Well I know this has been discussed before on the web on how ot make a night light, but I don't really need it to have the photo sensor. I want to use it as a pilot light on my workbench for my power supply 
Should I just use "X" resistor to drop the voltage and a single 1n4007 ? half wave? FWB?
Should I just use "X" resistor to drop the voltage and a single 1n4007 ? half wave? FWB?
0A3 on the power line
Its hard to easily calculate the discharge behavior when running off of an AC signal. I would get a rheostat (high-power potentiometer) and, starting with a high value (say 2K or so), and lower it until the RMS current flow is less than 40mA. A diode would be a good idea, since the reverse behavior of gas regulators is not well defined. A 1N4007 should be fine, but a bridge would not put any DC on the power line. The positive output of the rectifier should go to pin 5. This set-up will likely create RF noise, so if this is a problem, you could add a 0.01uF 1000V cap across the power line. I hope this helps.
- John Atwood
Its hard to easily calculate the discharge behavior when running off of an AC signal. I would get a rheostat (high-power potentiometer) and, starting with a high value (say 2K or so), and lower it until the RMS current flow is less than 40mA. A diode would be a good idea, since the reverse behavior of gas regulators is not well defined. A 1N4007 should be fine, but a bridge would not put any DC on the power line. The positive output of the rectifier should go to pin 5. This set-up will likely create RF noise, so if this is a problem, you could add a 0.01uF 1000V cap across the power line. I hope this helps.
- John Atwood
heh, I like that statement " It is hard to easily calculate..." I find that to be the rule sometimes..
Anyways, I had not thought about the RF noise problem, would this be induced into the power line and thus affecting my power supply that it will be plugged into?
I want to clarify, I am not using this as a regulator, only an indicator lamp or "night light" to be run off of 120v mains. The website that talks about the night light version reccomends a 5k dropping resistor "because the VR75 can only take 75v on the plate, any more and it will age quickly." I did not understand this statement because the data sheet says a minimum of 105v to start it, so I would not think 125v would be a problem.
So basically, the resistor is only a current limiter, right?
I find that to be the rule sometimes..Anyways, I had not thought about the RF noise problem, would this be induced into the power line and thus affecting my power supply that it will be plugged into?
I want to clarify, I am not using this as a regulator, only an indicator lamp or "night light" to be run off of 120v mains. The website that talks about the night light version reccomends a 5k dropping resistor "because the VR75 can only take 75v on the plate, any more and it will age quickly." I did not understand this statement because the data sheet says a minimum of 105v to start it, so I would not think 125v would be a problem.
So basically, the resistor is only a current limiter, right?
Use a 0.47 uF AC rated capacitor as a current limiter - it will pass about 20 mA at 120V, 60 Hz. Look for 'X' capacitors - they're designed for direct connection to the AC mains. A 3K resistor will have about 3 watts of heat to get rid of... the capacitor approximately zero. You can follow it with a full-wave bridge if you like - the glow pattern will be different from AC. Half-wave won't work.
If it does create any line noise, a series resistor of 100 Ohms or more and a .01 or larger line bypass (AC rated cap!) will probably take care of it.
If it does create any line noise, a series resistor of 100 Ohms or more and a .01 or larger line bypass (AC rated cap!) will probably take care of it.
Not sure why this is so hard.... start voltage is 100 volts... operating, i.e., regulating voltage is 75 and operating current range is from 5ma - 40ma.
Regulator tubes have an operating range to allow them to be used as a shunt regulator. The circuit using the regulated voltage will typically fluctuate it's current demands during operation (like regulated screen grids on a P-P output stage). The operating current range allows the tube to sink the extra current, hence providing the voltage regulation.
I would suggest running at no more than half the rated current, that way you don't toast it or burn it out too quickly. Also, put a 0.01uF cap in parallel with the tube so it doesn't have a tendency to oscillate or "motorboat" as you really don't have a formal DC supply powering it.
Regards, KM
Regulator tubes have an operating range to allow them to be used as a shunt regulator. The circuit using the regulated voltage will typically fluctuate it's current demands during operation (like regulated screen grids on a P-P output stage). The operating current range allows the tube to sink the extra current, hence providing the voltage regulation.
I would suggest running at no more than half the rated current, that way you don't toast it or burn it out too quickly. Also, put a 0.01uF cap in parallel with the tube so it doesn't have a tendency to oscillate or "motorboat" as you really don't have a formal DC supply powering it.
Regards, KM
ThSpeakerDude88 said:Well I know this has been discussed before on the web on how ot make a night light, but I don't really need it to have the photo sensor. I want to use it as a pilot light on my workbench for my power supply
Should I just use "X" resistor to drop the voltage and a single 1n4007 ? half wave? FWB?
Check out this link:
http://www.electronixandmore.com/project/2.html
I grabbed a few VR tubes to try but they were 0D3 and 0C3 - the latter is 105V, so I tried it with a series .47 cap on 120v. It glowed purple and drew about 6 mA. I lied - you CAN use a half-wave rectifer - but it goes in PARALLEL with the tube. Either way works - one has glow on the center electrode, the other on the inside of the plate. It draws about 13 mA with the shunt diode. If you use one, you should add a series resistor to limit surge current - 100-200 Ohms should do.
No, the capacitor is still in series, limiting the current. On one half-cycle, the diode conducts, and the capacitor charges. On the other half-cycle, the diode is reverse-biased, ands the capacitor discharges through the gas tube.
A surge on the power line COLUD zap the diode, so a current limiting resistor is a good idea. Of course if the diode shorts, the worst that will happen is that the bulb goes out - the capacitor still limits the current.
A surge on the power line COLUD zap the diode, so a current limiting resistor is a good idea. Of course if the diode shorts, the worst that will happen is that the bulb goes out - the capacitor still limits the current.
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