I am trying to use a IRF7832 MOSFET {datasheet here} to control a short LED strip.
I have connected the LED strip between 12 V and drain and 0V to gate and source. The LED strip has 6 strings of 3 LEDs. As I understand pin 5-8 is drain, pin 1-3 is source and pin 4 is gate so I have pin 1-4 to 0V and 5-8 to the LEDs.
I expect this to turn off the LEDs but they are turned on very dim and I measure about 4.5 V between drain and source.
If I connect gate to 5V the LEDs turn on and I measure 2.5 mV across drain and source.
I have tried 2 different IRF7832's.
What am I doing wrong?
Best Answer
What you say you have done would be completely correct and would work if you had done it and if the FET was alive. So either -
The MOSFETs are dead or
You are doing something different to what you say you are doing.
This seems the most likely situation. Check painstakingly, step by step.
Murphy.
Drain to source leakage current at Vgs = 0 is 1 uA max.
Odds are your eyes are not that good :-).
Presumably you are assuming that pin 1 is indicated by a dot on the package.
It should not matter, but=t ensure that all 3 S leads are grounded and that all 4 D leads are connected to the LEDs.
IF LEDs are white you can expect about I = V/R = (12 - 3 x 3.3)/100 =~~ 20 mA when on.
I find IR data sheets to usually be quite reasonable.
They should indeed have noted it as an N Channel on page 1.
The diagram on page 1 shows it is an N Channel.
MOSFETs are inherently exquisitely ESD sensistive.
A MOSFET with no protection built in can be destroyed by 20 to 30 volts applied gate to source. You can induce that level of voltage just by opening a non-ESD safe bag that they are contained in.
However, modern MOSFETs usually have reasonable to good ESD protection built in BUT always regard them as "static sensitive". Once you have handled on without ESD protection it is always suspect subsequently.