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'[EE] Simple Mosfet question'
2007\03\14@193205 by Tal Go

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Hi All

Can you give me a short explanation why schematic (1) is the correct one?
and what happen when using schematic (2)?

Thanks


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2007\03\14@194820 by peter green

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> Can you give me a short explanation why schematic (1) is the correct one?
> and what happen when using schematic (2)?
sorry but we need more context. the first schematic is an open drain nmos switch/voltage to current amplifier, the second is a nmos source follower (similar to the bipolar emmitter follower).



2007\03\14@195708 by Marcel Birthelmer

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Tal,
I'll take a shot, but I might be wrong - the transconductance of the
FET is determined by the difference between the drain-gate and
drain-source voltage. In case two, the drain voltage is 24V.
Therefore, you'd have to drive the gate with about 25V (Vds + Vth,
threshold voltage = 1V). In case 2, the drain voltage is 24V - Vlamp,
which should be considerably lower. Therefore, the gate voltage can be
much lower.
I think both should work if you can drive the gate high enough,
though. But if your supply is equal to the drain voltage, it won't
work, since you can't get enough potential at the gate to exceed the
turn-on voltage.
- Marcel

On 3/14/07, Tal Go <spam_OUTtalgoTakeThisOuTspamusb.co.il> wrote:
> Hi All
>
> Can you give me a short explanation why schematic (1) is the correct one?
> and what happen when using schematic (2)?
>
> Thanks
>
> -

2007\03\14@195812 by Marcel Birthelmer

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from earlier:
threshold voltage = 1V). In case 2, the drain voltage is 24V - Vlamp,
which should be considerably lower. Therefore, the gate voltage can be
much lower.

Sorry, this should've been case 1.

>
> On 3/14/07, Tal Go <.....talgoKILLspamspam@spam@usb.co.il> wrote:
> > Hi All
> >
> > Can you give me a short explanation why schematic (1) is the correct one?
> > and what happen when using schematic (2)?
> >
> > Thanks
> >
> > --

2007\03\14@200037 by Sean Breheny

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Hi Tal,

Circuit (1) will cause almost a full 24V to be applied across the
light bulb. For circuit (2), the voltage on the bulb will be equal to
the gate voltage minus a little more than the threshold voltage
(usually about 4 volts). If you apply something like 35V to the gate,
then the bulb will see the full 24V.

The reason for this is that factor which determines the amount of
current flow from drain to source is the gate to source voltage. In
the first circuit, the source is grounded so the gate to source
voltage equals the gate voltage and it is easy to turn the fet fully
on. In the second case, there is a negative feedback effect (more
current through the fet causes the voltage drop across the bulb to
increase, increasing the source voltage, which decreases the gate to
source voltage difference). This causes the drain to source current to
reach an equilibrium (which is lower than just 24V divided by the
resistance of the bulb) if the gate voltage is less than the drain
voltage.

Sean


On 3/14/07, Tal Go <talgospamKILLspamusb.co.il> wrote:
> Hi All
>
> Can you give me a short explanation why schematic (1) is the correct one?
> and what happen when using schematic (2)?
>
> Thanks
>
> -

2007\03\14@200804 by M. Adam Davis

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Sounds suspiciously like a homework problem...  :-)

The N-Channel mosfet shown switches "on" when a positive voltage
difference exists between the gate (middle terminal) and the source
(bottom terminal).

When on, the mosfet is essentially a low resistance wire between the
Drain and the Source. (hand waving exists here).

If the voltage potential between the gate and source is less than the
minimum needed for saturation, it will operate in the linear region,
and the resistance of the mosfet will vary according to the potential.

So let's apply +24V to the gate of both circuits and see what happens.

Circuit 1:
The mosfet sees a 24V positive voltage difference between the gate and
the source.  It becomes a low resistance between the drain and the
source.

The light turns on.  Since the mosfet is a low resistance, and the
load is presumably a high resistance, then both the source and the
drain are close to ground in voltage potential - nearly all the
current is being consumed by the load.  The mosfet stays on as the
gate and drain are still 24V apart.

Circuit 2:
The Mosfet sees a 24V positive voltage difference between the gate and
the source.  It becomes a low resistance between the drain and the
source.

The light turns on.  Since the mosfet is a low resistance, and the
load is presumably a high resistance, then both the source and the
drain *** are close to 24V ***.

Thus, there is no longer a 24V difference between the gate and source.

The mosfet therefore starts operating in the linear region - instead
of switching on or off, it's balanced somewhere inbetween.  Where it
ends up depends on:
1) resistance of load
2) mosfet resistance in linear region
3) other characteristics of the mosfet and load

So the mosfet ends up drawing increased power and dumping it as heat.
If the load is reactive, they may start oscillating, perhaps to
destruction.

Please note that I'm not an analog engineer, so you'll have to verify
everything I've just said, as it may well be incorrect, though the
general idea should be sound.  Perhaps I've made your job harder.  :-)

-Adam


On 3/14/07, Tal Go <.....talgoKILLspamspam.....usb.co.il> wrote:
> Hi All
>
> Can you give me a short explanation why schematic (1) is the correct one?
> and what happen when using schematic (2)?
>
> Thanks
>
> -

2007\03\15@023156 by Vasile Surducan

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On 3/14/07, peter green <EraseMEplugwashspam_OUTspamTakeThisOuTp10link.net> wrote:
>
> > Can you give me a short explanation why schematic (1) is the correct one?
> > and what happen when using schematic (2)?
> sorry but we need more context.

Correct!  In the whole storry here is supposed the drive voltage is
applied between gate and ground which is not the only choice. In (2)
the command can be between gate and source as well .

Vasile

2007\03\15@075611 by Tom Sefranek

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Tal Go wrote:

> Hi All
>
> Can you give me a short explanation why schematic (1) is the correct one?

Correct in the sense that a logic level will switch on/off the Bulb.

> and what happen when using schematic (2)?

The source follows the gate minus the Vgs, so the gate would have to go
to 27 volts to fully turn on the bulb.

>
>
> Thanks


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