Electronics > Projects, Designs, and Technical Stuff
Simple DIY Linear PSU
kallek:
Sure it is enough. R18 is total 0.05 ohm so you can calculate power dissipation with 5 amp current by formula P=RI^2 = 1.25W. But as mentioned earlier, make sure that traces next to these resistors are quite wide. As well as other high current traces too.
Kleinstein:
With the usual voltage drop of some 200-500 mV one already has a power of 400 mW to 1 W at 2 A.
To get a stable resistance the temperature rise should be low. So the 2x2 W may be just OK for 2 A if the voltage drop is low (some limitations when the set current is low, may want a good OP for the current limit than).
For the resistor the permissible temperature rise is also different, depending on the type - some of the power wire wounds are allowed to go very high, so there rating is high, but the useful power as a shunt is still relatively low.
xavier60:
An important test for a power supply is how the output voltage responds to mains power loss.
Without an oscilloscope, this test will need to be done with a multimeter while input voltage(18V) is gradually decreased to zero.
One way is by temporarily connecting a very large capacitor across the 18V so that the voltage falls gradually when the mains is switched off.
The output should be set to 1V, then monitored as the 18V falls.
panoss:
Which value is 'very large'?
4700uF?
10000uF?
More?
And I should measure the time that it takes for Vout to go from 1V to 0V?
xavier60:
--- Quote from: panoss on March 16, 2020, 10:23:29 am ---Which value is 'very large'?
4700uF?
10000uF?
More?
And I should measure the time that it takes for Vout to go from 1V to 0V?
--- End quote ---
10,000uF will make it about 1v/s. You are looking for any increase of output voltage which is what we don't want to see.
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