I have 13.5 VDC output, and i want to power up a small radio that takes 4.5 VDC. I thought about adding 5/6 used 1.5V batteries in series, so it would be about 4.5 VDC for radio and about 8 VDC for batteries. Would that "battery voltage reducer" work? If not, how can i reduce voltage without using any complicated electronical stuff? (I am not an electrician)
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\$\begingroup\$ You are looking for a voltage regulator. Switching ones are more power efficient. \$\endgroup\$– kennyDec 6, 2012 at 13:37
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\$\begingroup\$ The most practical solution will be a device sold as a battery eliminator or universal AC adaptor. Often they have a switch allowing you to choose an output voltage from a selection of multiples of 1.5V. A quick Google shows lots of available devices in the under $10 range, some even set up for either 110V or 220V AC systems. The biggest problem you'll have at that point is making a connection to your radio, and possibly locking the voltage selection down so it doesn't get bumped. \$\endgroup\$– RBerteigNov 21, 2013 at 23:49
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4 Answers
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Your battery solution will not work and may be dangerous.
You need something like this; (search for "dc dc converter"; anything car orientated will be suitable, given that car battery voltage is somewhere in the 12-15V range)
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\$\begingroup\$ Isn't it possible to make my own regulator that would lower that DC output? I really don't want to buy anything since it's an old radio and i don't really need it THAT much. \$\endgroup\$– ojekDec 6, 2012 at 13:46
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\$\begingroup\$ Then it depends entirely on what electronic parts you have lying around already, which sounds like "none"? A 5V linear regulator would work for a ~$1 cost. \$\endgroup\$– pjc50Dec 6, 2012 at 14:06
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\$\begingroup\$ Well i have some old electrical train that was powered up by a transformer that could go from about 1.5 volt to 9V. If i would break up that transformer, disconnect electrical circuit from it and use it as my voltage regulator? Is there a slightest chance this could work? \$\endgroup\$– ojekDec 6, 2012 at 14:15
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You said "small" radio, but didn't say how much current it draws. If only a little, a 7805 voltage regulator may be all you need. Does the radio actually say on it that it needs 4.5V, are you getting this from the fact that it takes 3 1.5V cells, or something else? If this is battery operated, then most likely it will work fine on 5 volts. If you are worried about the exact voltage, use a adjustable regulator to make 4.5V.
Keep in mind that a linear regulator dissipates the difference in voltage times the current as heat. If the radio draws 100 mA, for example, then a 5V linear regulator would dissipate 850 mW. That's about the limit for a TO-220 package standing up from the board in free air. Put even a small heat sink on it and it should be fine. If the radio draws only 50 mA, then just a bare 7805 in TO-220 package is all you need. If 200 mA, then you should start to seriously consider a switching regulator.
answered Dec 6, 2012 at 14:20
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\$\begingroup\$ Yes, i am telling 4.5V based on how many batteries there go in. I want to power it up from a battery input. As i wrote above, is there a chance i would find that "7805" in any devices i already posess? Like phone chargers, laptops, etc? (Because i got some broken electrical things around here) \$\endgroup\$– ojekDec 6, 2012 at 14:31
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6\$\begingroup\$ @ojek: 7805 regulators are cheap and plentiful. After the nuclear annihilation, the only things left will be cockroaches and 7805 regulators. \$\endgroup\$ Dec 6, 2012 at 14:43
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Something like linear voltage regulator 7805 probably will be excellent for your case as it is very cheap, very simple to use, you need only 2 very cheap capacitors (something in the range from 22pF - 100nF will be OK for you) for "just in case" smoothing and optionally but highly recommended one rectifying diode as 1N4007 before 7805. You don't need a heat-sink for 7805 with 13-14 volt input. This setup will provide you with arguably the cheapest solution (bill of materials wise) and most simplistic one. Unused power will be dissipated as heat but it will be negligible and you shouldn't feel it.
P.S. I will edit my post little bit later to provide you with schematic to use.
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\$\begingroup\$ I have an old broken hi-fi stereo, old broken laptop, some broken phone chargers, and one tiny broken radio. Would i find that "7805" in that devices? If so i would take it off and build up what youre saying about with it. \$\endgroup\$– ojekDec 6, 2012 at 14:28
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\$\begingroup\$ Quite often if you salvage some kind of power supply to find more than one voltage regulator in it. Almost always they are mounted on heat-sink so it is plus. If you find a switching regulator use that for best efficiency and least thermal problem. \$\endgroup\$– zzzDec 6, 2012 at 16:17
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\$\begingroup\$ You would find the 7805 somewhere in the power supply (thats the PCB area where the battery or power cord is connected). \$\endgroup\$– 0x6d64Dec 6, 2012 at 22:31
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1\$\begingroup\$ Just one note from my (stupid) experience. Once I dismounted a power supply for the high quality parts in it. THEN I realized, that the complete, assembled, PS is immensely more worth than the sum of its desoldered elements! \$\endgroup\$– VoracDec 7, 2012 at 10:52
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2\$\begingroup\$ "You don't need a heat-sink for 7805 with 13-14 volt input" - This is outright wrong: The voltage alone does not determine heat generated in a linear regulator, the power output as heat is
P = V x I. So even if the current drawn by the 7805's load is a couple of hundred mA, you most definitely need a heat sink. \$\endgroup\$ Nov 22, 2013 at 5:33
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The problem with using the batteries is that they will always be charging. Draw it out and notice how current will enter the positive terminal of the battery. That's opposite to the way batteries are usually used and means the battery is charging, instead of supplying. Always charging means the batteries will eventually pop.
Unfortunately, to convert voltages you need complicated electronic stuff. The simplest solution is the linear regulator others have mentioned. It's almost as simple at your battery solution (one lead goes to the source and one goes to the battery). The only difference is that there is a third lead which needs to go to ground. Depending on how hungry your radio is, it also may get (too) hot, but that's electronics for you...
answered Feb 11, 2020 at 21:02