When do you need to heatsink?

[furrylaser]

I was searching around the thread, and I noticed a lot of people heatsink drivers. I couldn't find a definite answer for my question by searching, so here it is. At what mA would I need to heatsink it?




Thanks

 

[ARG]

What type of driver, what power source?

 

[furrylaser]

i was thinking of a LPC-826 at 500mA with a flexdrive. I will be using one 18650.

 

[lazeerer]

You dont need heatsink with a flexdrive up to 1A or around that.

Some drivers do not need heatsinking. It Just depends Really.

 

[Bionic-Badger]

The Flexdrive has efficiencies at about 90% or so, so only 10% of the power would be lost in the driver as heat, which isn't too much, even at "high" wattages (by low-power laser standards).

 

[Fiddy]

ive been heat-sinking flex-drives at 500mA and above.

from whats on the flexdrive manual:

For powering high current (>500mA) 1.8V diodes, a series shottky rectifier is required so that the output voltage is
2.2V or greater. When powering loads above 500 mA and below 3V, a heatsink on the black 5-pin chip on the back
of the driver is required. An electrically insulating thermal epoxy to the heatsink or casing is recommended.

the driver heatsinks i use are these:

 

[Isoleucine]

Instead of being slightly worthless and stating 'estimates' im going to APPLY math. Math is a LOT more useful these days. Lets apply some formulas.

AFAIK, most drivers and diodes have efficiency. Typically, the equation for power in watts IS

Watts = Voltage x Amperage

Say I'm using a 1 watt diode... aka the 445. (for simplicity's sake)
It pulls 1 AMP at 4.5 volts. The power is 4.5 watts. But it outputs ONLY 1 watt of power.
That means the 3.5 watts that are lost in the equation becomes heat, which needs to be dissipated away by a heatsink pretty fast. 3.5 watts is a lot of heat (power if you want to say).

Similarly, you can say the like for drivers. If I am using the MicroDrive or the Groove, (for simplicity) and Im feeding 4.5V to the diode, BUT I am imputting 9 volts, thats gonna be a conundrum to be sorted out, since the heat will quickly accumulate. if the diode's voltage is 4.5, and I feed it 6 volts through the driver, it will drop due to voltage drops in regulation, which will bring it down by 1.2 volts AFAIK form most common drivers leaving you with 4.8 volts. 0.3 Watts of heat is quite small and thus I can get away with not much of a problem. Using information from the groove's manual, you need to keep a dissipation of .75Watts of heat from the driver. If you feed it with an excess of Amps, you will generate more heat, coupled with voltage.

When you convert voltages, you lose some power in the conversion, resulting as heat. If you're using the laser momentarily, I dont think you'll need a heatsink, but for higher powered applications, you may need one.

Tell me if I'm wrong...

 

[BShanahan14rulz]

You could set it up and use one of your biological thermal conduction sensors to determine if the chip is heating up too much...

Isoleucine's idea is nice, assuming you can find the driver efficiency for the particular input power and load consumption power you will be running it at for the switching sepic or boost or buck types.

 

[Benm]

Well, your fingers arent that good a judge of thermal problems. These chips are often rated to operate at considerable temperature (70 celcius or more), so you are likely to burn your fingers on them even if within allowed range.

You could do the wet finger test, where the sizzling sound gives away a temperature over 100C, which is probably a point you really need to start cooling most components better.