I need to find a 82 volt DC power supply capable of running a 300 watt halogen bulb.
Now i figured that someone out there might have an old OHP that has a 82vdc power supply in it.
I just need the power supply not the bulb, fan or bulb holder.
PLEASE someone must have one... I can pay by paypal immediately!
I am in Canada so shipping by USPS would be cheapest.
Thanks, Gordon
Now i figured that someone out there might have an old OHP that has a 82vdc power supply in it.
I just need the power supply not the bulb, fan or bulb holder.
PLEASE someone must have one... I can pay by paypal immediately!
I am in Canada so shipping by USPS would be cheapest.
Thanks, Gordon
I need one as well. I do not understand exactly how to make this. I am a visual kind of person. Do you have or know where I could get schematics at?
simple circuit
You don't really need the transient voltage suppressor. You just wire a 5 Amp (or higher) 480 volt (or higher) silicon diode in series with your lamp. That means that the current flows from the hot side of your line cord, then through the diode, then through the lamp, and then back to the neutral side of your line cord. (EEs: please, I'm trying to keep it simple.)
The polarities don't matter: It is impossible to wire it backwards, since it runs on AC. You should be able to buy a suitable diode at Radio Shack for a couple of dollars.
If you want transient protection for the lamp, you can plug it into a power strip, or a socket that has a dimmer switch.
black wire------diode---------lamp------white wire
You don't really need the transient voltage suppressor. You just wire a 5 Amp (or higher) 480 volt (or higher) silicon diode in series with your lamp. That means that the current flows from the hot side of your line cord, then through the diode, then through the lamp, and then back to the neutral side of your line cord. (EEs: please, I'm trying to keep it simple.)
The polarities don't matter: It is impossible to wire it backwards, since it runs on AC. You should be able to buy a suitable diode at Radio Shack for a couple of dollars.
If you want transient protection for the lamp, you can plug it into a power strip, or a socket that has a dimmer switch.
black wire------diode---------lamp------white wire
right
The diode cuts out half of each cycle of the A.C., so you get pulsating D.C.
117 VAC is the RMS (DC voltage equivalent) of standard power in the US. It is really 117 / 0.707 = 165 volts peak-to-peak. If you cut half of each cycle, then that gives you 165/2 = 83 volts peak to peak with a low duty cycle. That is what an 82 volt projector lamp needs.
Not 82 volts DC! That would burn it out very quickly.
I think you can make your lamp run a lot longer if you include a standard dimmer switch (before the diode). Then you can turn it on and ramp up the voltage over a couple of seconds, instead of instantaneous. That first inrush of current through a cold filament is what burns out lamps.
The diode cuts out half of each cycle of the A.C., so you get pulsating D.C.
117 VAC is the RMS (DC voltage equivalent) of standard power in the US. It is really 117 / 0.707 = 165 volts peak-to-peak. If you cut half of each cycle, then that gives you 165/2 = 83 volts peak to peak with a low duty cycle. That is what an 82 volt projector lamp needs.
Not 82 volts DC! That would burn it out very quickly.
I think you can make your lamp run a lot longer if you include a standard dimmer switch (before the diode). Then you can turn it on and ramp up the voltage over a couple of seconds, instead of instantaneous. That first inrush of current through a cold filament is what burns out lamps.
why do you say the lamp can´t run at 83 volts DC? isn´t it what the specs of the lamp need?
The effective value of 165 peak half sinusoidal waves is 52 volts.
So you can consider like if it was 52 volts DC.
Now, the slow dutti cicle will decrease color temp something not welcome for an already low alogen color. (dimng triacs have same effect).
still would like to know about the 83 volts... spec GG.
The effective value of 165 peak half sinusoidal waves is 52 volts.
So you can consider like if it was 52 volts DC.
Now, the slow dutti cicle will decrease color temp something not welcome for an already low alogen color. (dimng triacs have same effect).
still would like to know about the 83 volts... spec GG.
my fault. The effective value of half sinusoidal waves is not 52. I mixxed it all with the average value. The effective value is as GG said 83 volts RMS. In any case, this is like if we had 83volts DC power suply. I don´t see why should be any problem with either systems (83 DC or half sinusoidals at 165 peak).
you have to integrate
The original circuit description from an OHP was: One diode and one transient absorber. The transient absorber had a high voltage rating, which means it would do nothing unless there was a noise spike. If the circuit included a big capacitor, then the current through a high current diode could charge the capacitor up to the peak voltage: about 166 volts above ground. But there is no capacitor!
That means the lamp sees a voltage that is zero 50% of the time. The 50% of the time it is on the half-cycle that gets through the diode, it sees the 166 volt peak. But the RMS voltage (the DC equivalent) is only 120 volts. (That is the integration of the area under the curve.) 120 volts RMS half the time is equivalent to 60 volts DC.
But cheap AC voltmeters don't measure RMS, they measure zero-to-peak by rectifying and then charging a capacitor. So you would measure 83 volts. That is probably why the lamp has that parameter value.
Exercise for the interested: Write a program to do a numerical integration of a half wave rectified 166 volt sine wave, over one full cycle.
