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I'm having trouble understanding a concept.

Let's just say that I have a simple circuit made using a 15V power supply.

If I have a linear 7805 5V regulator connected along the circuit, and then add an 1000 ohm resistor to the board, how would the output voltage change?

Is there a way to calculate this?

enter image description here

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  • \$\begingroup\$ You could start by first thinking what is the purpose of a voltage regulator, i.e. what voltage does it regulate? What is the voltage if there is a 100mA load? 50mA load? 5mA load? Assuming the voltage regulator is ideal (which they in real life are not). \$\endgroup\$
    – Justme
    Commented Oct 14, 2020 at 21:52

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The idea with a voltage regulator is that the output voltage will be constant no matter what the load is - up to the rated current of the regulator.

With 1 kΩ connected to the output of your 5 V regulator the current will be 5/1000 = 5 mA and since this is well within the current and power rating of your regulator the voltage will remain at 5 V.

Is there a way to calculate this?

Yes. Read the datasheet to work out what the maximum current the device can handle is.

You also need to watch the temperature rise. The regulator will "drop" voltage across it and this drop will be equal to \$V_{IN} - V_{OUT}\$. We can calculate the power dissipated from \$ P = VI \$ where \$V\$ is the voltage drop. So let's say you had 9 V in, 5 V out and were passing 100 mA then \$ P = (9-5)\times 0.1 = 0.4 \ \text W\$. If the regulator gets too hot it will go into thermal shutdown. You can improve the heat dissipation with a heatsink.

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  • \$\begingroup\$ So if I replace the 1 kΩ with let's say a 47kΩ resistor, the output voltage will still be a constant 5V correct? Thank you for your help \$\endgroup\$
    – Christian Mack
    Commented Oct 14, 2020 at 21:57
  • \$\begingroup\$ In theory, yes. In practice, some regulators lose regulation at very low loads like 47K, or no load at all. That's why they all have datasheets describing what they can do. With a 7805 you're probably OK though. \$\endgroup\$
    – user16324
    Commented Oct 14, 2020 at 22:05
  • \$\begingroup\$ @ChristianMack For an ideal voltage regulator, yes. For an actual 7805 which is not an ideal voltage regulator, a 47kohm resistor would be too high to allow for the minimum 5mA current draw specified for the regulator, so the output voltage could be unstable. The uA7805 regulator is specified that the voltage does not change more than 100mV with loads between 5mA and 1500mA. \$\endgroup\$
    – Justme
    Commented Oct 14, 2020 at 22:09
  • \$\begingroup\$ The cartoon shows one of the TO92 (I think) packages. It will handle only a couple of hundred milliamps so check the datasheet. \$\endgroup\$
    – Transistor
    Commented Oct 14, 2020 at 22:10
  • \$\begingroup\$ The cartoon also seems to read "NPN", so it might just be an artist's impression of a regulator, or a mock-up. \$\endgroup\$
    – Justme
    Commented Oct 14, 2020 at 22:29

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