To all the clever ones out there.
I recently obtained about 40 tiroidal transformers from obsolete computor monitors, and rectified they produce about 30 volt under load. I have in the past wired two in series and used them to power two ESP 60 Watt amps, and they performed well, voltage dropping from about 63 to 60 whith both amps working hard. The transformers have no power rating printed on them.
Is it possible to determine how much current I can safely pull from these transformers? I am presently building two bridged amps (ESP 60 watter) and would like to know if I need to have separate supplies, or if two of the transformers will survive.
I am also considering using 4 of them in series to provide a +60 - 60 volt supply for a 350 Watt amp I have built.
Any suggestions?
Dieterd
I recently obtained about 40 tiroidal transformers from obsolete computor monitors, and rectified they produce about 30 volt under load. I have in the past wired two in series and used them to power two ESP 60 Watt amps, and they performed well, voltage dropping from about 63 to 60 whith both amps working hard. The transformers have no power rating printed on them.
Is it possible to determine how much current I can safely pull from these transformers? I am presently building two bridged amps (ESP 60 watter) and would like to know if I need to have separate supplies, or if two of the transformers will survive.
I am also considering using 4 of them in series to provide a +60 - 60 volt supply for a 350 Watt amp I have built.
Any suggestions?
Dieterd
If the transformers dont have a VA rating marked on them, the only way to get a rough idea of what the VA rating probably is is to run them under a resistive load till the rails start to sag then work out the VA that was required... the VA rating of the transformer was probably near to this figure.
Thanks,
What % sag is acceptable, these transformers lose about 3 volts from 63 - 60 when working but it seems to stabalise at that. That represents about 5% is that ok.
What exactly is a resistive load, I presume it must be adjustable so that one can load the transformer progressivly untill it get hot or sags?
What % sag is acceptable, these transformers lose about 3 volts from 63 - 60 when working but it seems to stabalise at that. That represents about 5% is that ok.
What exactly is a resistive load, I presume it must be adjustable so that one can load the transformer progressivly untill it get hot or sags?
http://www.betschart-trafo.com/deutsch/menu2/2/rt.htm
here´s a list with VA ratings and weight for standard torodial transformers
william
here´s a list with VA ratings and weight for standard torodial transformers
william
Based on how they seem to stabilize with a 5% sag, you may consider the cutoff point to be where they begin to sag again down from the 5% plateau. Like the others have said, be sure to run it for a while to be sure it doesn't overheat.
A good resistive load would be power resistors. For example at 30 VAC, 100 ohm, 10 watt resistors work. You would parellel 10 or 20 of them, say. For test purposes, you could make a power resistor by wrapping steel wire, about 24 guage, around a concrete or cinder block. More wire will give more resistance. You should be able to wrap enough to give you up to about ten ohms. Don't let the windings touch each other. Prep the ends of the steel wire with with 20 guage stranded copper wire by wrapping it up the ends for about 2 inches. Then take about 4 inches of the steel wire and wrap it over the copper to keep the connections tight. Try 5 ohms first, and if that draws too much current, wind more off the spool around the block and move the connection to the new end. You should do this test outside if possiblle since the zinc or enamel coating on the wire can emit fumes when heated. But it may not get hot enough to do that. It is best to use uncoated wire. Don' forget to scrape any coating at the connection points
Here is another possibility. I have often used power transistors mounted on heat sinks as loads. You can run the power through a rectifier and filter it with a cap, then through a .22 ohm, 20 watt, or maybe a .1 ohm 10 watt resistor to get a shunt to measure voltage drop across so that you can calculate current using V=IR. Connect a resistor in series with a potentiometer from the collector to the base and then increase the load by lowering the resistance on the pot. If your meter measures DC current up to 20 amps, you can put it in place of the shunt reistor and measure the current directly.
A good resistive load would be power resistors. For example at 30 VAC, 100 ohm, 10 watt resistors work. You would parellel 10 or 20 of them, say. For test purposes, you could make a power resistor by wrapping steel wire, about 24 guage, around a concrete or cinder block. More wire will give more resistance. You should be able to wrap enough to give you up to about ten ohms. Don't let the windings touch each other. Prep the ends of the steel wire with with 20 guage stranded copper wire by wrapping it up the ends for about 2 inches. Then take about 4 inches of the steel wire and wrap it over the copper to keep the connections tight. Try 5 ohms first, and if that draws too much current, wind more off the spool around the block and move the connection to the new end. You should do this test outside if possiblle since the zinc or enamel coating on the wire can emit fumes when heated. But it may not get hot enough to do that. It is best to use uncoated wire. Don' forget to scrape any coating at the connection points
Here is another possibility. I have often used power transistors mounted on heat sinks as loads. You can run the power through a rectifier and filter it with a cap, then through a .22 ohm, 20 watt, or maybe a .1 ohm 10 watt resistor to get a shunt to measure voltage drop across so that you can calculate current using V=IR. Connect a resistor in series with a potentiometer from the collector to the base and then increase the load by lowering the resistance on the pot. If your meter measures DC current up to 20 amps, you can put it in place of the shunt reistor and measure the current directly.
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