The original circuit description from an OHP was: One diode and one transient absorber. The transient absorber had a high voltage rating, which means it would do nothing unless there was a noise spike. If the circuit included a big capacitor, then the current through a high current diode could charge the capacitor up to the peak voltage: about 166 volts above ground. But there is no capacitor!
That means the lamp sees a voltage that is zero 50% of the time. The 50% of the time it is on the half-cycle that gets through the diode, it sees the 166 volt peak. But the RMS voltage (the DC equivalent) is only 120 volts. (That is the integration of the area under the curve.) 120 volts RMS half the time is equivalent to 60 volts DC.
But cheap AC voltmeters don't measure RMS, they measure zero-to-peak by rectifying and then charging a capacitor. So you would measure 83 volts. That is probably why the lamp has that parameter value.
Exercise for the interested: Write a program to do a numerical integration of a half wave rectified 166 volt sine wave, over one full cycle.
;D. you are wrong here GG (OOOO finaly I caught you 😀) just joking 😀
the area under the shine is not RMS value, that same area converted to square would just be average volts value, (integrating a shine wave and making a square wave).
the 165 volts peak sine wave (all wave rectified lets say, so all is positive), would have an average value of 105 volts. But would have an RMS value of 117 volts (RMS).
Now a 165 volts AC full wave or 117volts DC will have same power disipation on the load.
Now consider the half wave rectification;
as you said it is 0 volts value half the time. We have half sine waves at 165 volts peak. The average volts value (integration of the area and squared peak value) is this time 53 volts. But the RMS value of half sines at 165 volts peak is 83 volts RMS.
So a half wave rectified sine at 165 volts peak or 83 volts DC source would have same power disipation on the load.
sometimes we confuse mean value and RMS value, 😀
the area under the shine is not RMS value, that same area converted to square would just be average volts value, (integrating a shine wave and making a square wave).
the 165 volts peak sine wave (all wave rectified lets say, so all is positive), would have an average value of 105 volts. But would have an RMS value of 117 volts (RMS).
Now a 165 volts AC full wave or 117volts DC will have same power disipation on the load.
Now consider the half wave rectification;
as you said it is 0 volts value half the time. We have half sine waves at 165 volts peak. The average volts value (integration of the area and squared peak value) is this time 53 volts. But the RMS value of half sines at 165 volts peak is 83 volts RMS.
So a half wave rectified sine at 165 volts peak or 83 volts DC source would have same power disipation on the load.
sometimes we confuse mean value and RMS value, 😀
here's a diagram i found:
and here's a link to most of the 120v to 82v diodes used in overheads:
http://www.mbelectronics.com/view.aspx?id=128
BadBoy 😎
http://www.mbelectronics.com/view.aspx?id=128
BadBoy 😎
oh yeah, in the diagram:
TS = Thermostat
R-1 = 1.2 ohm ceramic resistor (big one!)
CR-1 = 82v diode
"M" in circle = 120v fan
SW-1 = on-off switch
BadBoy 😎
TS = Thermostat
R-1 = 1.2 ohm ceramic resistor (big one!)
CR-1 = 82v diode
"M" in circle = 120v fan
SW-1 = on-off switch
BadBoy 😎
can someone check my math? i want to figure out how many amps in a circuit.
82V AC, 410W FXL bulb.
I = P/V
I = 410/82
I = 5 Amps?
i want to put a fuse in line so i blow fuses, not bulbs... 🙄 i'm going to use a dimmer switch for now, until i can get a diode. would a 5 Amp fuse do the trick? any suggestions?
thanks,
BadBoy 😎
82V AC, 410W FXL bulb.
I = P/V
I = 410/82
I = 5 Amps?
i want to put a fuse in line so i blow fuses, not bulbs... 🙄 i'm going to use a dimmer switch for now, until i can get a diode. would a 5 Amp fuse do the trick? any suggestions?
thanks,
BadBoy 😎
there are like 2 kind of fuses;
absolute peak ones, and time dependent ones.
the 5 amps calc is only for RMS amps value, the peak amps on that circuit is higher than 5 amps. (the peak value on your scheme, assuming half wave rectification by 1 diode is roughly 10 amps) so I don´t know if your 5 amps fuse will last very long. It all depends on what kind of fuse is.
I would go for 8 amps one personally. But you can try, they are cheap.
absolute peak ones, and time dependent ones.
the 5 amps calc is only for RMS amps value, the peak amps on that circuit is higher than 5 amps. (the peak value on your scheme, assuming half wave rectification by 1 diode is roughly 10 amps) so I don´t know if your 5 amps fuse will last very long. It all depends on what kind of fuse is.
I would go for 8 amps one personally. But you can try, they are cheap.
i'm not going to use a diode right now, just throttle the voltage back with a 120V AC dimmer switch. have to order the diode. i already blew 3 bulbs, and it's getting expensive!!! i would like to play it safe and get a fuse thats .5 - 1 amp below the max amperage.
thanls,
BadBoy 😎
thanls,
BadBoy 😎
cheaper solution?
