Cold Fusion
Senior Member
- Location
- way north
? looking for 120 to 480v transformer
- Thread starter Denis
- Start date
- Status
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100110-2245 EST
Without doing calculations correctly, but making some estimates.
Suppose the capacitor bank is 1000 mfd, and the charging resistor is 200 ohms. Then the time constant for charging is about 0.2 seconds. Thus, 95% charge occurs in 0.6 seconds.
I will estimate the average current for 0.6 seconds is 0.6*450/200 = 1.35 A. This 0.6 has nothing to do with the 0.6 seconds. If the flash rate is once every 2 seconds, then average current would be about 1.35*0.6/2 = 0.4 A and the rough transformer VA rating would need to be 180 VA. If the flash rate is once per 10 seconds, then the VA rating could be about 40 VA.
Thus, the transformer size will be greatly dependent upon the required flash rate. My use of 1000 mfd is probably way high. Thus, if a much smaller capacitor is used, then for given recycle time the transformer could be smaller. A 1000 mfd at 450 V is 0.5*10^-3*450^2 = 100 joules.
.
Without doing calculations correctly, but making some estimates.
Suppose the capacitor bank is 1000 mfd, and the charging resistor is 200 ohms. Then the time constant for charging is about 0.2 seconds. Thus, 95% charge occurs in 0.6 seconds.
I will estimate the average current for 0.6 seconds is 0.6*450/200 = 1.35 A. This 0.6 has nothing to do with the 0.6 seconds. If the flash rate is once every 2 seconds, then average current would be about 1.35*0.6/2 = 0.4 A and the rough transformer VA rating would need to be 180 VA. If the flash rate is once per 10 seconds, then the VA rating could be about 40 VA.
Thus, the transformer size will be greatly dependent upon the required flash rate. My use of 1000 mfd is probably way high. Thus, if a much smaller capacitor is used, then for given recycle time the transformer could be smaller. A 1000 mfd at 450 V is 0.5*10^-3*450^2 = 100 joules.
.
That's fine if the source is pure DC. In this case it isn't.
For much of the time the output voltage from the rectifier is lower than the voltage and no charging takes place at those times.
A couple of pictures:
Not sure how this one will post but it shows the voltage after about one second.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100111-1956 EST
Besoeker:
Thanks for the plots. I substantially under estimated the charge time. I did not want to bother to do an actual calculation. and should have thrown in a fudge factor.
My guess at 1000 mfd may not be too far off. I took a look at an old capacitor bank for a drawn arc stud welder using a capacitor discharge energy source. This bank was about 4" * 24" * 10" and consisted of 6 - 5200 mfd 200 V capacitors.
.
Besoeker:
Thanks for the plots. I substantially under estimated the charge time. I did not want to bother to do an actual calculation. and should have thrown in a fudge factor.
My guess at 1000 mfd may not be too far off. I took a look at an old capacitor bank for a drawn arc stud welder using a capacitor discharge energy source. This bank was about 4" * 24" * 10" and consisted of 6 - 5200 mfd 200 V capacitors.
.
chris kennedy
Senior Member
- Location
- Miami Fla.
- Occupation
- 60 yr old tool twisting electrician
complete unit
Sorry but I have to ask again, what is that?
SAC
Senior Member
- Location
- Massachusetts
"photographic strobe unit"
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100112-0931 EST
Besoeker:
Suppose I had fudged the exponent to 0.636*t/RC how would the two charge curves differ?
Additionally if we adjusted the above fudge factor so that the 95% charge points were equal, then how would the DC exponential curve compare with the full wave charge curve? What is the fudge factor for this equality point?
.
Besoeker:
Suppose I had fudged the exponent to 0.636*t/RC how would the two charge curves differ?
Additionally if we adjusted the above fudge factor so that the 95% charge points were equal, then how would the DC exponential curve compare with the full wave charge curve? What is the fudge factor for this equality point?
.
I think the biggest difference is the lack of discontinuities when the charging voltage is level DC.
In both cases the capacitor being charged would start at zero and end up at 450Vdc - one as a smooth curve, the other as a series of steps.
Wrong image.....new one if I make the edit in time...
Last edited:
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100112-1925 EST
The DC charging curve is an exponential. The shape of a smoothed version of the AC charging curve will differ somewhat from the DC curve. If R is adjusted so that the DC and AC charging curves intersect at 95%, then how do the curves compare?
I believe the early part of the AC curve will overshoot the DC curve. Is this correct
.
The DC charging curve is an exponential. The shape of a smoothed version of the AC charging curve will differ somewhat from the DC curve. If R is adjusted so that the DC and AC charging curves intersect at 95%, then how do the curves compare?
