What is the science behind a dummy load?

[Fiddy]

no worries

 

[djQUAN]

you can use an ammeter to measure current but it's more risky due to the ammeters usually are connected via banana plugs which in time can become intermittent. A running driver with the load suddenly disconnected and reconnected in an instant could mean a blown laser diode. A resistor soldered inline with the diode string prevents this by having strong soldered connections and just measuring the voltage across it lets you safely measure and be able to calculate the current through that circuit.

 

[Cyparagon]

You can get ammeters for $3 shipped and solder it instead of using your regular DMM. The diode/resistor/DMM is just the simplest way to make a load - not necessarily the best.

 

[Wolfman29]

I'd just like to point out that NO4H99 isn't entirely wrong. Electrical energy is a type of potential energy, which is extracted from the battery then converted to heat, kinetic energy, stored electrical or magnetic potential energy (capacitors or inductors), electromagnetic energy, etc. But in essence, electricity really is sort of a potential, even though current, which is the typical way electrical energy is transferred, technically uses kinetic energy and a Coulomb potential.

From Wikipedia, "Electrical energy is energy newly derived from electrical potential energy."

But no, if a laser diode is a resistor, then resistors don't follow Ohm's Law =p

 

[NO4H99]

1) Laser diodes are not like resistors at all.
2) Resistors do not convert from potential energy.
3) Resistors do not convert energy to kinetic energy.
4) Resistors do not convert energy to light.
5) Laser diodes do not convert from potential energy.
6) Laser diodes do not convert energy to kinetic energy.
7) Electric circuits do not have kinetic energy.
8) Photons themselves may technically have kinetic energy, but light is electromagnetic energy.

This was just a terrible explanation, and you should be ashamed of yourself.



9) Electrons cannot change their charge.



10) Voltage is needed to "pass the junction," not energy.

Sorry, but I gotta defend the post

1) If you look at my last post (#10), I did not claim that they were the same, just similar. In that post I actually outlined the practical differences between the two.
2) Resistors do convert from the potential energy of a circuit (∂PE=q*∂Potential Voltage)
3) Resistors can convert to kinetic energy in various forms (heat energy etc)
4) I never said that resistors convert energy to light, I said that some diodes do.
5) Laser diodes do convert from the potential energy of a circuit (see #2)
6) Laser diodes do convert that potential energy into kinetic energy (in the form of electromagnet radiation [commonly referred to as light])
7) I never claimed that an electric field has kinetic energy. I said that it has potential energy (derived from the electric potential and electric field).
8) Light absolutely does have kinetic energy (E=hf, individual photons have energy, etc.) Why would you think that it doesn't?
9) I said that electrons can have different amounts of energy, not different amounts of charge. A perfect example of this is in the photoelectric effect.
10) More energy is needed to pass the junction, but you are correct in that this extra energy comes from extra voltage.

With all that said, I am sorry if any of my comments were misleading - especially comparing a resistor to a diode in my first post.

Wolfman29: if you look at my last post (#10 i think) I did say in the first paragraph that resistors follow ohm's law while diodes do not.

Hope this clears everything up :thanks:

 

[Zeebit]

:bumpit:

I can't find any precision 1 ohm resistors in all the local shops so I'd like to use my DMM instead.

We know that boost drivers do not like to be run without a load but how about buck drivers? I'm using the AX2002 driver from DX. Can these be hooked to a test load with a DMM instead of the 1 ohm resistor?

 

[The Lightning Stalker]

AFAIK, the resistor is not necessary at all. It is only there to facilitate current measurement. Just add
diodes to compensate for the lower voltage drop.