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Nixie Watch, Part 1: Development Board

Step 4Power supply & Battery

Power supply & Battery
The power supply has been the most difficult and evolving part of the design. Nixie tubes require high voltage, easily generated with an inductor coil as in my nixie switch mode power supply:
( http://www.instructables.com/ex/i/B59D3AD4E2CE10288F99001143E7E506/ )

Things get harder when we want to run from batteries at a low voltage. There are several dedicated boost converter chips that can be used to generate the required voltage, David Forbes uses the LT1308B. I avoided using such a chip because it adds an additional (surface mount) component. It also promotes a vendor specific design that doesn't contribute to understanding what actually goes on in our SMPS. Any uC can drive my SMPS without major modifications.

I used the nixie power supply from the previous instructable as a starting point. One input (100uF/16V) and one output filtering capacitors (1uF, 250V, high-temp) are used. At such a low load (1 tube at ~ 1mA) nothing more is required. The supply voltage sense resistors are eliminated – the watch will not measure the supply voltage, it will be hard-coded with ideal values. The rest of the design is exactly as presented in the SMPS instructable: IRF740 FET, BYV26C diode, 100uH inductor, and TC4427A FET driver. The 16F684 uC drives the SMPS.

The rub -
My first draft design used a small 12 volt alkaline battery (V-23). I though -WOW- I'll use this small cell with a zener to provide 5.1 volts for the PIC and feed the SMPS directly from the battery @ 12 volts. Nope. The V-23 is probably a stack of 10 x 1.2 volt button cells, but I haven't opened it to be sure. These cells cannot provide enough current for the circuit – even without a nixie tube installed. Voltage from the battery fell to under four volts at any significant amount of current. The IRF740 FET needs a bare minimum of 5 volts to switch at all. Efficient switching is more in the order of 10 volts. At four volts the SMPS FET is never actually switched, thus no high voltage is generated.

The fix -
I looked over David's design and noticed the CR2 3 volt lithium battery he uses. A quick googling showed that this cell is intended to power a camera flash – that means plenty of current is available. This cell is much bigger, but should have a longer life than the V-23.

The rub (part II) -
The CR2 is only a 3 volt battery – we have the same problem as before, nowhere near enough voltage to switch the FET. I was feeling awfully confident that I could come up with a solution. I tried to 'boot strap' the FET – feed a little of the SMPS output voltage to a simple 15 volt resistor/zener regulated supply, then use that voltage to switch the FET through the TC4427A FET driver. No go. Adequate voltage didn't accumulate, the FET didn't switch.

The fix -
Where do we go from here?
1. Redesign the power supply using the guts of a disposable camera flash. This could work, but I would like the option of buying new, standard components.
2. Give up and use two batteries. The supply works fine when using the CR2 for supply and the V-23 (12 volt) battery just to switch the FET. 2 x 3 volt coin cells work well, and would fit nicely under the larger CR2 battery.
3. Use a DC/DC charge pump to double the CR2 output to 6 volts. This is an appealing option. Only a single battery is needed, but an additional (surface mount) IC, 2 capacitors, and 2 diodes are required.

This development board was made to evaluate the last two options. It can test various button cell batteries to determine their usefulness as a FET driver supply. It can also be used to test the suitability of a DC/DC charge pump as the FET drive supply. The charge pump has its own caveats that might rule out its usefulness (dubbed: the rub part III). From the Microchip TC7660 datasheet:

( http://ww1.microchip.com/downloads/en/DeviceDoc/21465b.pdf )

The voltage thus created...becomes (2 V+) – (2 VF), or twice the supply voltage minus the combined forward voltage drops of diodes D1 and D2.

So the CR2 @ 3 volts and 2 good diodes with an (ideal) 0.6 drop:
4.8 volts=(2*3)-(2*0.6)

Thats might just barely do the job. Perhaps different diodes might help, or cascading two charge pumps. This all makes dual batteries seem like a small inconvenience for the space and complexity saved.
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Author:ian(DangerousPrototypes.com)