[Solved] Help with a L7805 Linear Voltage Regulator
So I've got 12 volts coming into my breadboard via a barrel jack. I've tried connecting this gut up a few times and I'm getting nowhere near 5 volts.
In Bill's video, he recommends a 2.2uF capacitor on the input and a 100uF on the output. I don't have a 2.2uF capacitor, so I used a 10uF on the input and a 100
uF on the output. I was still getting fluctuating low voltages.
Is there something I'm not understanding, or am I just wiring this wrong?
I had miswired my barrel jack. Oops!
Cool! I was just about to post about this voltage regulator. Are you using a circuit like this?
@yurkshirelad Yes! But I used different capacitors, 10uF and 100uF respectively.
Have you noticed if the voltage regulator needs a heat sink? Or is it not getting warm?
@yurkshireladI haven't gotten much use out of it yet, got distracted by some 3D printing. It was getting warm, not hot.
A lot of electronics beginners get trapped with these linear regulators. You may be surprised to find how little output current you can supply your load, without a heatsink. E.G. 100mA. The particular package type you have is a TO220 and if you add a heat sink you can provide more output current IF the voltage drop across the regulator is not huge. The 7805 will overheat if it's maximum Power dissipation is exceeded. It's all in the Watts! So 12 volt input minus the 5 volt output is 7 volts. PWR (watts) = V*I So if your load needs 1 amp then Pwr = 7 watts. (Hot). The approx. minimum input volts is usually 3 volts higher than the output, so IF you had an 8 volt input then the watts = 3v*1A = 3 watts. Now if you only needed half an amp (500mA) that would be 3 volts by 0.5 A = 1.5 watts. Etc. This link is a bit tough to take in all the thermal calcs. but you get to see some heatsink pictures down the bottom. Cheers mate!
Try Using a transformer with a smaller output voltage then when rectified to dc the power loss through the regulator will be smaller and use a big a heat sink as you can fit
Hope This helps