DC 3V battery backed power supply for Gas Water Heater

Hi,

My Gas water heater uses two D-size cells 1.5 V (x2) = 3V DC to ignite the gas when water tap is opened. battery is used only to trigger the gas igniters when water tap is opened, gas ignites, the gas solenoid is kept powered, 7seg displays water temperatures. When the tap closes circuit is powered off.

I have very frequent use of heater and it consumes battery quickly needing frequent replacement. I am looking to build a 220V AC to 3V DC power supply circuit which will stay on 24x7 but I want to use it in combination to the two D sized cells. when 220V AC available it must power the igniter and cells power source stays isolated. But when AC mains power goes out this circuit should change over to the cells power source automatically.

I want to make circuit solid state (avoiding the use of relays) and the Mains power supply should consume least current when on mains AC since it would be powering 24x7. Batteries would be used only in case of power outage. At the moment I do need to automatically recharge the batteries when on Mains. Currently I am using non-rechargeable type batteries. so recharging is not a requirement.

The problem is quite simple and there are many possible solution so please suggest your's

ADDING MORE:
I need exact same solution for my micro controller implementation of arduino uno. I want it to be running 24x7 on Mains 220v with a battery backup and auto switch over to battery when mains is out. Changeover should be like computer UPS, I mean microcontroller must not reset during power sources changing battery to mains or mains to

waiting for suggestions.

Thanks

The simplest solution is two diodes. Place one in series with each power source and wire them to combine the current to the load. No current will be drawn from the battery unless the AC supply fails and no current from the AC supply will be fed back to the batteries. There will be a small voltage drop but if you use a Schottky diode it will only be ~0.4V. The change over is instant so nothing should reset.

Brian.

Hi,

You don't give current rating.

There are dedicated ICs for that. With built in ideal diodes.

Klaus

betwixt said:

The simplest solution is two diodes. Place one in series with each power source and wire them to combine the current to the load. No current will be drawn from the battery unless the AC supply fails and no current from the AC supply will be fed back to the batteries. There will be a small voltage drop but if you use a Schottky diode it will only be ~0.4V. The change over is instant so nothing should reset.

Brian.

Click to expand...

Yes that's too simple and seems to work. One question. as you say each power source with a series diode, combining their cathodes to a node which connects power to load circuit. consider the case when mains power is present. will the load circuit not get power from both of the sources that time, because both diodes are forward biased ?

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KlausST said:

Hi,

You don't give current rating.

There are dedicated ICs for that. With built in ideal diodes.

Klaus

Click to expand...

I haven't calculated yet but it would not exceed on Amp I believe. guess how much current would a 3V DC based gas igniter would consume, I guess would be less than an amp.

in case of microcontroller power, my second requirement, it would also be less than an amp, say 500mA

Hi,

LT4413 as an example device.

Klaus

KlausST said:

Hi,

LT4413 as an example device.

Klaus

Click to expand...

LT4413 is what is needed. Let me see if it is available here. in case if not, can we a solution based on common components npn/pnp/fet/mosfet which would be easily available?

I had an oven like this. You pushed a button in and it turned on a gas pilot burner and switched a pair of D cells to the thermal element which then glowed red hot and lit the gas. After a few seconds of the pilot being lit it opened the main valve and the main burners came on.
The problem I had was with the battery contact resistance which makes me think that the current is a lot higher then 1 A, say 7A. If faced with this problem I would upgrade the wiring to minimise voltage drops replace the D cells with Nicads and just use a low current constant current charging circuit. As 20 usesages per day st 20 seconds a time is 400 X 7 A/S = 2800 As or 2800/3600 Ah. (.65) . For a 24 hr charging this is about 33 mA.
Frank

Hi Frank,

7A continously for 20 seconds. I estimated much smaller value.
7A even as peak inductor current i considered as too much.
But to be true, i never tested such a device...


Klaus

Yes that's too simple and seems to work. One question. as you say each power source with a series diode, combining their cathodes to a node which connects power to load circuit. consider the case when mains power is present. will the load circuit not get power from both of the sources that time, because both diodes are forward biased ?

