# Bridge rectifier circuit

The full wave bridge rectifier shown provides a DC output voltage of +/- 15 volts on which 1V peak to peak ripple is allowed. The supply should be capable of providing 1A DC current to its load resistor R find:

• The required rms voltage that must apear across the transformer
• The required value of C and R
• The average DC current
• The peak DC current through the diode

I don't get what is the relation between the DC output voltage and Vs in another word how can I get benefit from +/- 15 volt to calculate Vs?

Also (Vpeak = Vo + Vr) according to this law can I consider the Vo as 15 and Vr as 1? Is the ripple voltage value in this law for peak to peak or I have to regard it as 0.5? And what about the current of 1A, is it the average DC current?

I don't get what is the relation between the dc output voltage and Vs

Firstly, the output voltage is not "+,- 15V" - it is +15V dc (relative to ground/earth).

Secondly, the "+" and "-" labels on the transformer secondary winding are meaningless in this context - the secondary voltage is alternating and labeling them as "+" and "-" is confusing.

If your output dc voltage is +15V then, in simple terms, the secondary ac voltage has a peak value of about 15V. If your secondary voltage was less then your dc output voltage would be less.

Diode volt-drop - diodes aren't perfect conductors of electricity in one direction; they do "lose" about 0.7V and this means for a +15V dc output, the ac voltage feeding the bridge is more like 16.4V peak.

To calculate what the transformer secondary ac RMS voltage is, you divide the calculated peak voltage (16.4V) by $\sqrt{2}$ and this is 11.6V RMS.

Ripple - if you didn't have a load resistor, the capacitor would become charged to 15V and remain at that voltage. However, the resistor discharges the capacitor over every half cycle of the AC secondary voltage - this is what causes ripple voltage: -

To calculate ripple you need to know what R and C values are and here's another picture with a more detailed analysis: -

• Thanks alot ; but could you please advise how do you calculate the ac voltage feeding the bridge? I thought it will be 17.4 rather than 16.4 for this : Vp= Vo+Vr= 16 then Vs= Vp+2Vd= 16+1.4= 17.4 ! – user30717 Oct 21 '13 at 8:12
• @user30717 the ac peak is 16.4v i.e. two diode drops on top of the 15V output. Ripple voltage doesn't reduce the peak pinnacle value - see the bottom diagram in my answer - the peak remains the peak but the load discharges the capacitor between peaks. Imagine there were no capacitor at all - the peak would still be the peak but the ripple would be the full rectified voltage. – Andy aka Oct 21 '13 at 8:49
• Yes , I agree with you, but could you please advise how can I get the values of R & C ? Can I use Id average= IL (1+ 3.14159265359 (Vp/2Vr)^(1/2)) and assuming IL is equel to Vo/R ???? And by this I can find R??? – user30717 Oct 21 '13 at 22:53
• R is what the defined lowest resistance of the maximum load is ie if it's a 15watt transformer producing 15volts then the R value to use is 15 ohms. The 2nd picture contains the formulas to calculate C for a given ripple voltage expectation. – Andy aka Oct 22 '13 at 8:49
• Ok but I didn't have N1and N2 for the transformer,nor R or C values? But I have the current and ripple voltage so accoring to this is the current provided to me represents Id average or the load current? And according to the law of Id average (I've posted in previous comment), Do I have to use the peak to peak value ? Or I have to regard it as 0.5? Appreciate your cooperation ..... – user30717 Oct 22 '13 at 23:56