# Reduce voltage of DC adapter without reducing current

S

#### suputnic

Jan 1, 1970
0
I have instant hot water heaters which take 2 X 1.5 batteries = 3V. I
want to replace the battereies with DC adapters we have. The adapters
have various voltages. I tried to drop the voltage by adding a resistor
to the circuit, but then the current drops too low for the unit to fire
up. Is there anything I can do? The measurements are:

Batteries: 3.36V, 22.2mA

DC Adapter, no resistor added: 5.9V, 22mA (Why does the current not
increase with the increased voltage?)

One unit is running OK using the DC adapter without a resistor, but I
think it would be risky to use the higher voltage adapters without
reducing the output voltage to near 3V.

J

#### John Fields

Jan 1, 1970
0
I have instant hot water heaters which take 2 X 1.5 batteries = 3V. I
want to replace the battereies with DC adapters we have. The adapters
have various voltages. I tried to drop the voltage by adding a resistor
to the circuit, but then the current drops too low for the unit to fire
up. Is there anything I can do?
Batteries: 3.36V, 22.2mA

DC Adapter, no resistor added: 5.9V, 22mA (Why does the current not
increase with the increased voltage?)

---
The heater may have an internal input voltage regulator.
---

---
The resistor you added was too large in value.

If you have a 5.9V supply and a load which wants to see 3.36V at
22mA, then you need to select the [series] resistor to drop the
difference between the supply voltage and the load voltage at the
current the load needs to operate, like this:

Vs - Vl 5.9V - 3.36V
R = --------- = -------------- = 114.4 ohms
Il 22.2mA

110 ohms is a standard 5% value and since your heater seems able to
stand an overvoltage, should work quite well. The resistor will
dissipate:

(Vs - Vl)²
P = ------------ = 0.059 watts,
R

so a standard 110 ohm +/- 5%, 1/4 watt carbon film resistor would be
fine.

Another way to do it is to place a Zener diode in series between the
supply and the load so that the Zener drops the voltage difference.

What you do to select the Zener is to subtract the load voltage from
the supply voltage and find a Zener closest to that voltage on the
requirements.

For example, in the case you gave:

Vz = Vs - Vl = 5.9V - 3.36V = 2.54V

and Il = 0.02A

So what you're looking for is a 2.5V Zener with a 20mA test current.

Consulting:

http://www.fairchildsemi.com/ds/1N/1N5222B.pdf

the 1N5222B is a near-perfect fit, and you can also select the
Zeners you need for the other supplies that way, the only caveat
being that for a test current of 20mA the highest voltage Zener
you'll be able to use in the 1N52XXB family is the 12V 1N5241B,
15V.

You could also use a Zener as a shunt regulator and, if you're
interested, I'll explain how to do that.
---
One unit is running OK using the DC adapter without a resistor, but I
think it would be risky to use the higher voltage adapters without
reducing the output voltage to near 3V.

---
It depends on what the control input to the heater looks like. From
your description I'd guess that there's a more-or-less constant
current source in there driving an optoisolator which is driving a
TRIAC or an SCR which is doing the _real_ work. But yes, there are
limits, and if you don't know what they are you could easily hurt
something in there.

R

#### redbelly

Jan 1, 1970
0
suputnic said:
I have instant hot water heaters which take 2 X 1.5 batteries = 3V. I
want to replace the battereies with DC adapters we have. The adapters
have various voltages. I tried to drop the voltage by adding a resistor
to the circuit, but then the current drops too low for the unit to fire
up. Is there anything I can do? The measurements are:

Batteries: 3.36V, 22.2mA

DC Adapter, no resistor added: 5.9V, 22mA (Why does the current not
increase with the increased voltage?)

One unit is running OK using the DC adapter without a resistor, but I
think it would be risky to use the higher voltage adapters without
reducing the output voltage to near 3V.

There is something weird about these measurements. The battery
measurement is giving 0.075 Watts, which is not much for an "instant
hot water heater".

Did you do the measurement with the heater immersed in water? If not,
I wonder if the heater unit can sense this (perhaps with a temperature
sensor that gets above the boiling point of water when the unit is not
actually in water?), and goes into a "low current" mode when this
happens?

