power transformer current

I confess, that I did only a rough hand calculation, may be not very exact. AndrzejM kindly has drawn some typical waveforms found in the circuit. Assuming a load independant transformer voltage, the capacitor is always charging to the sine peak value (reduced by rectifier voltage drop) and (in a simplified view) discharging linear until it falls below the momentary transformer voltage. The linear falling ramp has a gradient of Idc/C. With a smaller capacitor, the intersection point with rectified sine voltage occurs more early, thus the ripple is less than doubling when using a halved capacitor value.

To calculate all these rectifier parameters complicated methods are used. Most of them involved graphical representation, not the exact equations. You need to know all the transformer parameters in order to determine the internal impedance of the transformer as the source, then you need count rectifying diodes parameters, sometimes even wires that connect rectifier to the filtering capacitor. Of course, all capacitor parameters must be known. Beside this you need to keep in mind what starting current will flow through the circuit. With big capacitors it is easy to get huge current surge that will destroy or dramatically reduce the diodes and transformer life. It is really complicated issue, but fortunately, in most of real low power applications we can use simple empirical formulas for rectifier design. Probably the best way is to go to any small transformer manufacturer web site and check their application notes. There are a lot of them on the web. You will find simple formulas that will give you good result especially for this vendor's transformers. But when you design something really powerful, e.g. welding machine, you need all these complicated design tools that will give you excellent result with all the small details.

hi all

thank you very much
i must say i have learned a lot from you guys
and i understand it now
regards

Good discussion and few good remarks. I must explain myself. At my first post I declared:

I do not want to analyze problem in deep. I just want to show you a direction you can follow and do a precise analyze.

Click to expand...

Discussion concentrated on main capacitor value (Ok I can agree wit 10 000 uF and with 1 000 uF, both values has their pro and contras, I personally like low ripple at main capacitor so 4700uF for me is the value I prefer). However there is a significant mistake on the drawing: The voltage at capacitors were drawn twice , once as it were without capacitor (half of sine) and on second drawing correct voltage with capacitor.

Finally I will calculate the ripple.
The load current is constant, because the 7805 gives the constant voltage at its output, so the load takes constant current and 4700uF capacitor is discharged with constant 1A current. Hence the voltage drop at capacitor is linear (not exponential)
From definition capacitor voltage is equal U= Q/C (Q= charge)
Q= I * time = 1A *8 msec ( not 10 ms , because for 2 ms capcitor is charged from rectifier ).

U = Q/C = (I*t)/C= 1A * 0,008 s / 0,0047 F = 1,7 Vpp

(1,7V peak to peak . Ripple value is often given as rms value of ac component).

The final remark.
To avoid the power on surge a thermistor can be included at transformer primary.
When cold, thermistors resistance is high and limits current surge , during normal work it worms up so its resistance is lowering and losses in thermistor became insignificant.

hi pal
thanks for following up

i got what you meant
cool
cheers

Added after 52 minutes:

hi
about the NTC thermistor
is it correct to connect it in the drawing ?
thanks

Yes, thermistor configuration your drawing is OK .

The power on surge gets interesting, when the device has a certain risk to blow a fuse at your mains supply. For transformer below some 100 VA, it isn't a topic, to my opinion. Toroidal transformers are more critical because of their hard saturation characteristics.