Erm, right. I should probably know this, but...

J

Jamie

Guest
I apologise for the HTML, if that's not liked in this group.

Mosfets...

If is was to take say, this (http://us.st.com/stonline/books/pdf/docs/2950.pdf) mosfet, and connect like this (excuse the drawing, oh and it'll only work in Courier, or some other fixed-width font like that):

4.5V 18R
-----------------\/\/\/\------
| |
\ |Source
Switch \ ____|____
| Gate | |
------------------| Mosfet |
|_________|
|
|Drain
____|____
| |
| Load | 200mA (bunch of LEDs in parallel)
|_________|
|
|
0V |
------------------------------

Now as far as I know, say the switch is open (ie. gate is at 0V) then no current will flow between the source and the drain, and the LEDs won't be lit, right? And if the switch is closed (gate at 4.5V and higher than the source voltage) then the mosfet will be 'on' and current will flow through the mosfet and the LEDs will light up, yeah? Or not? 'Cause I soldered it all together and the LEDs just stay lit, no matter what I do with the gate voltage. It's all getting abit annoying.

Anyways, any help would be appreciated,
Jamie
 
Not only is usenet a HTML free zone, binaries are only allowed on
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--


Michael A. Terrell
Central Florida
 
Oh yes! - good catch. I didn't check the PDF. Also, (ideally) he should put
the source to ground and the load and current limit resistor on the drain
side, and then to 4.5V. This will maximize Vgs and ensure that it is
sufficiently greater than threshold voltage.

Steve B

4.5V
___________resistor_________LEDs________Drain__________Source________GND

|

|

|
 
Right, hang on. Your gonna have to explain that one again. I'm confused.

Jamie
"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:t120lvovtocs654e8rruf1hkn599talm9s@4ax.com...
On Sat, 30 Aug 2003 00:52:01 +0100, "Jamie"
jizzproductions@hotmail.com> wrote:

I apologise for the HTML, if that's not liked in this group.

Mosfets...

If is was to take say, this
(http://us.st.com/stonline/books/pdf/docs/2950.pdf) mosfet, and connect like
Click to expand...
this (excuse the drawing, oh and it'll only work in Courier, or some other
fixed-width font like that):
4.5V 18R
-----------------\/\/\/\------
| |
\ |Source
Switch \ ____|____
| Gate | |
------------------| Mosfet |
|_________|
|
|Drain
____|____
| |
| Load | 200mA (bunch of LEDs in parallel)
|_________|
|
|
0V |
------------------------------

Now as far as I know, say the switch is open (ie. gate is at 0V) then no
current will flow between the source and the drain, and the LEDs won't be
Click to expand...
lit, right? And if the switch is closed (gate at 4.5V and higher than the
source voltage) then the mosfet will be 'on' and current will flow through
the mosfet and the LEDs will light up, yeah? Or not? 'Cause I soldered it
all together and the LEDs just stay lit, no matter what I do with the gate
voltage. It's all getting abit annoying.
Anyways, any help would be appreciated,
Jamie


This is an n-channel fet; so the substrate diode (the freebie
intrinsic s-d diode) is foreward biased, so it will always be on.

John
Click to expand...
 
The whole switch thing was just for simplicity, the gate is actually connected to the output from an OR gate.

Jamie
"Steven Bastien" <sbastien@optigain.com> wrote in message news:vkvut1m0dg1na8@corp.supernews.com...
Jamie,

Use a SPDT switch so that you can ground the gate when in the off state, or use a 1K pull down resistor on the gate with your SPST switch. MosFETS have very high input impedance into the gate, and if left open as you have it, they can float in the on state (especially after they are already on). See diagrams below.

Good Luck,
Steve B


Option 1:

4.5V 18R
-----------------\/\/\/\------
| |
\ |Source
Switch \ ____|____
| | Gate | |
| ----------------| Mosfet |
| |_________|
| |
| |Drain
| ____|____
| | |
| | Load | 200mA (bunch of LEDs in parallel)
| |_________|
| |
| |
0V | |
------------------------------

Option 2:

4.5V 18R
-----------------\/\/\/\------
| |
\ |Source
Switch \ ____|____
| Gate | |
------------------| Mosfet |
| |_________|
| |
/ |Drain
\ ____|____
/ 1K | |
\ | Load | 200mA (bunch of LEDs in parallel)
| |_________|
| |
| |
0V | |
------------------------------
 
On Sat, 30 Aug 2003 04:21:14 +0100, "Jamie"
<jizzproductions@hotmail.com> wrote:

Right, hang on. Your gonna have to explain that one again. I'm confused.

Jamie
Click to expand...
Take a look at the little schematic on the first page of the fet
datasheet. There's a parasitic substrate diode from source to drain.
Basicly, a n-fet is intended to work with the drain more positive than
the source. In your circuit, it's "upside down" so the diode is
foreward biased all the time, so you can't turn the puppy off.

So, use a p-channel fet, or, to use the same fet, rearrange things,
putting the load in the drain circuit like Steve suggests. (Nice
linear ascii schematic!)

John




"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:t120lvovtocs654e8rruf1hkn599talm9s@4ax.com...
On Sat, 30 Aug 2003 00:52:01 +0100, "Jamie"
jizzproductions@hotmail.com> wrote:

I apologise for the HTML, if that's not liked in this group.

Mosfets...

If is was to take say, this
(http://us.st.com/stonline/books/pdf/docs/2950.pdf) mosfet, and connect like
this (excuse the drawing, oh and it'll only work in Courier, or some other
fixed-width font like that):

4.5V 18R
-----------------\/\/\/\------
| |
\ |Source
Switch \ ____|____
| Gate | |
------------------| Mosfet |
|_________|
|
|Drain
____|____
| |
| Load | 200mA (bunch of LEDs in parallel)
|_________|
|
|
0V |
------------------------------

Now as far as I know, say the switch is open (ie. gate is at 0V) then no
current will flow between the source and the drain, and the LEDs won't be
lit, right? And if the switch is closed (gate at 4.5V and higher than the
source voltage) then the mosfet will be 'on' and current will flow through
the mosfet and the LEDs will light up, yeah? Or not? 'Cause I soldered it
all together and the LEDs just stay lit, no matter what I do with the gate
voltage. It's all getting abit annoying.

Anyways, any help would be appreciated,
Jamie


This is an n-channel fet; so the substrate diode (the freebie
intrinsic s-d diode) is foreward biased, so it will always be on.

John
Click to expand...
 
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