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N96/II/Qns 4 

In the circuit of Fig 4.1, the 4 diodes, A,B,C,D are connected to form a bridge rectifier. The diodes may be assume to be ideal.
 

(a)(i) On the Fig. 4.1, mark with a + sign the positive output terminal 
    (ii) State which diodes do not conduct when input terminal X is positive with respect to input terminal Y 

(b)The input terminals X and Y are connected to the secondary coil of an ideal transformer. The primary coil is connected to a 240Vrms alternating supply. The input to the bridge rectifier is to be 12Vrms. 
  
(i) what is the value of the ratio [no. of turns  on secondary coil] / [ no. of turns on primary coil] ? 
(ii) an ideal diode does not conduct when a potential difference is connected in one direction across the diode. The max. potential difference which can be applied in this direction across a real diode before it conducts is known as the breakdown voltage of the diode. Calculate the min. breakdown voltage of the diode in Fig 4.1 if the circuit  is to function correctly.

  
Solution: 
 
Diagram: 
 

(ii)Diodes B and D  

(b) 
(i)(Ns)/(Np)= (Vs)/(Vp)= 12/240= 0.05 

(ii)If the input to the bridge rectifier equal 12 Vrms, then the max. output potential 12 x 21/2 = 17V. 
Diodes B and D must not conduct when terminal X is positive with respect to input terminal Y. 
The diode must be able to withstand at least 17 V in the reverse bias connections. Therefore min. breakdown voltage of each diode = 17 V. 
 

 
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