Power Switching with mosfet

[Jon Slaughter]

I only have an n-channel enhancement mosfet(2N7002) and I'm trying to use it
to switch power to an IC.

I can only bring the gate voltage about 1.7V above the V_DS. I'm not
switching the entire voltage to the IC.

Some question I have, What is the difference between a high side mosfet
driver and a low side? A low side brings the drain within a few mV above
ground but the high side brings the source within only several hundred mV
below the drain. Shouldn't it be symmetric?

i.e., whats the difference between

V
|
Mosfet
|
Load
|
G

V
|
Load
|
Mosfet
|
G

I'm calling the first one a high side and the second a low side, I think
thats the correct terminology. Both mosfets are the same as is the load.

The low side works fine but the high side doesn't. It seems that in either
cause it should be the same. V_DS is the same in both cases? Obviously this
isn't true but I can't seem to understand why?

The way I'm looking at it is basically as two resistors in series so the
voltage drop across each should be exactly the same.

So whats wrong?

Thanks,
Jon

 

[D from BC]

I only have an n-channel enhancement mosfet(2N7002) and I'm trying to use it
to switch power to an IC.

I can only bring the gate voltage about 1.7V above the V_DS. I'm not
switching the entire voltage to the IC.

Some question I have, What is the difference between a high side mosfet
driver and a low side? A low side brings the drain within a few mV above
ground but the high side brings the source within only several hundred mV
below the drain. Shouldn't it be symmetric?

i.e., whats the difference between

V
|
Mosfet
|
Load
|
G

V
|
Load
|
Mosfet
|
G

I'm calling the first one a high side and the second a low side, I think
thats the correct terminology. Both mosfets are the same as is the load.

The low side works fine but the high side doesn't. It seems that in either
cause it should be the same. V_DS is the same in both cases? Obviously this
isn't true but I can't seem to understand why?

The way I'm looking at it is basically as two resistors in series so the
voltage drop across each should be exactly the same.

So whats wrong?

Thanks,
Jon

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

High Side:
(Vgs >Vdd )

Low Side:
(Vgs < Vdd )


D from BC

 

[Winfield]

I only have an n-channel enhancement mosfet(2N7002) and I'm trying
to use it to switch power to an IC.

I can only bring the gate voltage about 1.7V above the V_DS.
I'm not switching the entire voltage to the IC.

Some question I have, What is the difference between a high side
mosfet driver and a low side? A low side brings the drain within a
few mV above ground but the high side brings the source within only
several hundred mV below the drain. Shouldn't it be symmetric?

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.

 

[Winfield]

I only have an n-channel enhancement mosfet(2N7002) and I'm trying
to use it to switch power to an IC.

I can only bring the gate voltage about 1.7V above the V_DS.
I'm not switching the entire voltage to the IC.

Some question I have, What is the difference between a high side
mosfet driver and a low side? A low side brings the drain within a
few mV above ground but the high side brings the source within only
several hundred mV below the drain. Shouldn't it be symmetric?

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.

 

[Tam/WB2TT]

I only have an n-channel enhancement mosfet(2N7002) and I'm trying to use
it to switch power to an IC.

I can only bring the gate voltage about 1.7V above the V_DS. I'm not
switching the entire voltage to the IC.

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.

Some question I have, What is the difference between a high side mosfet
driver and a low side? A low side brings the drain within a few mV above
ground but the high side brings the source within only several hundred mV
below the drain. Shouldn't it be symmetric?

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

 

[Paul Hovnanian P.E.]

I only have an n-channel enhancement mosfet(2N7002) and I'm trying to use it
to switch power to an IC.

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.

 

[D from BC]

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

High Side:
(Vgs >Vdd )

Low Side:
(Vgs < Vdd )


D from BC

Ooops ...let me fix that..

High Side:
(Vg > Vdd )

Low Side:
(Vg < Vdd)


D from BC

 

[Jon Slaughter]

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,

http://www.innovatia.com/Design_Center/High-Side Drivers.htm

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

 

[John Devereux]

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,

http://www.innovatia.com/Design_Center/High-Side Drivers.htm

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?

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.

 

[Jon Slaughter]

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).

http://en.wikipedia.org/wiki/Threshold_voltage

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?

 

[Jon Slaughter]

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.

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

Thanks,
Jon

 

[Jon Slaughter]

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.

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

 

[Tam/WB2TT]

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.
.......................

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

 

[Tam/WB2TT]

http://en.wikipedia.org/wiki/Threshold_voltage

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?

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

 

[John Devereux]

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

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).

 

[Jon Slaughter]

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...

 

[Jon Slaughter]

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?

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.

 

[neon]

it depends where the loading will be sourcing or outputs.

 

[Tam/WB2TT]

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

 

[John Devereux]

[...]

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...

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!