1
\$\begingroup\$

I am trying to use a IRF7832 MOSFET {datasheet here} to control a short LED strip.

enter image description here

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.

schematic

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?

|improve this question
\$\endgroup\$
  • 1
    \$\begingroup\$ A link to the datasheet of the FET would help so we can see things like off state current and maximum D-S voltage. No, I'm not going to look it up. That's your job. \$\endgroup\$ – Olin Lathrop Sep 21 '12 at 10:55
  • \$\begingroup\$ @Olin - I'm not sure you want to see the datasheet. IMO those from IR are pretty bad. This one doesn't even say whether it's an NMOS or PMOS FET, which should be in the title. \$\endgroup\$ – stevenvh Sep 21 '12 at 10:58
  • \$\begingroup\$ @stevenvh I managed to find something in the datasheet that suggests it's a N-channel. Page 7, Fig. 18: for N-Channel HEXFETPower MOSFETs \$\endgroup\$ – m.Alin Sep 21 '12 at 11:02
  • \$\begingroup\$ @steven: The reason I want the datasheet is to check how much current can be expected at 0 gate voltage, and also what the maximum D-S voltage really is. It's unlikely that 12 V with 3 LEDs in series is abusing the FET, but that's one thing I'd check. Most likely the 0 G-S voltage leakage is lighting the LEDs a little, but again, that's why we need the datasheet to see what should be expected. \$\endgroup\$ – Olin Lathrop Sep 21 '12 at 11:13
  • \$\begingroup\$ @m.Alin - Yes, and you can see it in the figures and graphs too, of course, but page 7, then you're almost at the packaging information. \$\endgroup\$ – stevenvh Sep 21 '12 at 11:14
1
\$\begingroup\$

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.

|improve this answer
\$\endgroup\$
1
\$\begingroup\$

I had the very same problem just now, and it turned out to be caused by ESD damage. To do a simple test, try measuring current flowing through the gate (you can use a standard multimeter for that). There should be no measurable current going through - the transistor is voltage controlled; if you see any current on the gate, that means you need to get another MOSFET.

|improve this answer
\$\endgroup\$
0
\$\begingroup\$

The IRF7832 is an enhancement FET, which means that you have to apply a positive voltage relative to the source to the gate. \$V_{GS(th)}\$ is maximum 2.3 V, so you'll need a higher voltage than that to be sure of 250 µA drain current. A volt higher should give you enough current to light the LEDs.

The current with the gate grounded is leakage current, and should be as low as possible. For most FETs that would be around 1 µA at 25 °C, but it increases exponentially with temperature. The datasheet gives a value of 150 µA at 125 °C, and that should be enough to get some light out of the LEDs. Are the FETs considerably hotter than 25 °C?

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ Just curious, why do you think the LEDs do light (very dim) with the gate tied to GND? \$\endgroup\$ – m.Alin Sep 21 '12 at 11:04
  • \$\begingroup\$ @m.Alin - that should be leakage, but it must be rather high to get some light out of the LEDs. The datasheet gives a value of 1 uA at 25 °C, but 150 uA at 125 °C. Anything in between is guessing. \$\endgroup\$ – stevenvh Sep 21 '12 at 11:20
  • \$\begingroup\$ Leakage change should about double every 10 degrees C. So 25C - 125C = 100C. 2^ (100/10) ~= 1000 so you'd expect 1000 uA leakage a 100C if it's 1 uA at 25C. Actual figure is not vastly different than that. \$\endgroup\$ – Russell McMahon Sep 21 '12 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.