I believe the early part of the AC curve will overshoot the DC curve. Is this correct
.
info
info
it's a high-power flash 200 watt sec. recharge 10 sec +/- 20%
I have read that if the capacitors recharge too quickly, the ready light might not come on
One idea I came up with is to hook it up to a 20-500v Electrophoresis power supply, after I replace the burnt diodes
the bridge chosen is rated for 1,000v 35A # 641-1380 from digikey
I prefer to order from them, quick shipping from MN to IA
performance data
4 x 600uf caps in parallel
the complete manual can be found here
description of capacitor pack
schematic of capacitor pack
strob unit schematic
power supply componets
I was busy building a workbench and running romex, and working the snow
info
it's a high-power flash 200 watt sec. recharge 10 sec +/- 20%
I have read that if the capacitors recharge too quickly, the ready light might not come on
One idea I came up with is to hook it up to a 20-500v Electrophoresis power supply, after I replace the burnt diodes
the bridge chosen is rated for 1,000v 35A # 641-1380 from digikey
I prefer to order from them, quick shipping from MN to IA
performance data
4 x 600uf caps in parallel
the complete manual can be found here
description of capacitor pack
schematic of capacitor pack
strob unit schematic
power supply componets
I was busy building a workbench and running romex, and working the snow
broadgage
Senior Member
- Location
- London, England
This would appear to be a photographic flash unit that was intended to be powered by a high voltage dry battery.
I very much doubt that the battery could supply 550ma,(except perhaps very briefly) that would be about 250 watts ! from a portable zinc carbon battery?
The rating is almost certainly 550ma/h, that is a battery capacity of just over half an ampere hour, and is a common capacity for high voltage dry batteries.
A line powered unit to replace such a battery can be very low powered since it is required to charge a capacitor, not to supply a continous load.
I would aim for an output of about 425 volts since the dry battery had a voltage of 450, which would fall a bit with use.
It might be unwise to exceed the 450 volts nominal of the dry battery.
It would appear that a transformer voltage of about 300 volts is required, since after rectification this would charge the capacitor to about 425 volts.
A resistance of perhaps 1,000 ohms would limit the initial charging current to about what the battery would have provided.
No great accuracy can be expected since we are trying to simulate a dry battery, the voltage and internal resistance of which are both rather variable.
Alternatively it might be simpler to obtain a 450 volt battery.
Or to connect 50 batteries each of 9 volts, in series.
I very much doubt that the battery could supply 550ma,(except perhaps very briefly) that would be about 250 watts ! from a portable zinc carbon battery?
The rating is almost certainly 550ma/h, that is a battery capacity of just over half an ampere hour, and is a common capacity for high voltage dry batteries.
A line powered unit to replace such a battery can be very low powered since it is required to charge a capacitor, not to supply a continous load.
I would aim for an output of about 425 volts since the dry battery had a voltage of 450, which would fall a bit with use.
It might be unwise to exceed the 450 volts nominal of the dry battery.
It would appear that a transformer voltage of about 300 volts is required, since after rectification this would charge the capacitor to about 425 volts.
A resistance of perhaps 1,000 ohms would limit the initial charging current to about what the battery would have provided.
No great accuracy can be expected since we are trying to simulate a dry battery, the voltage and internal resistance of which are both rather variable.
Alternatively it might be simpler to obtain a 450 volt battery.
Or to connect 50 batteries each of 9 volts, in series.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100113-0940 EST
Besoeker:
Thanks.
I am using the 95% point as a reasonable full charge value because this is about 90% of full energy charge. Your 52.3 ohms for the resistor is a ratio of resistances of 200/52,3 = 3.82 .
There are two resistance values in the provided diagrams in the charging circuit, 750 ohms and 200 ohms. This ratio is 3.75 .
There are clearly some connector labeling problems.
I suspect the 750 ohm resistor is used for charging from the battery, and 200 ohms when charging from AC.
The time constant for 750 ohms and 2000 mfd is 1.5 seconds. This means about 4.5 seconds to 95%. This would be consistent the 10 second recharge time.
Denis:
I think you need some relabeling of the pins in your post #1.
In the AC circuit the peak initial current is 450/200 = 2.25 A. Four MR760 diodes in a bridge would be more than adequate. Personally I would probably use four 1N5627 diodes.
Assume 50% efficiency and 200 watt-seconds becomes 400 for input energy. Suppose you flashed every 5 seconds than the input VA is about 80. So a 100 VA transformer should be adequate. Being less conservative you can use a 50 VA transformer.
To get a nice package you may find the easiest path is to have one wound for you. Input 120 V output 320 V.
You might consider a Signal Transformer STP-0100-2115 with a special secondary. The package size is 4.5" OD, 1.5" high, and 3#.
.
Besoeker:
Thanks.
I am using the 95% point as a reasonable full charge value because this is about 90% of full energy charge. Your 52.3 ohms for the resistor is a ratio of resistances of 200/52,3 = 3.82 .
There are two resistance values in the provided diagrams in the charging circuit, 750 ohms and 200 ohms. This ratio is 3.75 .
There are clearly some connector labeling problems.
I suspect the 750 ohm resistor is used for charging from the battery, and 200 ohms when charging from AC.
The time constant for 750 ohms and 2000 mfd is 1.5 seconds. This means about 4.5 seconds to 95%. This would be consistent the 10 second recharge time.
Denis:
I think you need some relabeling of the pins in your post #1.
In the AC circuit the peak initial current is 450/200 = 2.25 A. Four MR760 diodes in a bridge would be more than adequate. Personally I would probably use four 1N5627 diodes.
Assume 50% efficiency and 200 watt-seconds becomes 400 for input energy. Suppose you flashed every 5 seconds than the input VA is about 80. So a 100 VA transformer should be adequate. Being less conservative you can use a 50 VA transformer.
To get a nice package you may find the easiest path is to have one wound for you. Input 120 V output 320 V.
You might consider a Signal Transformer STP-0100-2115 with a special secondary. The package size is 4.5" OD, 1.5" high, and 3#.
.
- Status