I've got a linear step-up transformer (120V AC input) which has an output that under load nominally supplies 230V AC @ 100mA. But with no load it floats to as high as 280 V. I need to switch the load in/out, but for "reasons" I need that no-load voltage to never exceed, say, 240V. Line voltage AC is sort of like a foreign country to me, where I don't get the natives' taste in hats. What's the easiest, cheapest way to do this? Seems like Zener diodes + resistor would have to be very large and probably a crazy way to do it? I don't really want to load that output when it's not being used for its intended purpose.
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1\$\begingroup\$ Please explain these "reasons" of yours. \$\endgroup\$– HearthCommented Mar 14, 2019 at 18:34
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\$\begingroup\$ Hearth, we're sort of tied to use a small, low-cost relay to switch the load in/out, and it is not UL rated to switch more than 250V AC. \$\endgroup\$– Drone601Commented Mar 14, 2019 at 18:41
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\$\begingroup\$ If the relay is rated for 250V AC, I wouldn't want to use it on 230V AC. That's cutting it really close. Why not put the relay on the primary of the transformer instead? \$\endgroup\$– HearthCommented Mar 14, 2019 at 18:49
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\$\begingroup\$ Relays rated for 277 VAC are common, though a simple contactor will have the lowest cost. \$\endgroup\$– user105652Commented Mar 14, 2019 at 18:52
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\$\begingroup\$ I'm having trouble finding a 12V DC coil relay rated for 277V that will also fit in the space allotted. And again, 280V > 277V. But I'll keep looking. I know many relays -- even 250V rated ones -- have manufacturer max ratings that well exceed even 277V. But it's a UL approval thing. @Heath, the transformer also has other outputs which are used for other tasks, so I can't just switch the input. \$\endgroup\$– Drone601Commented Mar 14, 2019 at 19:06
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2 Answers
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- Use a relay with higher voltage tolerance. You are cutting it too close with your specifications.
- Add a 1uF 400V capacitor to your secondary (perhaps in series with a small <100Ω resistor to attenuate resonances). This should introduce a load current in the secondary, while not consuming any real power beyond the induced secondary and core transformer losses.
Do note that a capacitive load introduces a leading power factor, under the wrong conditions this could actually increase the voltage of the transformer. A purely inductive load would clearly work better, but it would not be practical for this application (>10H would be required).
answered Mar 14, 2019 at 19:29
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One option is to leave a 1K ohm 100 watt resistor connected at all times, so when your load is OFF, the resistor will prevent the voltage from peaking so high. However, this needs to be a ceramic wire-wound resistor. At 240 VAC it will dissipate 57 watts, so it will need to be wrapped in chicken-wire to prevent touching it.
Running at 60% of full rated power, it should last a long time. Also 1K 100W 5% resistors are common, so should have a low cost. Even 300W 5% resistors of the same style cost about $33 USD. Check Digikey, Newark, Mouser and Allied for cost on a per-each basis.
This load resistor will drop it down from the high peak voltage, but it is unknown what it will be with this 1K load connected. Think of it as connecting a 60 watt light bulb that is always ON.
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\$\begingroup\$ Thanks Sparky. I'm looking into using a wirewound resistor, though experimenting with lower currents in the hopes I can find something that loads the transformer juuuust enough to make it happy while keeping the resistor dissipation down to something slightly cooler than, say, the average blue-white star. Still hoping there's an elegant physics workaround that can make a load without...er...a load. I guess. :) \$\endgroup\$– Drone601Commented Mar 14, 2019 at 19:32
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\$\begingroup\$ It is possible that a 2K ohm resistor will provide enough load, also Edgar Brown has a viable load option. Without data on the transformers wattage or VA rating or output current, we can give you only ballpark values. \$\endgroup\$– user105652Commented Mar 14, 2019 at 20:13