# Driving a filament

I have a filament that needs 1V, and is rated at 50mA at 1V. I would like to use 5V supply for it, and I am wondering, how to approach this. 50mA at 1V means that the resistance is 2k, as per Ohm's law.But it changes, does it not, as it heats up?

1.) I thought about adding a 8k resistor in series with it, so it acts as a voltage divider, giving 4V over 8k resistor, and 1 V over filament. However, that would be 20% efficient, as most power would be dissipated across 8k resistor.

2.) I thought about a buck converter, but I am not sure, if that is a good option

My question is: What is the best way to go down from 5V to 1V in this case, when the load current is so small?

The filament is in a numitron tube.

• At less than 1 mA, it probably doesn't pay to obsess over efficiency. (and most buck converters waste more than that in their internals)
– user16324
Commented Feb 6, 2018 at 13:58
• 0.5 mW isn't going to be much a filament. Commented Feb 6, 2018 at 13:59
• I remember Numitrons taking about 25mA at 5V, are you sure you have your figures right? Commented Feb 6, 2018 at 14:08
• You are right, it is 50mA Commented Feb 6, 2018 at 14:31
• How many of these filaments are you looking at driving? Commented Feb 6, 2018 at 15:17

What to do really depends on your power budget and physical size of the circuit.

The brute force approach is to linearly drop the 5 V to 1 V. You say the current is 50 mA, so that will take a total of 250 mW. 50 mW of that will heat the filament, and the other 200 mW are wasted as heat.

If 200 mW wasted power and extra heat to get rid of isn't a big deal, then a 80 Ω or so ½ W resistor is all it takes. That is certainly cheap and simple.

If this is battery operated, or a extra 200 mW of power usage and/or heat to get rid of matters, then use a switcher.

Depending on how the filament is connected inside the tube and what the tube is used for, you might be able to drive the filament directly with 5 V pulses. You can easily switch fast enough so that the filament temperature effectively doesn't change over the on/off parts of each pulse period. You can therefore consider the filament a fixed resistor once it gets up to temperature. You say it is supposed to draw 50 mA at 1 V, so looks like 20 Ω when hot.

Note that power into a resistor is proportional to the square of the voltage across it. A steady 5 V will drive the same resistor with 25 times more power than 1 V. The duty cycle with unfiltered 5 V pulses should therefore be 1/25. For example, 1 µs at 5 V and 24 µs off repeatedly should drive the filament nicely. At that 40 kHz switching speed, the filament temperature will be quite constant.

Another thing to consider is driving multiple filaments in series if you are using multiple tubes anyway. This may be inappropriate if the filament is also the cathode. It depends on details of the tube and how you use it. Perhaps having the cathodes of 5 tubes varying over the 0-5 V range doesn't matter in your application.

REF

I am more inclined to believe 25mA steady state but at what voltage drop? Let's reverse engineer something we know works and may confuse some. Here they drive filaments like LEDs with no resistors needed using Vdd=5V.

I don't know the cold temp resistance of the filament, so let me compute it.
We know it glows red, so the color temperature and the PTC of the conductor determine the resistance ratio from hot,Rh to cold, Rc. You can see the normal bulb around 3000'K on the Planckian locus curve and we know the Rh/Rc=10x but these tubes ( made by RCA after the Burroughs Nixie Tube) are just glowing red or around half the temperature or half the resistance ratio = 5x.

The above link shows Numitrons driven by 5V CD4511B CMOS which as we know all CD4xxx CMOS has high RdsOn that depends on Vdd (3~18V) and Vout

That means if you use this chip , it has an internal R already. But what is the short circuit power or I²R @ 5V, 25mA? where R= RdsOn of the CMOS. about 58 Ω ±50% Let's look up the V/I=ESR=RdsOn on the datasheet @ 25mA and 5Vdd

The R of a 5mm LED is high (>10k) below 2.5V and above 3V is ~ 16 Ohms and cold resistance. resistance drop from quadratic to linear. These are often driven with 20mA as rated for 5mm LEDs.

Luckily — since the tube current requirement at about 23 mA per segment is fairly low and it is a seven-segment device operating at about five volts — we can drive them with a standard BCD to seven-segment decoder designed for seven-segment LED displays with a max driving output of 25 mA. The CD4511 also has a data latch built in, so it was chosen for this design. ( linked in 1st line)

I just found the Numitron datasheet.
Electrical specs . . . . . DR21xx . . DR22xx
Segment voltage . . . . . . . 4.5V . . . 2.5V
Segment Current . . . . . . . 24 mA . . . 14 mA

Est. R(hot) . . . . . . . . 188 Ω . . . 179 Ω
Calc Power . . . . . . . . . 108 mW ... 35 mW

## Conclusion

Depending on which Lumitrons, the max power dissipation is limited to MTBF and model version but current seems to be either 15 or 24 mA at 100kh MTBF. Hot filament resistance are both around 180 Ω which means cold resistance is 1/5 or 36Ω so the CMOS driver of 58 Ω+/50% @ Vdd=5 and Voh=3.55 should be adequate.

## Let's check Ohm's Law again

5V/(36Ω(cold)+ 58 Ω(RdsOn)) = I = 53mA (cold surge on)
5V/(180 (hot) + 58 Ω = 21 mA (hot)

(53mA is a tad high and might fuse the micron gold bondwire in the CMOS but then again possibly not, but close enuf for government work, but you've been warned)

Configure a linear current source. 40mW wasted won't be a problem for any linear voltage regulator (yes voltage regulator configured as a current source).