Does anyone know anything about these photocopier HV power supplies? Has anyone successfully used one?
I have 2 on hand a pin out for only 1 of them. I tried to get it working using a simple voltage divider to measure its ouput, but i could not get it working. I am thinking it may need a certain load to operate correctly.
They both take a 24V DC input and have several outputs including 6.3KV, but they do not say wether or not this output is AC or DC. Not hard to figure out using a scope but i can't get anything out of the one i tried to use.
I also have a simple power supply that outputs 300V DC but stepping it up seems to be more trouble then its worth.
Any thoughts?
thanks,
furly
I have 2 on hand a pin out for only 1 of them. I tried to get it working using a simple voltage divider to measure its ouput, but i could not get it working. I am thinking it may need a certain load to operate correctly.
They both take a 24V DC input and have several outputs including 6.3KV, but they do not say wether or not this output is AC or DC. Not hard to figure out using a scope but i can't get anything out of the one i tried to use.
I also have a simple power supply that outputs 300V DC but stepping it up seems to be more trouble then its worth.
Any thoughts?
thanks,
furly
They are usually DC. Pinouts are all proprietary- i.e. there is no standard.
Those supplies are usually only good for a few tens to hundreds uA out. Your resistive divider may be killing the output. You need a very high resistance divider. Try a 10M and a 1M resistor in series. When you measure the voltage don't forget to factor your meter's resistance in Total R = 1M (resistor) || 10M (typical meter) + 10M (resistor)
Expect >600V, so if your meter can't handle it, use 20M in series with 1M to drop the voltage. You should probably try to use HV resistors, but if you don't have them (who does?), off the shelf parts should work get you results that are close enough.
I_F
Those supplies are usually only good for a few tens to hundreds uA out. Your resistive divider may be killing the output. You need a very high resistance divider. Try a 10M and a 1M resistor in series. When you measure the voltage don't forget to factor your meter's resistance in Total R = 1M (resistor) || 10M (typical meter) + 10M (resistor)
Expect >600V, so if your meter can't handle it, use 20M in series with 1M to drop the voltage. You should probably try to use HV resistors, but if you don't have them (who does?), off the shelf parts should work get you results that are close enough.
I_F
Yeah this supply is rated at 800uA max at 6.3 KV.
I was using lower values in my divider (100K & 100), so yes, I guess it could not supply enough current. When I hooked it up to the scope, the signal was all noise which shows something like this was probably the case.
I will give the higher resistors a try and hopefully that will solve the problem. I don't have any HV resistors but given the low amount of current it should be ok, at least for a test run.
Also, interesting point on the meter. I have never considered taking into account its resistance even though it will be practically negligible.
I was using lower values in my divider (100K & 100), so yes, I guess it could not supply enough current. When I hooked it up to the scope, the signal was all noise which shows something like this was probably the case.
I will give the higher resistors a try and hopefully that will solve the problem. I don't have any HV resistors but given the low amount of current it should be ok, at least for a test run.
Also, interesting point on the meter. I have never considered taking into account its resistance even though it will be practically negligible.
The meter resistance is usually negligible, but when measuring very high resistance sources you have to consider it.
If you measure voltage across a 1K resistor, the total resistance becomes 1K||10M which is essentially 1K. Measure voltage across a 10M resistor with a 10M meter and the resistance becomes 10M||10M= 5M. Your measurement will be off quite a bit if you don't take the meter resistance into account.
I_F
If you measure voltage across a 1K resistor, the total resistance becomes 1K||10M which is essentially 1K. Measure voltage across a 10M resistor with a 10M meter and the resistance becomes 10M||10M= 5M. Your measurement will be off quite a bit if you don't take the meter resistance into account.
I_F
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