| Electronics > Repair |
| Slow start circuit 115VAC to 230VAC conversion |
| (1/2) > >> |
| ballen:
Putting this into "repair" though it is really "modification". I have a Parasound HCA-1200 amplifier from the 1990s. I bought it in the USA but now live in Germany. The amp is set up for 120V 50/60Hz, and I have used it in the past with a 240V/120V transformer. I'd like to eliminate that conversion transformer and if possible power it directly from the mains. While the amplifier has no provision to switch input voltages, I just took a look at the large (1kVA) torroidal transformer that provides DC. The good news: the primary comes out as four wires, with two pairs in parallel. So I think it is likely that there are two primary coils, currently being run in parallel, and that if I put them into series then I can drive the transformer directly from 230V. The amplifier has a soft-start circuit. When initially turned on, there is a 10 ohm 30W resistor in series with the primary. Then, when the DC voltage (from a separate low-current secondary winding, time constant set by a 22uF capacitor and a 220k resistor, so a couple of seconds) has risen to a reasonable level a relay closes and shorts out the 10 ohm resistor. I assume this is to keep the input surge current down and also to provide a gentler start to the amplifier itself. My question: if I am correct that if there are two primary windings, currently in parallel, and that all I need to do is put them in series, then is it also correct that I should replace the 10 ohm 30W resistor with a 20 ohm 30W resistor, to maintain the soft start? (If so, I will probably get another 10 ohm 30W resistor and put it in series, thus ending up with a 20 ohm 60W resistor.) Snapshot of the relevant bit of the schematic is attached. The resistor in question is R05. Note that this schematic does not show the primary as made of two separate windings. Thanks, Bruce |
| inse:
In principle the conversion should not be a problem, just make sure you have the polarity of the primary windings correct. With 10 Ohms of inrush limit you usually should not trip a 16A breaker, I would try it with the existing value. |
| Psi:
--- Quote from: ballen on October 27, 2024, 07:38:05 pm ---My question: if I am correct that if there are two primary windings, currently in parallel, and that all I need to do is put them in series, --- End quote --- Yep, but as others have said make sure you get them in series around the right way. |
| ballen:
Inse, Psi, thanks for your replies. Regarding the correct polarities of the two primaries, yes, I'll be careful to ensure that the series connection keeps the relative direction of current flow in the two windings the same as it is now. I'll use the standard trick of digging out an incandescent bulb and wiring that in series with the (new) primary on first power-up. If I have it right the bulb will stay dim, if I got it wrong the bulb will light. Inse: Coming back to the resistor question, since the transformer is rated 1000kVA that's about 8A continuous in the primary, which means that as designed the 10 ohm resistor will limit total inrush current to about 12A. If I leave that resistor as is, then the inrush current might be as much as 24A, assuming for the moment that I neglect the transformer self-inductance. That is high for a typical household 16A circuit, no? |
| inse:
With a guesstimated resistance of 2Ohms of the primary in series with 10Ohms, the initial current will be 20A for few seconds at max. This shouldn’t trip the breaker. But to be sure, go for 20 Ohms. |
| Navigation |
| Message Index |
| Next page |