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#1 |
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Account Disabled
Join Date: Mar 2003
Location: Vancouver
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How do I make a standby power supply that will power the remote control reciever and a relay (about 50mA total)? My main goal is efficiency. I see that Digikey carries the BPS 1/4 to 2W supplies which are 50% minimum regardless of load, but that's $10 + as much for shipping, so $20. There's got to be a cheaper way to do a copule of watts power supply that will do that current at about 9V.
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#2 |
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diyAudio Member
Join Date: Dec 2003
Location: Austin
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Go to a pawn or resale shop and find a wall wart.
Integrate the guts or even the whole thing into your design. /edit Don't you have something like this laying around? I usually have at least a couple of random wall warts @ the house. Ask your friends and neighbours and coworkers. Someone is bound to have one.
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#3 |
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Account Disabled
Join Date: Mar 2003
Location: Vancouver
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How much power does a typical wall wart draw from the mains? Doesn't the wallwart draw power even when there's no load?
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#4 |
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Proud Union Member
diyAudio Member
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Yes, but proportional to load of course. If you're needing say, 50ma at 12v, that's only .6 Watts. Practically nothing.
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#5 |
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Account Disabled
Join Date: Mar 2003
Location: Vancouver
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Hmm. I hooked up a small transformer ripped from a broken wall wart (about 5 to 7 VA) to the mains and with a DMM set to AC current, it measures almost the same current drawn by the primary (about 25 mA) whether there's a load on the 9V secondary or not. Plus the transformer gets fairly warm, without any load. I don't know the phase between voltage and current so I can't calculate power. But it seems a good deal of power is going into heating the transformer. Is there a way to measure the power use without an actual power meter?
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#6 |
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Proud Union Member
diyAudio Member
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W=VA So, 120 X .025 =3W The transformer shouldn't get "hot" without a load. Maybe very slightly warm, but not hot. Try it without the secondary connected to the rectifier and filter section. If the load is dramatically less there may be something going on in that section of the circuit.
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#7 |
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Account Disabled
Join Date: Mar 2003
Location: Vancouver
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No, it's not hot, but quite warm (maybe 50* C). Probably it's just a crappy transformer with lots of parasitic losses.
Any way to roughly measure power without a power meter? Maybe some simple circuit that can be made? I guess I could just measure the heat given off but that's pretty hard to do without a lab. |
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#8 |
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Proud Union Member
diyAudio Member
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If you can measure how much current the primary is drawing from the line (you said 25mA in the last post and I assumed you were speaking of primary current), just multiply that in Amps by your line voltage. That will give you power, or watts. My appologies if this is known.
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#9 |
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Account Disabled
Join Date: Mar 2003
Location: Vancouver
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That is not correct, because it assumes that the current and voltage sine waves are in phase. It is possible to have them 180* out of phase, in which case there is no power drawn (power factor 0).
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#10 |
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diyAudio Member
Join Date: Dec 2003
Location: Austin
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Wall warts get warm. Yes. It is a large reason for a large electrical bill for someone who uses a lot of them. If you want a transformer to not draw current at load-idle, get a switch for the mains. Wasteful little transformers are a way of life for the modern small-electronics user. For grins, have you measured the idle current of a bigger transformer that you may have laying around for comparison?
If this is standby for a 500W dissipation monster, then a couple of watts at idle is relatively speaking ~peanuts~. If it really bothers you that much, make up a little PWM supply.
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