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We are trying to make an antique bell work I'm told it needs 12v .1 / amp or 100milli amp Our current power system is 24 volt 20 amp can you please help us step this down . We were going to use a common 24 to 12 volt resistor to drop the voltage but unsure of how to drop the amps down to .1

Lundin
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Brian
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  • But having said that, if you just need 12A 1.2W PSU, just buy something closer to that rating (SMPS are dirt cheap nowadays). A 480W PSU is just plain overkill... If you already have the supply, any 2W or so DC/DC converter module will do. – Tom Carpenter Aug 18 '16 at 02:52

2 Answers2

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Since you state you are going to use a resistor, I'll directly answer you about the current limiting question you asked.

If the bell works properly at \$12V\$ and it actually draws \$100mA\$ when attached to a \$12V\$ source, then you actually can use a resistor and it will still only draw the \$0.1A\$ and the bell will still only consume \$1.2W\$, as before.

To drop \$24V\$ down to \$12V\$ for the bell you would use a \$\frac{24V-12V}{100mA} = 12\Omega\$ resistor. However, you will need to also dissipate another \$1.2W\$ in the resistor. So I'd recommend getting a \$5W\$ resistor for this purpose (better safe than sorry.) \$10W\$ resistors are also available and may not be that much more expensive.

There are other ways, but given you are only wasting an extra \$1.2W\$, you may be right and there may be less incentive to make things overly 'difficult' here. The risks are modest, as you are dealing with low voltages and low wattage. However, this is an antique device, so you will want to be sure about your estimate of the required current. The resistor makes a poor voltage regulator. But it is easy and may work fine. I'd recommend verifying the current required at \$12V\$ before using a resistor, if possible, and certainly double checking the voltage across the bell afterwards.

So, in this circumstance the resistor itself doesn't do all the current limiting work. It does exactly half of it. Your bell does the other half. The \$10A\$ rating for your power source is almost certainly just a compliance rating and means it can handle up to that amount, but not more.

The risk you are taking here is that the bell's rated current draw is assumed to be known, fixed and unvarying. But it is a bell. And the current may vary. If so, the resistor does a very poor job regulating the voltage and you may expose the bell to a varying voltage exceeding what it can well tolerate. This is probably why you may want to consider other options offered here.

jonk
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    I would discurage this pratice unless you can guarantee it draws 100 mA at all times. A 7812, resistor and zener or DC/DC would be my recommendation. – winny Aug 18 '16 at 10:44
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I wouldn't use a resistor for this, it would be terribly inefficient and the bell most likely doesn't draw exactly 0.100 A at 12 V so you can't really use Ohm's law in this case. Google "24 V to 12 V converter" and get some ideas. Keep in mind you don't need one rated exactly 0.100 A, rather that is the minimum rating.

Jim
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    "Terribly inefficient" is probably fine for something that will be on for a few seconds per day. Power lost in a resistor would be 1.2 W. – Transistor Aug 18 '16 at 11:19
  • That's a good point. Lost track of the bigger picture. – Jim Aug 18 '16 at 13:02