Try doing the measurements with the heater immersed in water.

Mark

P

#### Phil Allison

Jan 1, 1970
0
"redbelly"
There is something weird about these measurements.

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

The battery measurement is giving 0.075 Watts, which is not much
for an "instant hot water heater".

Did you do the measurement with the heater immersed in water?

** ROTFL .....

The OP is an idiot.

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

....... Phil

R

#### redbelly

Jan 1, 1970
0
Phil said:

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

Well, shi-ite ...
Never mind!

S

#### suputnic

Jan 1, 1970
0
Thanks for all replies. I was only using a 47 ohm resistor, and still
the current dropped too low. I tried to measure resistance between the
positive and negative terminals when the burner was firing and when
off, but it justs gives an open circuit reading. I just found out that
the power cables feed into some sort of inverter? It (the white box)
says

Input: DC 3V
Output >= DC 12 KV

Not sure if this changes anything.

require a resistor to stop blowing up zener, and this circuit doesn't
seem to like any resistor.....

S

#### suputnic

Jan 1, 1970
0
Yes this inverter(?) does power the spark igniter.

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

S

#### suputnic

Jan 1, 1970
0
Actually it only seems to ever draw 22mA, so I suppose the appropriate
combo of reverse biased zener diodes and forward biased silicone diodes
in series would reduce the voltage to the right level, without using a
protective resistor.

E

#### ehsjr

Jan 1, 1970
0
suputnic said:
Actually it only seems to ever draw 22mA, so I suppose the appropriate
combo of reverse biased zener diodes and forward biased silicone diodes
in series would reduce the voltage to the right level, without using a
protective resistor.

It's doubtful that precise voltage is needed.
4 or 5 1N400x diodes in series between the adapter
& igniter should work. If you want precision, you
can use an LM317 voltage regulator:
(View in fixed font)

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

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

Ed

S

#### suputnic

Jan 1, 1970
0
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:

0.5A ??
-------------------------------------------------------
Results
worked but igniter kept sparking after ignition

10V Zener diode in series
3.3V Zener diode in series
-------------------------------------------------------
Results
3.3V to the igniter as required
46mA current, too high should be 22mA
Did not work

10V Zener diode in series
-------------------------------------------------------
Results
4.7V to the igniter, acceptable
68mA current, too high should be 22mA
Did not work

3.3V Zener diode in series
386 ohm resistor in series
-------------------------------------------------------
Results
5.3V to the igniter, acceptable
18.8mA current, too low should be 22mA
Did not work

3.3V Zener diode in series
330 ohm resistor in series
-------------------------------------------------------
Results
5.6V to the igniter, acceptable
20.2mA current, too low should be 22mA
Did not work

3.3V Zener diode in series
267 ohm resistor in series
-------------------------------------------------------
Results
6.2V to the igniter, acceptable
22.9mA current, close to required 22mA
Did not work

suputnic said:
Actually it only seems to ever draw 22mA, so I suppose the appropriate
combo of reverse biased zener diodes and forward biased silicone diodes
in series would reduce the voltage to the right level, without using a
protective resistor.

It's doubtful that precise voltage is needed.
4 or 5 1N400x diodes in series between the adapter
& igniter should work. If you want precision, you
can use an LM317 voltage regulator:
(View in fixed font)

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

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

Ed

J

#### John G

Jan 1, 1970
0
suputnic said:
I have instant hot water heaters which take 2 X 1.5 batteries = 3V. I
want to replace the battereies with DC adapters we have. The adapters
have various voltages. I tried to drop the voltage by adding a
resistor
to the circuit, but then the current drops too low for the unit to
fire
up. Is there anything I can do? The measurements are:

Batteries: 3.36V, 22.2mA

DC Adapter, no resistor added: 5.9V, 22mA (Why does the current not
increase with the increased voltage?)

One unit is running OK using the DC adapter without a resistor, but I
think it would be risky to use the higher voltage adapters without
reducing the output voltage to near 3V.

