I have an Acoustic Image Focus guitar amp that has a blown power amp board. I am replacing it with an IcePower 50ASX2BTL. The IcePower has 2 auxiliary power outs of +/- 25V DC unregulated with a 200mA max current draw. I am going to use these outputs to a step down transformer to power the pre amp at 12 and 6 volt. What I need to know is the value of any current limiting resistor that will be placed between the IcePower aux power out, and the step down transformers to keep the voltage at 25v and limit the draw to 200mA max. The IcePower has 400mA fuses on the power out, and they require the current draw to stay below 200mA. Excuse my ignorance if I am not approaching this correctly. I’m coming up with a 120 ohm 5 watt resistor, and I would wire this in line between the IcePower and the step down transformer.
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Current limiting resistor question for a project
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You can't step down DC voltage with a transformer. I'd probably build circuits with dropping resistors and regulators.
Edit: I'd hate to see a brand new amp module go up in smoke!Last edited by The Dude; 04-12-2024, 11:48 PM."I took a photo of my ohm meter... It didn't help." Enzo 8/20/22
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Excuse my lack of knowledge on this subject. These are what I propose to use to step down the voltage, and they are converters . Regarding the current draw of the preamp- I don’t know. I’ve been in contact with the son of the Acoustic Image founder (who recently passed away) and he doesn’t have that info. Schematics currently are not available. If the current available is insufficient, the preamp just won’t power up, correct? Am I able to limit the current to 200 mA between the IcePower and these converters using a current limiting resistor?
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If there's insufficient current then the preamp is likely to run out of headroom and distort. A resistor in the supply drops voltage according to load. Without knowing what the demands of the preamp are it's possible that a resistor that would fully protect the supply would affect the preamp sound adversely. You really need to know what the current demand is under quiescent and maximum signal conditions.
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I believe it's a gain stage with gain of 1 and the minus means inverting. So, basically it's an inverter in this case.
Edit: You are basically making a single push/pull amp out of 2 amps. Each amp is doing a half cycle- one the negative and the other positive. So, one of the two amps needs to be fed an inverted signal, much like the phase inverter in a tube amp.Last edited by The Dude; 04-19-2024, 02:34 AM."I took a photo of my ohm meter... It didn't help." Enzo 8/20/22
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