What parts can I use to make 110V DC from 4 D-size alkaline or NiMH batteries?

Question

I need help with choosing DC/DC converter parts.

I want to build D-size battery powered lantern working with typical E14 or E27 compact fluorescent lamps.

Here is my awesome idea :)

It will be supposed to work on D-size NiMh or Alkaline batteries.

Typical compact fluorescent lamps work normally with DC (there is rectifier bridge at the input anyway). I was experimenting with some old compact lamps and most of them start at 80V DC. All lamps I tried work normally at 100V DC.

I've been messing around with some low voltage easy to use converters (like step-down LM2574), I have oscilloscope, I'm able to tune-up simple converter efficiency for specyfic load needs. However I'm not inverter expert, I don't know integrated chips to build such thing.

Converter requirements:

• output power at least 10W about 4W due to D-size alkaline capabilities

• efficiency at least 80%

• minimum input voltage 4V (3.2V would be awesome)

• output voltage 110V (it may be 230V)

• not complicated (i need to build this in a week - gift for sister)

The question is:

Are there any easy to use parts to build affordable DC/DC 3...7V to 110V converter?

It may be 4V, I wrote 3V because its possible to deep-discharge alkalines to 0.8V, thats 3.2V with 4 cells.

I know that LED lamps are more efficient, I'm considering using them too, but I want to stick to "typical home light sources".

I saw alkaline D discharge charts, it looks like single D size Duracell Procell can deliver 1W for 5 hours.

GP 700DH NiMh battery with 7Ah capacity can deliver much more power (14A discharge for 30 minutes or 7A for 1hour).

Answer 1

Yes this is possible, but not practical. Since you don't need isolation (the whole thing, batteries and all is insulated), a basic boost converter could in theory do this. 100 V out will mean you have to select parts carefully.

At that voltage you're not going to find a integrated buck converter chip, but you should be able to find one that can drive a low side switch. The 100 V is then divided down and fed back into the chip, which should have a fixed internal reference voltage to compare this feedback input to. The chip therefore never sees the high voltage directly.

However, the real issue is the current this will draw from a few batteries. You want 10 W out. Let's say the boost converter will be 85% efficient, so that means it will require 10 W / 85% = 11.8 W in. At 6 V from the batteries, that will be 2 A. A few D cells aren't going to live long with that kind of drain.

Answer 2

Im answering my own question to say, that "my awesome idea" was bad.

• I thought I can drain more energy from D-size alkaline (they are able to deliver just 1W for reasonable amount of time), here is Duracell Procell D-size datasheet
• I realized converter will be more complicated as I thought (to get reasonable efficiency)
• I realized that cost will be bigger, and for that cost I can make more effective LED lamp instead of CFL

I changed my mind and I will put some 1W leds into that lantern. I will be able to control current for specyfic types of batteries, dim light etc. and overall lantern cost will be similar to inverter with CFL.