VFD filament driver - Page 1
Forum section: electronics / beginners

author: jimon
author_rank: Regular Contributor
timestamp: July 08, 2015, 07:15:21 pm
content: Well, it's reasonably easy to drive anode and grid, but not so trivial to drive the filament (or maybe it's just me being a noob)My goal is to drive filament efficiently (I want to create battery powered device). So filament in my VFD is 10 Ohm, it needs around 30mA, so power dissipation should be P = 0.03^2 * 10 = 9 mW. My trouble is how to get this 30 mA without wasting too much power.I'm using 9v battery atm, so first guess was just to : 9 / 0.03 = 300 Ohm resistor ... minus 10 for filament. So I just placed 290 Ohm there. But it's no good because we dissipate 0.03^2 * 290 = 0.261 W of energy for nothing.Then I checkedhttps://www.noritake-elec.com/display/vfd_operation.htmlthis document says that it's the best to drive filament with AC. If you drive it with DC you will get luminosity gradient because of gradient of potentials between filament and grid/anode.So I though about it a while, it's not trivial to make DC-AC conversion efficiently. So ...Let's PWM this !I did really simple circuit which is driven by 1KHz PWM with low duty cycle (<10%).So it works just fine, I have 180 mA through mosfet and filament, but only for 0.1 ms. So power dissipation here (maybe I'm wrong in calculations) = I^2 * (Rf + Rds) * 0.1 = 0.18^2 * (10 + 2) * 0.1 = 38.88 mW. Which is not bad compare to perfect 9 mW. And because of low duty cycle, most of the time filament should have ground potential, so luminosity should be consistent. If we calculate Vrms = sqrt(0.03888 * (Rf + Rds)) = 0.218 V. Which will give us average current = Vrms / R = 18 mA.Two questions :1) Is it bad mode for driving filament ?2) Are there more efficient solutions ?

author: bktemp
author_rank: Super Contributor
timestamp: July 08, 2015, 08:15:51 pm
content: Quote from: jimon on July 08, 2015, 07:15:21 pmIt works, I have done that, but you will hear the PWM frequency.QuoteUse a P mosfet, this will reduce the losses (and increase the peak current to 0,9A).9mW sounds wrong. I have never seen such a small VFD. Did you measure the cold resistance of the filament? It will be much higher when hot (around 40 ohms).

author: jimon
author_rank: Regular Contributor
timestamp: July 08, 2015, 08:39:16 pm
content: Quote from: bktemp on July 08, 2015, 08:15:51 pmI guess I can easily do 30+ KHz, so not big of a problem there.Quote from: bktemp on July 08, 2015, 08:15:51 pmJust ordered bag of BS250P's, somehow I haven't had any P FET's in my stashQuote from: bktemp on July 08, 2015, 08:15:51 pmJust checked now, Keithley 2015 says cold resistance is 10.74 Ohm. This VFD is SAMSUNG SSED HNM-05SS62T (more info herehttp://www.picaxeforum.co.uk/entry.php?142-LCD-LED-OLED-assorted-displays-lucky-bags-at-Maplin-2-(N01CG)-Kemo-Electronic-S043). Just bought it in assorted bag in local electronics store.

author: bktemp
author_rank: Super Contributor
timestamp: July 08, 2015, 08:51:31 pm
content: From the size it looks more like 2-3V filament voltage instead of 0.3V. 2V/41ohms=48mA -> 96mW. That sounds much better than 9mW.When beeing multiplexed the segments need more than 9V or the display will be quite dimm.

author: jimon
author_rank: Regular Contributor
timestamp: July 08, 2015, 09:27:07 pm
content: Quote from: bktemp on July 08, 2015, 08:51:31 pmWell, I decided to recheck this, Keithley 2015 says it's 10.74 Ohm (in 2 wire measurement method, I have good fluke leads there), cheap UNI-T UT33d says it's 12 Ohm.Then I decided to try more active test, I have two banana<->alligator leads. I've checked total resistance of them ~= 2.0 Ohms (with both DMM's). I connected filament directly to a lab power supply with this leads. Total resistance should be = 2 + Rf. I've started to rise voltage slowly from 0v. And I got 50 mA at 0.6 V (lab power supply only have 1 digit after dot), which means that Rf should be around 10 Ohm.

author: Andy Watson
author_rank: Super Contributor
timestamp: July 08, 2015, 09:36:31 pm
content: Quote from: jimon on July 08, 2015, 09:27:07 pmThe filament will be a different resistance when it is hot.

author: Zero999
author_rank: Super Contributor
timestamp: July 09, 2015, 11:06:07 am
content: Quote from: jimon on July 08, 2015, 07:15:21 pmThat's a perfectly valid way of efficiently reducing the average voltage but the effect will be the same as using a DC voltage because the current still only travels in one direction only.If AC is required, you need a push-pull output and connect it via an AC coupling capacitor or a bridge.QuoteMaybe, maybe not. It depends on the response of the eye. Try it out!

author: jimon
author_rank: Regular Contributor
timestamp: July 18, 2015, 01:39:51 pm
content: Quote from: Andy Watson on July 08, 2015, 09:36:31 pmI was wrong, rechecked when it's hot, give's me 50 mA at 2 Volts, so it's around 40 Ohm when it's hotQuote from: Hero999 on July 09, 2015, 11:06:07 amAfter some experiments I figure out that DC driver (PWM'ed or just constant current) indeed give luminosity gradient, but it's almost impossible to notice on this VFD.Quote from: bktemp on July 08, 2015, 08:15:51 pmSo I finally got BS250P's, and then realized that it's not suitable for this application (Rds = 14 Ohm, when other pmos are <0.2 Ohm). I'll order some other pmos when I figure out how to actually drive them efficiently.My current idea is in attachment (I've just picked random pmos, not sure which one to use). R1 value is low because of input capacitance of pmos, I want to drive this at 50Khz 5% duty cycle, and with bigger R1 it just doesn't charge fast enough. It looks like I only waste around ~= 15v * 15mA * 5 / 100 = 11 milliwatt, which is quite good IMHO.How can I improve it further ?PS. I've calculated power and it feels like it's much easier just to put tons of AAA batteries in parallelIn pmos\pwm case we do 360 mA with 5% duty cycle at 15 volts, which gives as 0.360 * 5 / 100 * 15 = 0.27 watt. If it's just 1.5v batteries then we do 1.5^2/40 = 0.056 watt, which is 4.8x times lower ! I don't understand why difference is so big

author: Andy Watson
author_rank: Super Contributor
timestamp: July 18, 2015, 01:58:46 pm
content: I have a WaveTek 271, it uses a 555 to generate a 50% duty cycle square wave. This drives the VFD filament via a 100uF series capacitor - seems to work reasonably well. One caveat with these displays is that the filament is also the cathode so you have to ensure there is no significant DC component across it - otherwise the digits do not illuminate evenly.