. Neither one is a good thing. Why don't you try it, open up one or the other manually, and restore with a clip lead. Why not just hold the chip in reset? It is improbable that the chip manufacturer intended for you to control power to the chip independent of other circuits.

Tam

Tam

Well it depends on the circuit arrangement. But imagine you feed it from a power supply with an earthed 0V (A USB cable, perhaps). Your mosfet cuts the zero volts and the circuit turns off. But then you connect a RS232 cable, and the circuit powers up again, using the 0V connection in the RS232!

You most cases, you need a P-channel for high side. and N-channel for low side. so that would mean, the high side S (source) would be at the

• rail end low side (S) at the - rail etc...

Just think upside down :) On the high side you would pull to common to turn it on and low side, apply + to turn it on.