0
\$\begingroup\$

For conversion of AC into pulsating DC we used Rectifiers circuits. But I read somewhere (from a well-known site) that the circuit does not work in ideal condition if I used my first circuit. The second circuit is the correct circuit as per their view ideally in ideal condition.

As I used the first circuit many times but on breadboard only so it adds the resistance of wires so might be it works.But I do not understand the reason behind for that extra resistance actually. Please share your knowledge regarding this. It will be a great help in clearing my concepts on this topic. enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ A fine radiator of magnetic field interference is your first circuit. Home-entertainment installers know the 120Hz surges of microwave ovens, with 2000 volts charging the magnetron's power supply, generate 800,000 volt/second slewrates at input to rectifier; the 0.026 volt turnon (if just a single diode junction) scales up the edge by 40X to 32,000,000 volt /second edges. The peak current is? 100 amps, thus dI/dT is 3.2 Billion amps/second, worst case. Thus R1 is key to EMI control. \$\endgroup\$
    – analogsystemsrf
    Commented Oct 16, 2017 at 16:49
  • \$\begingroup\$ Alright, that means it is application specific actually. R1 is not required in all of the circuits. Until now I am only using the rectifier in power supply systems only. \$\endgroup\$
    – user152036
    Commented Oct 17, 2017 at 4:42

1 Answer 1

Reset to default
1
\$\begingroup\$

In an actual circuit, the effect of R1 is always present, whether or not there is a separate R1 installed.

If you simulate the top circuit with a zero impedance voltage source, diodes with no residual resistance, and a capacitor with zero ESR, then the current that flows during the charging pulses will be infinite, which is of course non-physical. When simulating, you should add an estimate of the transformer winding resistance as R1, as well as using diode models which include residual resistance, to get sensible charging current magnitudes. We can usually ignore capacitor ESR compared to those other two resistances.

Given the transformer resistance, wiring resistance, and the robustness of typical rectifier diodes to pulse loading, it's unlikely that you would ever need to add an explicit R1 to such a rectifier circuit to reduce the size of the charging current pulses. But I'd never say never.

\$\endgroup\$
5
  • \$\begingroup\$ Thanks for your input Neil_UK. But I would like to ask if the impractical value of current (infinite)is somehow generated then this circuit is unable to work?? Is it not give DC at the output if infinite current will flow through the circuit? I am just asking it in a theoretical point of you. \$\endgroup\$
    – user152036
    Commented Oct 16, 2017 at 9:32
  • \$\begingroup\$ One more thing to add- Is any component fails when infinite current will flow? Maybe that's the reason why circuit does not work at infinite current. \$\endgroup\$
    – user152036
    Commented Oct 16, 2017 at 10:10
  • \$\begingroup\$ @user152036 An infinite current never flows. I've addressed both theoretical and practical circuits in my existing answer. The circuit will 'work' to produce a DC output with or without R1. However, if the diodes are too weedy, they may fail with the high current pulses. The correct response to diode failure would be to use a bigger diode, or a diode with a better specified pulse behaviour, than to add an R1. Sometimes, R1 is provided as an NTC thermistor, to minimise inrush on switch-on. \$\endgroup\$
    – Neil_UK
    Commented Oct 16, 2017 at 11:54
  • \$\begingroup\$ Ok I think inrush current is the main reason behind that R1. Apart from inrush current, I am not observing anything which will stop working of this circuit. So to make a standard and practical rectifier a low-value R1 should be used !!! Is I am right ? \$\endgroup\$
    – user152036
    Commented Oct 16, 2017 at 11:58
  • \$\begingroup\$ @user152036 a 'standard' rectifier will not have a physical R1 provided. There's usually more than enough residual resistance in the transformer, the wires and the diode residual series resistance to limit the charging pulse current magnitude to the repetitive peak current handling of the diode. The diode will also have sufficient non-repetitive peak current to handle the switch-on inrush (see 1N540x diodes, 3A average, 200A non-rep peak), usually with a time-delay mains fuse to avoid blowing on inrush. To see an R1, NTC or fixed, would be unusual and non-standard. \$\endgroup\$
    – Neil_UK
    Commented Oct 16, 2017 at 13:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.