Click to expand...

No, the current will only flow from the source with the higher voltage. If the AC source is slightly higher it will not consume battery power at all. If you make your power supply 3.4V to compensate for the diode drop, you get 3V out while on AC and 2.6V out when on battery back-up.

Check which kind of igniter you have, there are two types. One as mentioned uses the power to operate a heating element and needs a lot of current at low voltage, the other works on a different principle and generates sparks to ignite the gas. The spark igniter uses far less current as it only has to generate a high voltage for discharge maybe twice a second. They detect ionization of the gas to tell when the gas has ignited and then switch off the HV inverter. Tell us which type you have, the spark type is easy to identify because it will go 'tick tick tick tick' until the flame appears.

Brian.

betwixt said:

Check which kind of igniter you have, there are two types. One as mentioned uses the power to operate a heating element and needs a lot of current at low voltage, the other works on a different principle and generates sparks to ignite the gas. The spark igniter uses far less current as it only has to generate a high voltage for discharge maybe twice a second. They detect ionization of the gas to tell when the gas has ignited and then switch off the HV inverter. Tell us which type you have, the spark type is easy to identify because it will go 'tick tick tick tick' until the flame appears.

Brian.

Click to expand...

Yes my ignitor is HV inverter type. it does exactly as you mentioned above. it gives HV sparks near burners repeatedly then when flame, it switches off sparks and gas-in solenoid is energized.

Still I insist please give me some seminconductor based solution. I am not satisfied with two series diodes although it may work. I need similart UPS like operation for my Arduino supply too

Thanks for posting

Ignore my posting. :-(
Frank

You can't get more semiconductor based than a diode!

Brian.

All right, still trying as time available.

at the moment replacing Cells.

I checked with Amp tester. On battery when igniting it takes approx full battery current my tester shows approx 3Amp and going beyond my tester limit so can't figure out exact current. This high current demand is only when three spark ignitors are working, the moment gas solenoid is opened as gas is available, sparks stop and solenoind stays on. At this time only 50 mA approx are consumed

I am ready to build a AC 220V power source with little higher volt then batteries so batteries turn as backup without consumption.

but please suggest me how to build a 24/7 220V AC to 3.5 or 4V DC power source with above current demands. If using ready available parts it would be easy to build

please suggest

The simplest:
Transformer-->bridge rectifier-->capacitor.
It will be unregulated voltage, but I don't think you require regulation in this application.

Hi, Guys, my Gas heater igniter is still working on two D-Size cells.

Please have a look in this schematic


can somebody modify component / rating for my application as I mentioned earlier?

I need only 3VDC Un-Interrupted output on battery + Adaptor. (that 5VDC regulated output mentioned in circuit is not needed)

Moreover at the moment I do not have rechargeable batteries, so removing D3 will be enough to disconnect batteries from being charged on AC ?

The diodes must be rated for 10A continuously then they will have a voltage drop of 1V to 1.5V each. So the ignitor will get only about 1.5V or less when the throw away battery is brand new and less as it runs down.
Rechargeable Ni-MH battery cells are about 1.2V each then 3 or 4 can be connected in series. The bridge rectifier will have a voltage drop of 2V to 3V. Then a 5V or 6V at 10A transformer can be used but it will not fully charge the battery.

My take on this is a little different. Put in some good rechargeable battery's. For the exact type look up there characteristics. Make certain they are made for high current short duration use.

Then look at there trickle charge rate. Get a 3.3 to 5v max power supply and put a resistor in series with the output to limit the current to the trickle charge rate. I say no more than 5v because if you pull the battery's out with the power supply connected the voltage will rise up to your power supply voltage and may cause harm.

I would go one step further.

Use a 3v rechargeable lithium cell recharged by a very small solar panel with a shunt regulator.
It the gas water heater is outside, as most are, it simplifies a lot of things.
And it will not even need direct sunlight to charge at only a very few milliamps which is probably all you need.