single power supply by centertapped transformer and bridge rectifier.

[Adwaittronics]

I am constructing a fixed power supply of 5v and 12v,I have a centertapped transformer with following rating 230v 50hz at primary and 24v-0v-24v ,12v-0v-12v at secondery .I gave 230v at primary of the transformer and used only 24v-0v-24v rating supply at o/p.
out of the three wires of 24v-0v-24v i connected 24v to one i/p of DB107 bridge rectifier and center tap wire(0v) to second i/p of bridge rectifier and i kept the third wire(24v) hanging (not connected to anything).
I checked o/p voltage of bridge rectifier by multimeter in dc mode by connecting +ve probe to +ve of bridge and -ve probe to -ve of bridge,i got 24v dc and when reversed polarity i got -24v dc ,until this ok but when i change multimeter on AC mode I am getting 50v AC and 0v with reverse polarity.
I am not understanding why am i getting 50v ac at bridge rectifier o/p ?

 

[chuckey]

If there is no smoothing on the DC (no capacitor fitted) then the output from the bridge will be a series of pulses, all going to +24V, but returning to zero in between. The fact that the voltage at any instant can be going + or minus (as it returns from + 24V to 0V), can be interpreted by the meter as AC. The fact that it appears as 50 V is due to the sensitivity of the meter being different between DC and AC.
Frank

 

[FvM]

The result can be expected by a "poor mans" AC meter made with a half-wave rectifier.

The measurement is more or less meaningless anyway. For a real power supply design, you'll firstly add a filter capacitor, raising the DC voltage to Vpeak (about 33 V). Then select suitable voltage regulators.

With 33 V input, you'll burn a lot of power in linear voltage regulators.

 

[Audioguru]

If you use a center-tapped transformer then for a positive DC output you use two rectifier diodes, not a bridge rectifier module that has 4 rectifier diodes.
For 12VDC output the transformer should be about 18V center-tapped then the filtered output is about 9V x 1.414= 12.7V - 1V= 11.7V when loaded since the peak voltage of the sinewave is 1.414 times higher than its RMS voltage and a rectifier diode produces about 1V drop when it charges the filter capacitor..

 

[jiripolivka]

You did not tell us what load current you need. For 5VDC and 12 VDC it may be easier for you to find another suitable transformer with say 10-15 V AC output which can be easily regulated to 5 and 12 V DC.
Using your higher voltage from the rectifier, you will lose a lot of power by dropping the voltage through regulators to 5 and 12 V.

 

[crutschow]

If you connect a bridge to the 12V-12V windings you will get about 15V DC filtered from the bridge.

The voltage at the center-tap will look like a full-wave rectifier output (from the two rectifiers going to the bridge common terminal) and will be over 7V DC filtered.
You can get a little more voltage if you use Schottky rectifiers.

That should give you regulated +12V and +5V if you use low-dropout regulators.

 

[crutschow]

I miscalculated on my previous post.
With a 12V-0-12V winding the bridge will give a filtered output of about 29V and the center tap filtered output will be about 14V.