Author Topic: New to electronics  (Read 4226 times)

Tony3dd

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New to electronics
« on: September 24, 2013, 05:10:48 PM »
Hi all, I'm new to this hobby, and just in the beginning learning stages. I just purchased the book Make Electronics, a bread board, a Klein mm1000 DMM, and just ordered a Korad power supply. Never dealt with these DC power supply's before, and just want to make sure I understand this. If I set the volts to say 6.3, and hook the outputs to a 1K resistor, the circuit should pull about 6.3mA's. So I preset the volts to 6.3, the amps to maybe 100mA's, and the circuit will only draw the 6.3 mA's it need correct? So theoretically I could set amps to 7mA's, and the supply should power the circuit fine right?

Mr Eastwood

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Re: New to electronics
« Reply #1 on: September 25, 2013, 04:15:04 AM »
Hi all, I'm new to this hobby, and just in the beginning learning stages. I just purchased the book Make Electronics, a bread board, a Klein mm1000 DMM, and just ordered a Korad power supply. Never dealt with these DC power supply's before, and just want to make sure I understand this. If I set the volts to say 6.3, and hook the outputs to a 1K resistor, the circuit should pull about 6.3mA's. So I preset the volts to 6.3, the amps to maybe 100mA's, and the circuit will only draw the 6.3 mA's it need correct? So theoretically I could set amps to 7mA's, and the supply should power the circuit fine right?

I believe so.
Hey! Frisbee! Far out!

dr_p

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Re: New to electronics
« Reply #2 on: September 25, 2013, 05:33:03 AM »
Yes, that's correct, because what you are setting is current LIMIT, so anything above 6.3mA will work, since your load only draws 6.3mA.

However, if you set the current limit to 5mA, the power supply will not output the set voltage anymore (6.3 Volts), because 6.3V would lead to a 6.3mA current, so it only outputs U=Iset * R(load)=5mA * 1Kohm = 5V. So the power supply will lower the output voltage until the current limit is no longer exceeded.

Tony3dd

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Re: New to electronics
« Reply #3 on: September 25, 2013, 05:49:44 AM »
Thanks for the detailed explanation!