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ArcticMyst Security by Avery

Power Supply

Joined
Apr 4, 2011
Messages
596
Points
28
You do have a test load, Right?

Yes, I have... many parts for test loads. haha When I ordered 6 1n4001 diodes, what I didn't notice, is each item was for 10 pieces... so I have 60 diodes haha. So I had to buy a bunch of 1 Ohm 1 Watt resistors to go with it.

From what I've read, I measure mV across the resistor to see what current is running into (or will be running into) the diode. I know I haven't done all the research I can on this part yet, and I don't fully understand how or why that works. I assume that it works because the 6 diodes(in this case) in series generates a similar demand to that of a diode and the resistor causes 1mV to be ~1mA for measuring purposes.

I know the current coming in is simply "supplied" from the power supply and the circuit takes whatever it can - the power supply does not force feed current into the circuit.

Please correct me if I'm wrong.
 





Toke

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Joined
Jul 25, 2010
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It is a bit tricky, the power supply creates a voltage difference between + and - .

A resistor is a conduit for current, the higher the resistance the small the hole for the current. The voltage drop across the resistor increase linearly with the current. Diodes act a bit different, as once it reaches it's Vf the voltage drop increase little with current.

Your current regulating lm317 is different from the power supply in that it regulates the voltage in an attempt to reach the set current. Or to be more accurate it is trying to get 1.25V between the output and the adj pin. The only way to do that is to get the correct current through your setting resistor, e.g. 1.25A for a 1 Ohm resistor.
 
Joined
Apr 4, 2011
Messages
596
Points
28
It is a bit tricky, the power supply creates a voltage difference between + and - .

A resistor is a conduit for current, the higher the resistance the small the hole for the current. The voltage drop across the resistor increase linearly with the current. Diodes act a bit different, as once it reaches it's Vf the voltage drop increase little with current.

Your current regulating lm317 is different from the power supply in that it regulates the voltage in an attempt to reach the set current. Or to be more accurate it is trying to get 1.25V between the output and the adj pin. The only way to do that is to get the correct current through your setting resistor, e.g. 1.25A for a 1 Ohm resistor.

Ahhhh that makes a lot more sense why test loads are built like that. Thank you for explaining that!

I'll actually be able to understand what's happening in my circuit now. :)

One other question I had... you mentioned you were using Qucs for the circuit simulation. I've tried others, but they didn't have a lot of certain parts/diodes/IC/resistors etc. Does Qucs have a lot of those or do you have to manually program the specs in?

:thanks:
 

Toke

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Joined
Jul 25, 2010
Messages
1,099
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One other question I had... you mentioned you were using Qucs for the circuit simulation. I've tried others, but they didn't have a lot of certain parts/diodes/IC/resistors etc. Does Qucs have a lot of those or do you have to manually program the specs in?

There are lots of components, you just have to set the specs for each.
e.g. a resistor is 50 Ohm by default, you click on it and type in the desired value if different.

It took me a while to locate the right menu for all the components, and I have still not gotten it to simulate a circuit even though I have a nice drawing of one.
The program runs under linux (Ubuntu)
 
Joined
Apr 4, 2011
Messages
596
Points
28
It's been a while, but here's an update... I don't know why this didn't even occur before, but running LM317s in parallel are actually giving me good control over voltage from 2v - 10v.
I am able to successfully calibrate with my dummy loads and lase with:
16x Sony 650nm LD at ~300mA
A140 445nm LD ~1200mA
Dummy Load for 3w 808nm ~3000mA (no lase of course... this is a dummy load)
 
Joined
Sep 12, 2007
Messages
9,399
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113
You don't need a dummy load for linear regulators. You can just short the output with an ammeter.
 
Joined
Apr 4, 2011
Messages
596
Points
28
You don't need a dummy load for linear regulators. You can just short the output with an ammeter.

I'm only regulating the voltage at the moment, so wouldn't I need the load to imitate the draw of amps? Or am I missing something...? Once I have another regulator to limit amps, I can see how I wouldn't need a dummy load...

haha well... there is still much testing to do :)
 




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