I have no idea where you live but sometimes it is better to have a
battery ignitor because if the power is off (Storm etc) you may still
want to use the heater.
My daughter has a heater which has no provision for batteries and she
was without hot water fo a week after a big tree brought down the power
lines.

M

#### [email protected]

Jan 1, 1970
0
suputnic said:
I have instant hot water heaters which take 2 X 1.5 batteries = 3V. I
want to replace the battereies with DC adapters we have. The adapters
have various voltages. I tried to drop the voltage by adding a resistor
to the circuit, but then the current drops too low for the unit to fire
up. Is there anything I can do? The measurements are:

Batteries: 3.36V, 22.2mA

DC Adapter, no resistor added: 5.9V, 22mA (Why does the current not
increase with the increased voltage?)

One unit is running OK using the DC adapter without a resistor, but I
think it would be risky to use the higher voltage adapters without
reducing the output voltage to near 3V.

From your IP address, you appear to be from New Zealand. How's the
weather out there?

So, I'm guessing you want to be able to power your hot water controller
I right?

Why exactly would you want to do something like this?

Could you get two AA NiMH batteries, and charge them when they go down?

Another idea: get the correct size DC adapter, as close to 3.36V,
22.2mA as you can find. Is there a thrift shop near you? Here where I
live, at the local Goodwill, they charge US \$1 each for assorted used

Michael

M

#### Michael A. Terrell

Jan 1, 1970
0
suputnic said:
Actually it only seems to ever draw 22mA, so I suppose the appropriate
combo of reverse biased zener diodes and forward biased silicone diodes
in series would reduce the voltage to the right level, without using a
protective resistor.

Diodes are made of silicon, bathtub caulk is made of silicone. ;-)

--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

S

#### suputnic

Jan 1, 1970
0
So, I'm guessing you want to be able to power your hot water controller
I right?

Why exactly would you want to do something like this?

Beacuse we have many hot water heaters, and replacing batteries all the
time is expensive. I suppose I could try the rechargeable option, if
they last long enough. We have no power supply issues in New Zealand.

M

#### [email protected]

Jan 1, 1970
0
suputnic said:
Beacuse we have many hot water heaters, and replacing batteries all the
time is expensive. I suppose I could try the rechargeable option, if
they last long enough. We have no power supply issues in New Zealand.

Ah. I'd recommend Energizer NiMHs, 2500 mAh, 1.2V. They come in packs
of 4 or 8. We have about 20 of them scattered throughout our house...

J

#### John Fields

Jan 1, 1970
0
Beacuse we have many hot water heaters, and replacing batteries all the
time is expensive. I suppose I could try the rechargeable option, if
they last long enough. We have no power supply issues in New Zealand.

---
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.

E

#### Electromotive Guru

Jan 1, 1970
0
I have a lot of trouble believing these ratings for an "instan
water heater"....Maybe you should consult the manufacturer an
couldn't heat 10cc of water +10C in anything less than 4
seconds.....

P

#### Phil Allison

Jan 1, 1970
0
"Electromotive Guru"
I have a lot of trouble believing these ratings for an "instant
water heater"....Maybe you should consult the manufacturer and
couldn't heat 10cc of water +10C in anything less than 45
seconds......

** The heater is powered by gas - as the OP has now revealed.

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

........ Phil

J

#### Jasen Betts

Jan 1, 1970
0
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:

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.
+ ---+---Vin|LM317|Vout---+-----+---> To igniter
| ----- | |
| | | | +
[.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.

S

#### suputnic

Jan 1, 1970
0
From your IP address, you appear to be from New Zealand. How's the
Fine weather recently, but cold with the onset of winter
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.

Thanks for that information, I'll stop putting resistors in my circuit.
Because of that, his assessment of the current required to do the
job is flawed.

I haven't assessed the required current, only measured it as 22mA.
However I put on a 4.7V DC adapter today, it made the unit work. The
measured current this time was 32mA, this is a different unit though.

This attempt below, any ideas how to limit the current without
inserting a resistor? Why is it drawing too much current anyway? Is
that why it doesn't work?

10V Zener diode in series
3.3V Zener diode in series

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