Rox: You are right. I was thinking of the ratio between peak and RMS in terms of voltage, but it is really power. You can't do an integration of voltage. You have to do an integration of power.
I wonder why that website has so many different power diodes? Power diodes have two basic parameters: Peak Inverse Voltage and Average Forward Current. (There are others, but for this application they don't matter.) All of these 120 VAC=>83 VAC overhead projector applications should be able to use the very same 6 Amp 400 PIV diode.
If you are in a rush, go to Radio Shack and buy two of these:
Part# 276-1144 3 Amp 400 PIV $1.59 each
Then connect them in parallel to get a 6 Amp 400 PIV diode. (Parallel means you lay them side-by-side with the markings facing the same direction, and then twist and solder the leads together on both ends.) You will never overpower another lamp again.
But the dimmer is also a good thing to include, since lamps almost always blow when you fire them up. The cold filament draws more than the rated current until it warms up. So if you start them slowly (ie. 1 second to full power), then they should run a lot longer.
For a fuse, I would get an 8 Amp "Slow-Blow" fuse like Radio Shack's:
Part# 270-1030 8 Amp 250 Volt Slow Blow Fuse 4 pack/$2.39
You can't use a 5 Amp fuse to limit your lamp voltage effectively, it is a current-limiting device. Since the lamp resistance changes quickly as it warms up, all you will do is blow lots of fuses if you use a quick-acting fuse. With a 5 Amp slow-blow fuse, you would blow the lamps instead!
Build the right circuit (fuse, diode, and dimmer) and you should be very happy with it.
Rox: You are right. I was thinking of the ratio between peak and RMS in terms of voltage, but it is really power. You can't do an integration of voltage. You have to do an integration of power.
I wonder why that website has so many different power diodes? Power diodes have two basic parameters: Peak Inverse Voltage and Average Forward Current. (There are others, but for this application they don't matter.) All of these 120 VAC=>83 VAC overhead projector applications should be able to use the very same 6 Amp 400 PIV diode.
If you are in a rush, go to Radio Shack and buy two of these:
Part# 276-1144 3 Amp 400 PIV $1.59 each
Then connect them in parallel to get a 6 Amp 400 PIV diode. (Parallel means you lay them side-by-side with the markings facing the same direction, and then twist and solder the leads together on both ends.) You will never overpower another lamp again.
But the dimmer is also a good thing to include, since lamps almost always blow when you fire them up. The cold filament draws more than the rated current until it warms up. So if you start them slowly (ie. 1 second to full power), then they should run a lot longer.
For a fuse, I would get an 8 Amp "Slow-Blow" fuse like Radio Shack's:
Part# 270-1030 8 Amp 250 Volt Slow Blow Fuse 4 pack/$2.39
You can't use a 5 Amp fuse to limit your lamp voltage effectively, it is a current-limiting device. Since the lamp resistance changes quickly as it warms up, all you will do is blow lots of fuses if you use a quick-acting fuse. With a 5 Amp slow-blow fuse, you would blow the lamps instead!
Build the right circuit (fuse, diode, and dimmer) and you should be very happy with it.
"Then connect them in parallel to get a 6 Amp 400 PIV diode. (Parallel means you lay them side-by-side with the markings facing the same direction, and then twist and solder the leads together on both ends.)"
mmm, I don´t ever seen a circuit with paralelled diodes, this is not welcome to diodes. You can´t say wich of them is really working... I would go for 1 diode only, it is more expensive but is the correct way.
mmm, I don´t ever seen a circuit with paralelled diodes, this is not welcome to diodes. You can´t say wich of them is really working... I would go for 1 diode only, it is more expensive but is the correct way.
i'm sorry, i don't mean to be a pain! i just really need to know what the amperage is at 82V with a 410W bulb. 😕
i had it going earlier; the dimmer i'm using is a $4 one, not very good. i had a 3A fuse in, and it blew when the bulb lit up (@ 30V spike). i changed to a 4A fuse and took it up to 50V. it ran for about a half hour, then blew the fuse. i don't want to blow another bulb! not until i can order a few spares off the 'net.
another question... i have my meter across the leads of the lamp, thats how i'm measuring AC voltage now. is this true voltage? does the dimmer do anything funky that i don't know about? i tried measuring voltage through the dimmer, but i don't get anything unless there is a load.
alright, two questions! this diode, the 1N5404, it just limits current? not voltage?
i'm trying to learn, thank you guys for your help!
BadBoy 😎
i had it going earlier; the dimmer i'm using is a $4 one, not very good. i had a 3A fuse in, and it blew when the bulb lit up (@ 30V spike). i changed to a 4A fuse and took it up to 50V. it ran for about a half hour, then blew the fuse. i don't want to blow another bulb! not until i can order a few spares off the 'net.
another question... i have my meter across the leads of the lamp, thats how i'm measuring AC voltage now. is this true voltage? does the dimmer do anything funky that i don't know about? i tried measuring voltage through the dimmer, but i don't get anything unless there is a load.
alright, two questions! this diode, the 1N5404, it just limits current? not voltage?
i'm trying to learn, thank you guys for your help!
BadBoy 😎
- Status
- Not open for further replies.