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I am learning electronics on my own and I have a feeling a capacitor can solve an issue I have.

The datasheet for a part, which is powered by three lines (+12V, 0V, -12V) says that it draws 40 mA continuusouly, but that it can also draw up to 250 mA temporarily, for 0.1 second.

Well, let's say I just cannot afford to put on my board a power supply capable of giving 250 mA (I can go up to 80 mA). Can I use a capacitor to help supply the temporary surge?

I have thought of :

Capacity required : 0.250 * 0.1 = 0.0250 A.s (Coulombs) Let's say I decide to connect the capacity to the +12V and -12V (is it ok ? Should I have two or three capacitors ?), then, with a 50 V capacity, that makes :

0.0250 / 50 = 0.0005 F

So I would need a 500 µF capacity to do the job. Is that correct ?

But then, when the part draws 250 mA for 0.1 s, the capacity get discharged, and it temporarily draws more current on the power source to recharge. I can solve this using a resistor, but where to put it, and how to choose it?

I have thought of that :

schematic

simulate this circuit – Schematic created using CircuitLab

OR

schematic

simulate this circuit

What do you think ? Is it even a good solution ?

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  • \$\begingroup\$ What is the part? If it draws every other 0.1 second say, that has to be factored in. \$\endgroup\$ – Erik Friesen Jan 1 '14 at 14:29
  • \$\begingroup\$ It is a 4-20 mA current loop reciever. So I would expect it to only occasionnaly draw that much current (thought I do not understand why it would). \$\endgroup\$ – PeterG Jan 1 '14 at 15:18
  • \$\begingroup\$ What is the part? (part number and link to datasheet??) This isn't really making sense yet. Is it a loop powered transmitter or what? \$\endgroup\$ – Erik Friesen Jan 1 '14 at 18:49
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You're sort of on the right track, but you can't just put 50V across your capacitor without also putting 50V across your circuit, which would probably damage it.

I would start by saying that you probably want two capacitors, one between +12V and ground, and the other between -12V and ground.

Then, you need to decide how much you can allow each of the 12V buses to "sag" during the current surge. For example, if you could tolerate a 2V drop during the surge, then the capacitance needed would be 0.025 C / 2 V = 0.0125 F, or 12500 µF.

The resistor — if needed at all — should be positioned as shown in your first diagram, in series with the source, not in series with the capacitor. But the power source's built-in current limiting will probably be sufficient without the need for another resistor.

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    \$\begingroup\$ The situation is a bit better if you assume that the power supply continues to provide 80mA while the capacitor is discharging. Then you need 0.017C or a measly 0.0085F. \$\endgroup\$ – Joe Hass Jan 1 '14 at 16:18

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