A hot high side MOSFET

Thread Starter

gibby_z

Joined Aug 19, 2009
20
How can I drive a 700ma LuxDrive LED driver with a 510 MOSFET using it as a high side driver? The FET will be switched by a PIC at 5v. The LuxDrive is using 18v. I am not using PWM. I tried a few random schematics that only resulted in the FET getting hot as hell pretty quick.
 

beenthere

Joined Apr 20, 2004
15,819
The short answer is that an IRF510 is unsuited. For full conduction, Vgs has to be 10 volts or higher. You only have 5 volts to work with, and, with the load in the source, Vgs goes down as the FET starts to conduct.

Even as a grounded source, you will still need to arrange for 10 - 18 volts to appear on the gate to place the FET into full conduction.

Oops -should have mentioned that there are logic level FET's that the PIC can place into full conduction. I use through hole devices, so the VN10LP is the one I use. They typically are good for up to 30 volts and 200- 300 ma. Use one to pull the gate on the IRF510 to ground and turn it off.
 
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SgtWookie

Joined Jul 17, 2007
22,230
As Beenthere asserted, the IRF510 is not suited for logic-level control.

How can I drive a 700ma LuxDrive LED driver with a 510 MOSFET using it as a high side driver?
I don't understand why you would want to use a MOSFET to control the high side of a BuckPuck.

It would make a lot more sense to use a logic level MOSFET to control the input ground side of the BuckPuck.
The FET will be switched by a PIC at 5v. The LuxDrive is using 18v. I am not using PWM. I tried a few random schematics that only resulted in the FET getting hot as hell pretty quick.
OK, the first thing to do to eliminate any doubt about what you are doing is to post a schematic; that's worth a thousand words.

If you want a nifty logic-level N-channel power MOSFET, have a look at an IRLD014; it's in a 4-pin DIP package rated as capable of sinking up to 1.7A, but I personally wouldn't go beyond 1A without some kind of heat sinking.
Jameco.com, Mouser.com, and Digikey.com among other distributors sell these little jewels.
 

Thread Starter

gibby_z

Joined Aug 19, 2009
20
I dont have a schematic yet because I'm not sure how im going to do this yet but I'll get one started.

The mosfet needs to be high side because there is another buckpuck running in parallel. The buckpucks seem to share their ground with the LED ground so if one puck is grounded the both fire up.
 

SgtWookie

Joined Jul 17, 2007
22,230
How about a NTE 2890 logic level MOSFET
That part number is not valid.

Avoid using NTE parts whenever possible, as you will pay a hefty premium above manufacturer's prices.

NTE doesn't make their own parts; they have manufacturers make a run with NTE's labels and part numbers on them when the manufacturer is getting ready to discontinue a particular item. Once the supplies of the original manufacturer's item is consumed, NTE increases their prices on their remaining stock as much as tenfold - sometimes even more.

NTE has a part number cross-reference that only goes one way; from other manufacturer's part numbers to theirs. Once you install an NTE part, the link to the original manufacturer's part number is gone.

A couple of years ago, I went to purchase some LM3914N IC's while I was visiting my parents up north. A local store didn't have the original National Semiconductor part, but did have an NTE equivalent; NTE1508. They were asking over $10 each for them! I told them to forget it; went home and ordered 10 of the originals from a major distributor for the same price.
 

SgtWookie

Joined Jul 17, 2007
22,230
I dont have a schematic yet because I'm not sure how im going to do this yet but I'll get one started.
Your schematic is your road map. Without it, you're bound to be lost. ;)

If it's not documented, it doesn't exist.

The mosfet needs to be high side because there is another buckpuck running in parallel. The buckpucks seem to share their ground with the LED ground so if one puck is grounded the both fire up.
Now this is strange - yet not so strange.

OK, document away - it really helps to have a schematic.

If you don't have anything else, you could use MS Paint. Save it as a .png file. Use the "Go Advanced" button below the reply text box, then click the "Manage Attachments" button to select and upload your graphic schematic.
 

k7elp60

Joined Nov 4, 2008
562
While im asking, could an opto coupler trigger the FET?
No for the circuit you are talking about. One type of opto coupler is a IR diode and a phototransistor in one package. When the IR diode conducts it sends a beam to the transistor that turns it on. The main use is for isolation.
If you had a seperate powersupply with enough voltage for the gate to source voltage for the fet the opto coupler could turn on the fet.

As a side note. A number of years ago before P channel mosfets were abundant I used to use N channel mosfets on occasion for highside switches. I would use a voltage pump to boost the supply voltage(generally 12V) to 24 and that would turn on the N channel mosfets.
 

SgtWookie

Joined Jul 17, 2007
22,230
You could use an NPN transistor to turn on a P-channel MOSFET that had a Zener diode on it's gate and a pull-up resistor.

But before I post more details, I want to see your schematic.
 
If you can use P-channel FETs, there's an idea I've been thinking about for a while. Use a 7905 (negative!) voltage regulator to produce a logic "ground" that's 5V lower than your LED power supply. The PIC will run on the high side, and can turn P-channel devices on and off. SgtWookie's observations about logic-level FETs are still applicable, but you have a better chance of running even a -10V P-ch FET with the 5V swing of a PIC if the PIC sits on the high side with the FET.

I've actually built this circuit with an Atmel microcontroller switching a P-channel FET to control 12V to an incandescent automotive bulb. Works like a charm!
 

SgtWookie

Joined Jul 17, 2007
22,230
OK, here's what I was talking about:



The box labeled "V2" on the lower left represents your PIC output logic signal. It's signal can be viewed as (B), the green trace, below on the simulated O-scope display.

R1 limits the base current to Q2 and the output current from your PIC to under 5mA. Q2 could be a variety of small-signal NPN transistors; 2N3904, 2N4401, etc.

R3 limits the current sunk through D1 and R3 to around 20mA.

D1 is a 10v Zener diode; you could use a 12v Zener if you wished. The point of the Zener is to limit the maximum Vgs of the MOSFET.

R2 keeps the gate of the MOSFET Q1 pulled up to 18v when Q2 is turned off.

The output of the MOSFET can be seen below as the yellow (A) trace.
Rload represents your Buckpuck.
 

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Thread Starter

gibby_z

Joined Aug 19, 2009
20
I just noticed that the IRFI9520GPBF has a Gate-Source Breakdown Voltage of 20 V. My power source is a Nimh battery pack with a possible charger resh voltage of 24 volts. Will this effect what part I need to use? If it does what alternative mosfet can I use here?
 

SgtWookie

Joined Jul 17, 2007
22,230
No, it won't affect it.

See D1, the 1N4740 Zener diode between the gate and the source? That clamps the maximum Vgs to -10v. That's why it's there.
 

Audioguru

Joined Dec 20, 2007
11,248
Nobody commented on the schematic posted from Gibby with IRF510 N-channel Mosfets as source-followers and driven with only 5V from a PIC.
The output will be zero because the gate voltage must be 10V higher than the source voltage so the gate must be 32V, not 5V.
 

SgtWookie

Joined Jul 17, 2007
22,230
Audioguru,
That's precisely why we were steering Gibby towards using a P-channel MOSFET.

Much easier than fiddling around trying to make a high-side driver for an N-channel MOSFET.
 
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