really dumb question time

SUPER dumb question. defiantly not doing this, i just thought about it and now I'm wondering what would happen. what would happen if you just connected a battery straight to a diode? no driver.
thanks

Instant death of the diode.

that sounds like a bad time

Yeah. Diodes are current hungry, so unless it is a button cell that cannot physically supply enough current to kill the diode it will draw as much current as possible and fry itself.

It is like giving someone in a lot of pain an infinite amount of morphine with no limit. They will use it all and die. But if you have a nurse to administer it(the driver) they will be happy and stay alive.

thats a nice analogy. makes sense to me thanks

what if you connect a diode and driver to batteries? no host? that should work right? no diode destruction?

Provided the diode is heat sinked properly and the electronics aren't allowed to short out, yes.

adrunkhobo said:

what if you connect a diode and driver to batteries? no host? that should work right? no diode destruction?

Click to expand...

Let's extend the analogy a bit. If you house your patient outside rather than in a hospital made to be the correct environment(host) then your patient will fare far worse than if you give him shelter. Same with the diode. Diodes get hot, fast. The reason we put them in heatsinks is to whisk away the unwanted heat. They are delicate components, and if left to deal with the heat the die on the diode could crack or become otherwise damaged. If there is a host though, the heat doesn't build up as quickly because it is conducted away.

Direct drive is really not that simple. It can be done .. but it depends on what voltage the battery has, and how it will change with current .. and also what diode it is.

Diodes have mostly linear voltage-current response (in the open direction). If you slowly increase voltage, first there will be only little current flowing through .. they they will slowly start glow in LED mode, and from then they will start to lase. Since that with increasing voltage the current will raise linearly, the power of the beam will rise more or less linearly with power (voltage * current) .. and at some point the diode will burn.

There is some sweet spot for current, which also means sweet spot for voltage. Working voltage mostly depends of wavelength of the light. It's under 2V for IR diodes, and over 4V for blue diodes.

We mostly use current drivers for diodes though. Why ? If you put constant voltage to the diode, the diode will decrease it's virtual resistance as it will heat up. So while the voltage will be the same, the current will not be the same, it will raise. Power will rise .. and the diode might run out of control, we call it thermal runaway.

With current drivers there is no such problem. If the diode heats up, it will need less voltage for the same current .. so we will actually lower the power output, and the situation is stable.

So what if we really wanted to direct drive a diode ?
We want to have battery and diode matched together. The diode voltage sweet spot should be roughly the same as the battery output voltage. The battery voltage for sure must not be larger then safe voltage for the diode. Putting IR diode to 9V battery - instant death.

Most batteries wont keep their voltage under load. The voltage will drop as the current drawn from the battery will increase. And powerful diodes will take TONS of current from the battery. The voltage drop is quite significant for common batteries, not so much for Li-ions. This characteristic is actually good .. cause it may work as a safety against thermal diode runaway. But unless you properly measure the battery UV plot, it's really nothing to rely upon.

And since thermal runaway is always a threat, we should better not push the diode to it's limits, and we should cool it properly.

Now if we talk IR or Red .. those diodes are really flimsy, you look at them wrong way, and they are gone. You want something which can handle large voltages, large currents, large powers. This lowers chance your abuse will kill it. 445nm 1W or even 3W diodes .. those are tough beasts (at least in world of laser diodes). Their working voltage is around 4.5V .. they lase usually from 3.5V.

You can quite safely direct drive 445nm 1W diode from 3x AA batteries, or 3x AAA .. and you can do it with single Li-ion cell. I'm saying this cause I did it many times, diodes are as good as new. Does it mean you wont kill yours ? Can't tell .. it is $50 part .. it's your call. I'm just saying I did it and I'm not afraid doing it again.

Anyway .. biggest problem with these is that the battery voltage will also drop as the battery gets discharged. While voltage of unloaded battery does not change much, voltage of battery under few A current changes drastically. With driver you can pull like 5 times more energy from the battery, then without it. It's not a big problem for small laser as that direct drive built of mine .. or if you just want to test if the diode works.

Btw. there is a thread about one of direct drive built in my sig ..

As for no host, ie. no cooling - don't ever think about that .. diodes are simply not designed to run without cooling .. especially not the strong ones. 445nm can run for few seconds with module alone at lower power settings .. like 500mW .. but it gets hot real fast .. but hey, it's just $50 ..

thanks for the responses guys. not trying it, just had me thinking.

I actually built the kipkay maglite laser way back when, which is a direct battery to diode build. The diode is from a dvd burner, and I have no idea what the specs are. I have had this laser for 7 years and it still works fine. It runs off 2 AA batteries or 1 Li-Ion battery, and there's no heatsink. It meters around 150mw and it does burn. After learning how to build a real laser, I'm actually amazed this thing still works. I'm going to eventually get this into a real host with a driver.

So I guess it's just luck that this particular diode's "safe" voltage is at least 3-4 volts. As far as heatsinking, I guess I've just never run it to the breaking point.

Thanks for the detailed explanation DrSid.