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FrozenGate by Avery

Driving Laser with Arduino






0.7V is the BE or CB drop, 1.4V (two junctions in series) is the CE drop, even in saturation (at least for the 2N222A's I've got here). There's always at least a single diode drop in Bipolar technology since it is always a P/N (or N/P) junction involved. That's why FETs are preferred for switching until you're dealing with ultra high currents (power lost to heat per unit of current is linear for bipolars but exponential for FETs).

For example in IGBT's you take the threshold voltage, which is the voltage drop across the CE junction and multiply it by the current the switch is carrying (for Pdiss calculation when in Saturation only). If Vth is 2V, then at 10A you're burning 20W of power as heat. For a FET it would be Pd = I * I * RdsON. Assume a 0.1 ohm RdsON, at 10A you're burning 10W of power as heat. Now change that to 1000A. The FET calculation would be: (1000 x 1000) x 0.1 = 100,000W of waste. The IGBT calculation would be: 1000 x 2 = 2000W of waste.

When you're pulling a signal off from the collector or emitter you might only see a single diode drop depending on architecture of your circuit, but when using a series load there will be a dual diode voltage differential.
 
Sigurthr said:
0.7V is the BE or CB drop, 1.4V (two junctions in series) is the CE drop, even in saturation (at least for the 2N222A's I've got here). There's always at least a single diode drop in Bipolar technology since it is always a P/N (or N/P) junction involved. That's why FETs are preferred for switching until you're dealing with ultra high currents (power lost to heat per unit of current is linear for bipolars but exponential for FETs).
Click to expand...

When in saturation both junctions are forward biased, so it's not 0.7V + 0.7V, it's 0.7V - 0.7V (in theory). In practice the difference between the voltages is ~0.1V

Just take a look at any VCE (SAT) voltage on a datasheet. This 2N3904 lists the VCE(SAT) as 0.2V
 
The numbers I used are just typical anecdotal values. The theme still stands that accounting for a Vce drop is important. That being said I typically don't work with BJTs and haven't myself had any which have <1V Vce(sat) at appreciable Ic currents.

Here's a screencap from the datasheet for P2N2222A NPN BJTs (a variant of the 2N2222):
RFLPoDN.png


It shows 1V Vce for high current. The particular 2N2222 variants I have here are 1.4V Vce (old metal can). I just verified it using 12V Vcc, 6V Vb, 120mA Ib, 700mA Ic.
 
Ah, you are right :) but I think for OP's purposes somewhere around 0.4V would be more reasonable to expect than 1.4V
 
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Is it possible to be burning out transistor's with the current/voltage I'm working with here? I ask because I found that sometimes I measure the current/voltage to be 0 but it's solved by switching transistors.
 
senuba91 said:
Is it possible to be burning out transistor's with the current/voltage I'm working with here? I ask because I found that sometimes I measure the current/voltage to be 0 but it's solved by switching transistors.
Click to expand...

Yes, entirely. Some 2N2222 variants are only rated for 200mW (not mA!).

If you're going to be using the arduino to switch the laser module for something important or for a long period of time I would look into either changing to a low power red laser module instead of a wasteful dpss green, or using a logic level mosfet instead of a bipolar transistor.
 
You never said what kind of laser module you were
using. I just assumed that it was one of the 5mW
red ones (80mA) or a green one, that typically only
draws 300mA max.
 
Sorry, the laser module runs on 3V and 250mA. The laser output is 200mw at 532nm (green). I'll look into the MOSFET; I anticipate this'll be pulsing in excess of a few minutes. Would the MOSFET work relatively the same way as the NPN?

EDIT:

I set up the circuit with the three AA batteries in series and it worked! However, the laser was pretty dull compared to when I hook it up directly with the batteries. Also, the laser doesn't turn on quickly; it slowly lights up. Am I right in assuming the cause for this is 1) the transistor or 2) the power source not having enough current or voltage?
 
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senuba91 said:
Sorry, the laser module runs on 3V and 250mA. The laser output is 200mw at 532nm (green). I'll look into the MOSFET; I anticipate this'll be pulsing in excess of a few minutes. Would the MOSFET work relatively the same way as the NPN?
Click to expand...

The Logic level N channel MOSFET will work identically but you'll need one additional component; a discharge resistor that goes between Gate and Source pins, value should be roughly between 1K and 10K. This ensures the FET turns off when the arduino stops holding the output pin HIGH.

senuba91 said:
EDIT:

I set up the circuit with the three AA batteries in series and it worked! However, the laser was pretty dull compared to when I hook it up directly with the batteries. Also, the laser doesn't turn on quickly; it slowly lights up. Am I right in assuming the cause for this is 1) the transistor or 2) the power source not having enough current or voltage?
Click to expand...

It was dull because there probably wasn't enough voltage to fully forward bias the laser pump diode. It has nothing to do with current from the battery (unless the arduino can't source enough current to fully turn on the NPN, which is possible, but if that was so then switching batteries probably wouldn't have helped any), both types are capable of sending several amps of current. Either way, switching to the MOSFET removes the voltage drop across the transistor and ensures long life (250mA @ more than 2V is more than most 2N2222s can handle, and none are rated for 250mA @ 4.5V).

Just make sure you get a logic level n channel fet because standard power fets require 10V for the gate signal to fully turn on.
 
I've been searching for the right mosfet, but am having no luck. How would I know what specs would future my purposes?
 
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