driver replacement

These are all of the old posts from the first two years of the forum. They are locked.
Updated: 2005-03-28 by HoloM (the god)
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Thilo K.

driver replacement

Post by Thilo K. »

Hello. Maybe there´s a topic that is already dealing with the following idea, but I didn´t find it, so I post it by myself. If it has already been published somewhere, please tell me.

What is it all about ? I think, a driver is just for giving current to a diode between a voltage(Hope one can write it that way). Could you just use a 'stabilized' power supply unit to supply the diode without any kind of additional driver. 'stabilized' power supply unit means that voltage and current remain almost stable all the time.
If it works, which pins will have to be connected to plus, which one(s) to minus ?
Tom B.

driver replacement

Post by Tom B. »

I have seen some reports of bad experiences with using
lab power supplies to drive diodes but I have had good
results so far. The most common problems are with
over and undershoots of the supply voltage when
turning it on and off which can destroy the diode.
This risk can be reduced by placing capacitors and diodes at strategic points in the circuit to absorb
the bad stuff.

The North American term for "stabilized" is "regulated"
- same thing. Laboratory supplies usually have two
knobs, one to adjust the voltage output, and the other
to set a maximum current limit to protect the supply
and the circuit under test from damage if there is an
accidental short circuit. Usually the voltage regulation
is well specified, but the current limit is not, so
I prefer to use a resistor in series with the diode
to limit the current. When driven, the diode has a
so called "forward voltage" which can be obtained
from the diode data sheet. The resistor absorbs the
difference between the diode forward voltage and the
oower supply voltage. For a given power supply voltage
V the required resistor value can be calculated from Ohm's law as R = (V - diode forward voltage) / current (in Amps). For example with a desired power supply voltage of 5V, diode voltage of 2.75 V and required current of 0.08 Amps, the required resistor would be (5 - 2.75)/0.08 = 28.125 Ohms, the closest standard value
being 27 Ohms.


What is it all about ? I think, a driver is just for giving current to a diode between a voltage(Hope one can write it that way). Could you just use a 'stabilized' power supply unit to supply the diode without any kind of additional driver. 'stabilized' power supply unit means that voltage and current remain almost stable all the time.
If it works, which pins will have to be connected to plus, which one(s) to minus ?
Tom B.

driver replacement

Post by Tom B. »

Sorry, looks like I was editing my little
essay there and hit respond by mistake -
maybe I'll leave it at that for now until
I know if I'm even coming close to answering
the original question, whatever it was.
Thilo K.

driver replacement

Post by Thilo K. »

Thanks so far. First of all, i´m happy that it works as a matter of course.

You wrote "..can be reduced by placing capacitors.."
What kind of capacitors do you think of and where. Can you include a sketch into your next answer showing your circuit completely ?

Maybe you´ll include it, but I´d also like to know, which pins will have to be connected to plus, which one(s) to minus ?
Tom B.

driver replacement

Post by Tom B. »

Sorry, I don't know how to include images and my attempt at an ASCII drawing was hopeless. Anyway, here's a verbal description of a scheme I have used: Power supply (-) output is ground. Connect 1 Amp silicon diode between power supply + and -, with cathode (the end with band or other marking) connected to +. This partly absorbs negative transients on the supply output. Connect resistor to power supply (+) output. Connect other end of resistor and ground to four components, wired in parallel: 1) 1000 uF electrolytic capacitor, + side to positive voltage 2) 0.1 uF ceramic capacitor, 3) power Schottky diode, cathode connected to + side - this absorbs negative transients that were not absorbed by the first silicon diode. 4) the laser diode, cathode to ground, anode to the + side. You will need the laser diode data sheet to know which pins to connect. As pointed out elsewhere, you will need an optical power meter to safely adjust the diode current or be prepared to lose a few diodes. Lots more info here:
http://www.misty.com/people/don/laserdps.htm
Thilo K.

driver replacement

Post by Thilo K. »

I don't know a great deal about electronics so to solder the circuit I think a sketch is the only useful method for me. That means, a normal sketch where all cables lead to plus or minus finally and not to a 'ground', which leads to nirvana for me
Additionally, i´ll be able to build it faster, if there are given all properties of the parts representing the whole sketch. I.e., Values of resistors and so on - and maybe names to identify parts exactly.
Then again we have to cope with the problem of sending the sketch to me - why not via eMail ?
Here it is: sketch@das-hologramm.de

Greetings
Thilo K.
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