EIKO GTL-2 UV-C Germicidal Bulb

[magic_eyes]

Hey,

I am an engineering student trying to power a 10V 2W (and thus a .22 A) UV germicidal bulb. Our project is to make a solar powered water purifying bottle that utilizes this bulb. The solar panel will charge two 9v batteries in series outputting to a ~80 ohm resistor connected to the bulb and then back to the solar panel all in series. Upon trying to complete this circuit many times with no success of lighting the bulb at all I called the company, EIKO. They are confident the bulb is functioning and believe the error to be in my circuitry. The tech help stated that the bulb works over a discharge path. He said this may result in the bulb (or something in the circuit) providing infinite resistance. I have received schematics of the bulb itself to confirm this. Any ideas as to what I am doing wrong? Any help would be greatly appreciated. Thank you,

Ryan

[magic_eyes]

I just got the bulb to work. The key was just strength of connections and wiring the 75 ohm resistor in parallel with a capacitor.

[Ron H]

Originally Posted by magic_eyes

I just got the bulb to work. The key was just strength of connections and wiring the 75 ohm resistor in parallel with a capacitor.

Congratulations on getting the bulb to work.
The resistor is a huge power waster. Efficiency is very important in battery powered equipment. Unless your circuit is just for proof of concept, you should consider a switching regulator.

[magic_eyes]

Interesting. I'll need to do some research myself but where in the circuit would this go? Still in parallel with the capacitor? Or simply replacing that parallel system? My friend also recommended a super-capacitor which I need to look into. Thanks for the advice!

[Ron H]

The switching regulator would replace the parallel RC, supplying a regulated 10V. You might have to design it to put out a higher voltage impulse when first switched on, similar to what your RC circuit does.
Good switching regulators are more than 90% efficient.
BTW, how did you come up with 75 ohms?
(18V-10V)/200mA=40 ohms.

[magic_eyes]

Thanks for the info that sounds much better than what we have now.

The 75 ohms was from standard formulas(without the discharge path) with everything in series. 18v/(~75ohm) = 0.2 amps. Some trial and error was also involved. However the equation you just posted is probably the one we should use as I was having trouble duplicate my success getting 17 volts through the circuit.

[Ron H]

Originally Posted by magic_eyes

Thanks for the info that sounds much better than what we have now.

The 75 ohms was from standard formulas(without the discharge path) with everything in series. 18v/(~75ohm) = 0.2 amps. Some trial and error was also involved. However the equation you just posted is probably the one we should use as I was having trouble duplicate my success getting 17 volts through the circuit.

It's a 10V, 200mA (or is it 220 mA) bulb. Ohm's law should let you determine the series resistor value. However, I suspect that UV bulbs have a nonlinear V-I curve, so I don't really know what happens if you use a series resistor instead of a 10V low impedance source.

[magic_eyes]

We have some switching regulators in the shop on campus so I'll test them out. I think they should be perfect. Thanks again.

[vinnie50]

Hello magic_eyes and Ron H
I have just purchased the same GTL-2 UV bulb for one of my work applications and am also having trouble getting it to light up. I am using a 9V DC source with no series resistance. You had mentioned that you used a capacitor. What value die you use? Also, what voltage did you use? The spec says 10V +-2V, however, I received some info from EIKO that it should be 24V?? Any help would be appreciated

[magic_eyes]

I believe I used a 36 ohm resistor in parallel with a capacitor or arbitrary capacitance. However, I found a better solution to be using a voltage regulator with a resistance ratio of 7. We did this via a 70kohm wired from pin 1 to the common and then a 10kohm from the common to pin 3. This creates the discharge path necessary for the bulb to light up and provides enough resistance to protect the bulb.