I’ve recently been wondering if my IC would get fried if I have a 700 mA power supply? Since the IC only needs 20 mA.

Also I don’t understand resistance in Ohm’s law since I didn’t put any resistor. How can there automatically be resistance?

  • 2
    \$\begingroup\$ It depends if it is a constant current supply or constant voltage supply. We don't know what you have. Add as much info you can about your power supply and your IC. \$\endgroup\$
    – Justme
    Jul 4, 2023 at 19:17
  • \$\begingroup\$ you will need to look also to the power that can be supplied and the power your IC is consuming. Every IC consuming can be modeled as a resistor that consumes power. \$\endgroup\$ Jul 4, 2023 at 19:18
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    \$\begingroup\$ Your power supply can produce 700mA but it does not have to!. If you unplug the load it produces 0mA. The load takes 20mA of the available 700mA. It is ordering a pie and only eating one piece. You don't have to eat the entire pie. \$\endgroup\$
    – user338146
    Jul 4, 2023 at 19:27
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    \$\begingroup\$ @ronsimpson Not true if it is a constant current supply. It will try to push 700mA no matter what through the load until voltage has reached maximum. \$\endgroup\$
    – Justme
    Jul 4, 2023 at 19:33
  • 1
    \$\begingroup\$ @brhans That's some phone charger! \$\endgroup\$
    – Hearth
    Jul 4, 2023 at 21:41

2 Answers 2


There are two possibilities:

  1. You have a voltage supply with a current limit of 700 mA: in this case, the voltage will remain (about) constant whatever current you draw, as long as you don't exceed 700 mA. So if it is for example a 5V supply, then if your IC needs 20 mA at 5 V, then the power supply will deliver only 20 mA. If you connect five such ICs in parallel, then you will draw 5 * 20 mA = 100 mA, and the power supply will deliver 100 mA. If you connect 100 such ICs, you would need 100 * 20 mA = 2000 mA = 2 A: your supply can't provide such high current: at best, there will be some protection mechanism (automatic shutdown, lower output voltage until the current becomes acceptable, fuse burning), at worse the supply will catch fire. So if what you got is a voltage supply, and the voltage matches the voltage of the IC, then it will deliver only 20 mA, and your IC is perfectly safe.

  2. You have a constant current supply of 700 mA: in this case, it will try to deliver 700 mA whatever the load, increasing the voltage until the current reaches 700 mA (or the supply reaches it maximal output voltage). If this is the case, the voltage will increase until 700 mA flow through your IC: it will burn before reaching that point.

Current supplies are rather rare, and voltage supplies very common. But to be sure, I would advise you post either the datasheet of your supply if you have it, or if not a picture of the markings on the supply (and where it comes from).


I'd just like to add to the other answers by answering your question "Also I don’t understand resistance in ohm’s law since I didn’t put any resistor how can there automatically be resistance?" There is resistance. Your IC will definitely have some measurable resistance. Every non-superconducting component will have some resistance. If you want, you can put a multimeter across the terminals and find out how much resistance it has. So you can apply Ohm's Law here.

I don't know how much voltage your power supply is giving, and if it is even a constant voltage supply (see the other answer/s for details) but if it was, say ,9V you might apply Ohm's Law and find the resistance of your IC to be R = V/I | 450 = 9/0.02: 450 ohms.


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