Dear members of All About Circuits, in a question I need to "Design a linear power supply consisting of a transformer, bridge rectifier and smoothing circuit, for use in the UK, to provide a + 15 V output for a load drawing 2500 mA. The ripple voltage should be set at 5 %." This is what I have done so far. 1) Vp - (2*0.7)=15V -> Vp = 15 + 1.4 = 16.4 2) Determine the transformer Vp=16.4, hence Vrms=16.4/(√2)=11.6Vrms Turns ratio=n:1= 230Vrms/11.16Vrms ≈ 20 turns ratio 3)Value of the ripple voltage= 5% of 15V = 0.75Vr 4) Determine the frequency UK Mains = 50Hz, ripple frequency = 100Hz 5) To Find C= (Iload)/(2f Vr)= 2.5A/(2*50*0.75)=33mF Is this enough? Which figure shall I report as the VA rating? Do all these components make the power supply linear? I hope that someone can help me. Thanks in advance! All the best!
Your approach looks correct as I can see. You have calculated for the diode drop. And use 2xmains frequency in your calculations. Then it comes to VA rating things are kind of more complicated. As the transformer charge the filter cap in pulses. You need a transformer than can supply a higher ampere rate than the DC current supply spec. Here it is more common to work with rules of thumb. Here is link to a document describing those rule of thumb. http://www.audiofaidate.org/it/materiale/20_020_020_001_PowerTransformer_FilterRatings.pdf
Thank you very much! Also, once the transformer is calculated with those criteria, does it meet the requirements of being linear? Thanks again
What a 'Linear' power supply is is basically a normal transformer power supply . In other words 220V - > transformer ->rectifier - > Filter (and maybe a voltage regulator or whatever) A non-linear supply is a switching power supply which uses a slightly diffrent technology called PWM if i'm not mistaken . So basically if you are using a transformer to step down the voltage then it's a linear supply , so yeah... you made a linear supply.
This part is wrong: That only applies for true sine waves. In a bridge retifier design, the load current flows in short duration (high current) peaks which load the trasnformer windings. That "clips" the tops of the AC V waveform so that the peak voltage is nowhere as high as the value of 1.414 X VRMS. Transformers are specified using resistive loading, so the peak voltages are not clipped at the top. This is a complicated subject. This appendix explains it.
This is correct: 5) To Find C= (Iload)/(2f Vr) That tells you the magnitude of the voltage ripple across the filter cap.