Reduce voltage of DC adapter without reducing current

(replies to self)

No actually AFAIK the zeners don't resist current flow, so stiil not sure why the attempt mentioned above didn't work.

Lucky! It's been like a furnace out here, and it's gonna get hotter. Are you near Auckland? Christchurch? Someday I'd love to go skiing in South Island...

I would hav ethought that was obvious from the first post.... Why is the OP an idiot? He has a legit question... (And if it took you this long to figure it out, who is the idiot? :-) )

As for the OP's Q, why not use a regulator?

Look up part MC33269T-3.3G - 3.3V regulator, only needs a couple of caps, total cost < $1 US. WIll take nearly any reasonable DC in.

--Yan

Wasn't obvious to me. At first I thought the OP was talking about heating a single cup of water or something small like that. Didn't think about it deeply enough at first to realize a couple of 1.5V batteries wouldn't heat up even that small amount.

Mark

It just struck me as funny.... And I needed a laugh... :-)

I'm in Auckland.

Thanks but I don't think any circuit with resistors is going to work, see the post by John Fields.

Jasen Betts wrote:

Oh - I thought your DC adapter was 5.9 volts.

Here are some of the options I have tried: (good info snipped along with the misleading current measurements)

Ok - but your current readings are meaningless. Not your fault - it's the nature of the circuit. I believe the igniter is drawing *much* higher current, but for brief moments. The net effect is that your current measurements mislead you. What is meaningful is the works/does not work observation.

Many/most of these DC adapters have current capability well under 1 amp - and my bet is you need far more current than your adapters can provide. Try this: CAT# DCTX-5200 from

It will give you 5 volts at up to 2 amps. You can put a couple of 1N540x 3 amp diodes in series between the adapter and the igniter to put the voltage in the right neighborhood.

Ed

John fields knows his stuff, but look at the circuit.

you see the resistors aren't in the path that the current must take to power the igniter.

the only device blocking the current is the LM317.

I'm fairly sure John was referring to circuits where the resistors are in the supply path.

---

Please bottom post.

It's not that resistors keep it from working, I'm guessing that the current required from the source, during ignition, is so high that almost anything you put in series with the igniter is going to drop the voltage to the point where the arc won't strike.

Just considering that the spark will dissipate 12 watts at 12Kv, if it has 1 milliampere in it, means that the input current to the igniter must be greater than 4 amps at 3 volts during ignition. Easy for the batteries to supply, with their low impedance, but considerably harder for anything else.

So, it may also be that the source itself (the wall-wart) is incapable of supplying the short-term ignition current requirements of the igniter. At least the higher voltage ones. I recall you said the 5V one worked. no?

So what is it you're trying to do? Outfit a bunch of heaters with some junk wall-warts you've got hanging around?

-- John Fields Professional Circuit Designer

As do we all. Can't fault ya fer that.

Mark

That is correct.

I suspect this may be incorrect, because every wall wart I have tried lights the burners no problem, and most are rated at under 1 Amp.

The higher the adapter voltage, the faster it sparks and starts. However once I get up to 12V nominal adapters, it keeps sparking after ignition. I am going to use up all my 9V and less adapters to start with.

I am a bit concerned that overvolting the input by such an amount (9V nominal = 11V actual, compared to 3V specified) may shorten the units' life. Or will it just be dissipated as heat?

I may look into the voltage regulators and other circuits mentioned by others for the high voltage adapters, thanks all for your assistance.

John Fields wrote:

Thanks for your input, but as I say below every wall wart works no problem, and most are rated under 1 Amp. I don't think the diodes will work because they will have a resistance of a few ohms, and the resistance needs to be kept much lower than this as John Fields has noted in a couple of places.

ehsjr wrote:

Your reasoning is wrong. Diodes do not have a resistance of a few ohms.

You have two alternatives to try - the diodes and the voltage regulator circuit. Report back after you have tried them.

You can even get a kit from Dick Smith Electronics - K3594 that will give you 3V out with 12V in if you are reluctant to try to build one from scratch. With the confidence you gain from the $6.49 kit, you'll be able to build a regulator from the schematic I posted from scratch.

Try the diodes first - they are only 9 cents apiece from Dick Smith - Z3204. Figure 1 volt drop per diode. The exact drop will depend on the current drawn - these 1N4007 diodes drop ~1 volt at 1 amp.

Ed

Thank God!

I can't believe I had to read this far down before somebody menti> > Phil Allis>

Chaining forward biased rectifier diodes in series to the positive lead does the trick, but it's a bit of a messy solution. I still would like to know why adding a reverse biased zener diode to the positive lead does not work.

Oh I think I know why. I tried a 15.5V measured adapter and a 10V Zener diode, but under load this combination would drop to less than 3V. I will try a smaller Zener diode.