Reduce voltage of DC adapter without reducing current

[suputnic]

The OP's problem is that he doesn't understand that in order to bump

a few volts up to the 15kV or so required to strike the arc required
by the ignitor he'll need to keep the impedance of the substitute
supply down to what batteries look like, milliohms.

Because of that, his assessment of the current required to do the
job is flawed.

The current I measure is after ignition....I'm guessing that the
somewhat crypic comment

Because of that, his assessment of the current required to do the
job is flawed.

means that a very large current is drawn during ignition, and that the
reversed zeners and resistors are stopping this current from being
drawn. If so then back to the drawing board.

 

[suputnic]

means that a very large current is drawn during ignition, and that the

reversed zeners and resistors are stopping this current from being
drawn. If so then back to the drawing board.

(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.

 

[[email protected]]

Fine weather recently, but cold with the onset of winter

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...

 

[Captain Dondo]

"redbelly"





** Yep - they are average values of a short pulse current waveform.






** ROTFL .....

The OP is an idiot.

His heater is powered by gas.

The two batteries supply power to the ( electronic) spark igniter.

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?

http://www.onsemi.com/pub/Collateral/MC33269-D.PDF

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

 

[redbelly]

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? )

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

 

[Captain Dondo]

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.

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

 

[suputnic]

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'm in Auckland.

 

[suputnic]

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

No it doesn't need a precise voltage, batteries between 3.3V and 2.8V
make it go. In fact I tried a 15.5V adapter (12V nominal) today, and it
worked as well, but the igniter kept sparking after it was lit. Also
the current was about half an amp, luckily it seems OK. I will try the
chained diodes as a last resort, but I'd need over well over 10 in this
case. Here are some of the options I have tried:

15.5V adapter (12V nominal)
0.5A ??

first off 15.5V is kind of high to reduce to 3 using zeners,
and also it's not a regulated supply so the output voltage will
change with the supply voltage (which fluctuates somewhat)

if you can't get a lower voltage plugpack you'll need to use the lm317
circuit to reduce the voltage.

Adapter -----
+ ---+---Vin|LM317|Vout---+-----+---> To igniter
| ----- | |
| Adj [240R] |
| | | | +
[.1uF] +----------+ [1uF]
| | |
| [330R] |
| | |
- ---+---------+----------------+---> To igniter

That will give you about 2.97 volts out to the igniter.

yeah, that one.
maybe a 10uF tantalum instead of the 1uf.

 

[ehsjr]

No it doesn't need a precise voltage, batteries between 3.3V and 2.8V
make it go. In fact I tried a 15.5V adapter (12V nominal) today, and it
worked as well, but the igniter kept sparking after it was lit. Also
the current was about half an amp, luckily it seems OK. I will try the
chained diodes as a last resort, but I'd need over well over 10 in this
case.

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
http://www.allelectronics.com/

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

 

[Jasen Betts]

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

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

Adapter -----
+ ---+---Vin|LM317|Vout---+-----+---> To igniter
| ----- | |
| Adj [240R] |
| | | | +
[.1uF] +----------+ [1uF]
| | |
| [330R] |
| | |
- ---+---------+----------------+---> To igniter

That will give you about 2.97 volts out to the igniter.

yeah, that one.
maybe a 10uF tantalum instead of the 1uf.

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.

 

[John Fields]

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

---
22mA

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?

 

[redbelly]

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

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

Mark

 

[suputnic]

So what is it you're trying to do? Outfit a bunch of heaters with

some junk wall-warts you've got hanging around?

That is correct.

means that the input current to the
igniter must be greater than 4 amps at 3 volts during ignition.

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.

 

[suputnic]

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]

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.

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

 

[suputnic]

I tried a 12V nominal (16V measured) adapter with a 1N4148 small signal
diode attached forward biased to the positive lead. It fired up the gas
a few times then stopped working. When it did work it kept sparking
after the gas was lit. I managed to get a voltage masurement, it was
about 9V.

 

[suputnic]

Oops that diode is only good for 100mA I've found out. I'll have to get
some rectifier diodes and try them.

 

[almo]

Thank God!

I can't believe I had to read this far down before somebody mentioned
using a 3v regulator ))

 

[suputnic]

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

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.

 

[suputnic]

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.