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FrozenGate by Avery

FS: ***Selectable test load up to 3A (assembled or DIY kit)*** IN STOCK!






It only took 2 days to receive mine and this thing is top quality I am a very happy! it was a nice little present on christmas eve.
 
Hey, this might sound like a real dumbass question, but how do I know what jumper to use? I know the diode limits the voltage of the input and the resistors limit the current, but how do I know what jumper to use to replicate the load of a diode? is it per wavelength of the diode, or is it current rating of the diode? say I wanted to push a 445nm diode to around 1.8 amps, what location would the jumper sit. I'm new to test loads and I know a little bit about circuitry, but this one has me stumped. :/
Looking forward to your reply and I'm thinking of getting one of these. Also, the resistors are rated at a maximum of 3 amps, but would I be better of using a lower current test load to test low amperage driver settings?
 
Hey Roger no problem, basically you just want to get as close as possible to the voltage drop the laser diode. It depends which 445nm diode you'll be using some of them will drop 4.5V or more at that current. The diodes on the test load will drop about .7V-.9V depending on the current each so you just multiply the voltage drop and number of diodes to reach as close to the voltage drop of the laser diode.

For example you wanted to use your 445nm diode at about 1.8A

0.9V x 5 = 4.5V
0.9V x 6 = 5.4V

So you could use 5 or 6. There is a number representing the number of diodes on the side of the jumper pins.

And don't worry too much about using a lower amperage test load. As long as you know the voltage drop for each diode you should be fine. There's a voltage drop chart a few pages back that will show you how much voltage the diode will drop at a given current.
 
Thanks for the reply. Would I need to PM you to inquire about shipping costs to Australia or is it alright to ask here?
 
Jufran88 said:
Hey Roger no problem, basically you just want to get as close as possible to the voltage drop the laser diode. It depends which 445nm diode you'll be using some of them will drop 4.5V or more at that current. The diodes on the test load will drop about .7V-.9V depending on the current each so you just multiply the voltage drop and number of diodes to reach as close to the voltage drop of the laser diode.

For example you wanted to use your 445nm diode at about 1.8A

0.9V x 5 = 4.5V
0.9V x 6 = 5.4V

So you could use 5 or 6. There is a number representing the number of diodes on the side of the jumper pins.

And don't worry too much about using a lower amperage test load. As long as you know the voltage drop for each diode you should be fine. There's a voltage drop chart a few pages back that will show you how much voltage the diode will drop at a given current.
Click to expand...

The calculation is flawed. You forgot the voltage drop of the 1Ohm resistor.
First subtract the voltage drop from 4.5V. That is 4.5-1.8=2.7Volt
Then estimate the number of diodes for the 2.7V ~ 4 diodes !
 
Blord said:
The calculation is flawed. You forgot the voltage drop of the 1Ohm resistor.
First subtract the voltage drop from 4.5V. That is 4.5-1.8=2.7Volt
Then estimate the number of diodes for the 2.7V ~ 4 diodes !
Click to expand...

So, by this calculation, the Vf of the resistor is 1.8? I am confused..I understand how many diodes you're determining. I'm not sure exactly how you determined it.

Jufran, I would like one of your DIY kits, if you have one available. PM me if that's the case.
 
@blord: Ah, yes I forgot. :o


@scumbagatheist: Vf of resistor is calculated by Ohm's law (V=IR). Since resistance is 1 with 2% tolerance, V=I or Vf = I

It could be a little higher or lower depending on the actual resistance, but it shouldn't vary that much.

PM incoming. :beer:
 
Ok, but in Blord's calculation, he say subtract the Vf of the resistor (1.8) from 4.5 (presumably the diode's Vf) and has a sum of 2.7V. Why is the Vf of the resistor being subtracted from the diode Vf? I would have expected it to be additive, that is that each of the components in the circuit (diodes and resistor) are all contributing to the load on the driver. Is this not the case? And how did Blord get 1.8 in his calculation if the Vf of the resistor is apx 1V?

Thank you all for helping me learn this. It's better than a course as I can ask these kinds of questions. I think I'll sign up for one of the free online courses in electronics.
 
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