I'm looking at a constant current MOSFET circuit as kindly described here.

What are the key things to consider when selecting the MOSFET?

I'm looking to dissipate a maximum of about 700mA at 1.4V through the MOSFET.

 

First, the mosfet has to be rated to survive the voltage. This circuit will need about 6 volts minimum. (I don't think you can find a mosfet that won't survive 6 or 12 volts.)

Second, you will need to get rid of a watt worth of heat, and that is listed on the datasheet, usually. Sometimes they make it difficult by claiming some huge number of watts that can only happen with a massive heat sink and a fan. Then you have to look at the "Thermal Resistance" which might be something like Rth Ja, Junction to ambient = 25C/W. That means it will heat up by 25 degrees centigrade per watt. Then you look at the maximum junction temperature before it melts. That's usually around 175 C. So, you figure this thing is going to heat up by 25C and the room temperature is already 25C, so the junction temperature will be 50C. Well within the safety margin.

To make this simple, buy a TO-220 or TO-247 package mosfet. They are about the size of a penny, only fatter and they will survive a watt, all day, without any help from a heat sink.

ps, There are things called, "Logic Level" mosfets that have a lower need for gate voltage to turn them on, less than 5 volts. If you want to keep the required power supply to the lowest voltage possible, look for one of those. There is also the fact that most mosfet gates die if you put more than 20 volts on them (compared to the source) but this circuit is arranged so that is not a concern.

 

Many thanks, seems almost too simple. As I understand it, the MOSFET dissipates heat through Rds(on) and switching losses. With this circuit configuration, will any MOSFET be able to dissipate any amount of power (assuming it is voltage tolerant and doesn't melt)? Or is there a ceiling (set perhaps by the maximum switching frequency/loss)?

 

Many thanks, seems almost too simple. As I understand it, the MOSFET dissipates heat through Rds(on) and switching losses. With this circuit configuration, will any MOSFET be able to dissipate any amount of power (assuming it is voltage tolerant and doesn't melt)? Or is there a ceiling (set perhaps by the maximum switching frequency/loss)?

I think when this circuit in current limiting mode, the power dissipation of the mosfet is, P = Vds * Ids (that's the main part you will need to do the calculation, bases on your load current and supply voltage, and the voltage drop of Rsense, if you are not switching)

when the mosfet is fully on, P = (Ids)^2 * Rds(on)

 

To make this simple, buy a TO-220 or TO-247 package mosfet. They are about the size of a penny, only fatter and they will survive a watt, all day, without any help from a heat sink.

I'd like to add to this if and only if they are in freely convecting air of 25 degrees (C). If you put them in a sealed housing, like a small plastic box, they will heat up the air in there and you risk thermal runaway, especially if Rds is only a small portion of the total resistance of the circuit.

 

That circuit you posted doesn't do any switching, and yes, there is a limit. That's why I explained to you how to calculate the limit by using Thermal resistance (from junction to air) times the wattage.

 

Thanks all.

So does the circuit reach equilibrium by having the NPN transistor partially on, pulling down Vgs (to just above Vgs(th)) to increase Rds(on) until its resistance is just right? So the MOSFET basically acts like series resistance?

In terms of thermal runaway in a small case, as I understand it, Rds(on) increases with temperature, so the circuit would increase Vgs to compensate (keeps the resistance the same). So does thermal runaway simply mean overheat and melt/fail (as oppose to LED thermal runaway where the current rises with temperature)? Will there be any current change as temperature increases?

 

Thanks all.

So does the circuit reach equilibrium by having the NPN transistor partially on, pulling down Vgs (to just above Vgs(th)) to increase Rds(on) until its resistance is just right? So the MOSFET basically acts like series resistance?

In terms of thermal runaway in a small case, as I understand it, Rds(on) increases with temperature, so the circuit would increase Vgs to compensate (keeps the resistance the same). So does thermal runaway simply mean overheat and melt/fail (as oppose to LED thermal runaway where the current rises with temperature)? Will there be any current change as temperature increases?

I think you understand thermal runaway. It means that, as a component dissipates power and gets hot, the temperature coefficient of it or of some other component in the circuit causes it to dissipate even more power and get even hotter. This is a vicious circle, and can result in the destruction of the component.
This current sink is not subject to that sort of thermal runaway.