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Higher forward voltage = lower efficiency?

ryansoh3

ryansoh3

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Hi LPF,

I was just wondering if a laser diode/LED's forward voltage is high, is the efficiency worse?

I came up with this idea because a constant current driver adjusts the voltage to match the required current. However, if the forward voltage is higher, the driver needs to increase its voltage is well. Will this result in a higher current draw from the batteries? Just wondering. :thinking:

Cheers! :beer:
 





T

Things

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Not necessarily, this is why the unit Watts exists, as it's a unit of overall power.

For example, 1A at 10V is 10W, but 10mA at 1000V is still 10W. 100uA at 100kV is still 10W :)

But yes, it depends on the driver. Boosting to a higher voltage will require more current.
 
ryansoh3

ryansoh3

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Things said:
Not necessarily, this is why the unit Watts exists, as it's a unit of overall power.

For example, 1A at 10V is 10W, but 10mA at 1000V is still 10W. 100uA at 100kV is still 10W :)

But yes, it depends on the driver. Boosting to a higher voltage will require more current.
Click to expand...

I understand. But a diode with higher Vf will consume more watts, right? (same current but more volts)
 
T

Things

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Nope, for current limited devices like laser diodes, current goes down as voltage goes up (Driven, of course, not if you connect it to an unregulated supply). In theory higher Vf is actually more efficient as there are lower losses in wires, but for what we're dealing with they're negligible.
 
BShanahan14rulz

BShanahan14rulz

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I would have thought it went the other way around, i.e. the voltage is used to bump up the electrons, how much voltage is determined by the molecule's band gap, and the extra voltage is basically soaked up in the material as heat? Looks like GaN(III) band gap is around 3.4eV
 
Cyparagon

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The driver always has a current sense resistor. It compares the voltage across the sense resistor with a set point. If the current it senses is marginally lower than the set point at any given time, it will raise the output voltage and subsequently the output current. Voltage technically controls current.

Efficiency is simply (power in)/(power out)
(power in) is V * I so:
efficiency = Vf * I /(power out)

You can't say efficiency is a function of forward voltage. You'd be ignoring input current and output power.
 
ryansoh3

ryansoh3

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Cyparagon said:
Efficiency is simply (power in)/(power out)
(power in) is V * I so:
efficiency = Vf * I /(power out)

You can't say efficiency is a function of forward voltage. You'd be ignoring input current and output power.
Click to expand...

Hmm... That's true, but

1. The input current is going to rise as a result of more power consumption of the driver (due to higher Vf but same current, say 1.8A), is it not?

2. Power output from the laser diode is a tricky subject. If 1.8A is applied to all M140's, does the Vf have anything to do with it's output measured via an LPM?
 
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