ripple current calculation?
I have assembled a timer with relay.
The supply is a nominal 9Vac bridge rectified and minimal smoothing.
The scope shows ripple on the DC feed of 12V to 14V (flat topped saw tooth) when driving the 555 timer and it's few components.
The supply drops to 3V to 13.5V (10.5V of ripple) when the relay pulls in.
The relay is 400r and the timer draws about 5mA.
How does one calculate the ripple current in the first and only smoothing capacitor? Iripple=Vpk/[relay resistance] + timer current?
hi Andrew, you might find Power Supply Designer good for your calculations.
Wondered how to calculate ripple current myself, however found this:
Under calculation examples there is formulas that seems comprehensive.
C=Iload / 8 x F x Vout
Iload is in Amps
F is in Hz
Vout is desired ripple in V
So, C= 1Amp / 8 x 60Hz x 1Vripple = 2 083uF
Calculating ripple current is not easy. In the end we used a current clamp on the big power amps to measure it.
the 22uF allows the circuit to work correctly and after running for an hour still felt cold.
10uF does not work, so I will adopt 33uF or 47uF (I have a few in stock).
the ripple current for a range of 22uF varies between 37mA and 50mA and that seems very close to the running current that the circuit consumes.
4fun's link shows a calculation method, but, it is more complicated than I expected so I will try a few numbers and see what pops out.
Thanks for the replies.
|All times are GMT. The time now is 08:15 AM.|
vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2014 DragonByte Technologies Ltd.
Copyright ©1999-2014 diyAudio