0
\$\begingroup\$

I'm trying to do a problem for class and I'm having trouble on clarification and where to start.
enter image description here

enter image description here

I understand how the turns ratio works, but I don't know how to find it here. I can't find anything in my book or in handouts; I only find them when there are resistors on both sides.

I'm also confused on what VL and Vp are. I think Vp is the peak of the whole wave but I don't know what VL is showing here. Vrpp is also confusing me. I just need some clarification and guidance.

Thanks

\$\endgroup\$
3
  • \$\begingroup\$ \$V_L\$ is the load voltage. \$V_p\$ and \$V_s\$ are the primary and secondary voltages. \$V_{r\ PP}\$ is the peak-to-peak voltage on \$V_L\$. \$\endgroup\$ Dec 12 '17 at 18:53
  • \$\begingroup\$ Ohhh, that makes more sense. So since the voltage across the load is 10, the turns ratio would then be 1:11? Or would I have to take into account the .7 per diode used then use that voltage? \$\endgroup\$
    – John
    Dec 12 '17 at 18:59
  • \$\begingroup\$ You need to take diode drops and ripple into account. They made it a bit more difficult by specifying RMS output voltage rather than average. \$\endgroup\$ Dec 12 '17 at 20:03
1
\$\begingroup\$

Since Vp is given as 110 Vrms, it must be the RMS voltage of the power source, not the peak of anything. VL is given as 10 V rms, so it must be the RMS value of the voltage at that point including Vr(pp), the peak to peak ripple voltage. You mostly need to work backwards from VL.

\$\endgroup\$
2
  • \$\begingroup\$ So how would I go about working backwards, would I have to take the load value and bring it back through the two diodes and take that value to find the secondary peak and turns ratio? \$\endgroup\$
    – John
    Dec 12 '17 at 19:15
  • \$\begingroup\$ First find the rectified but unfiltered peak value and filter capacitor value that will give the filtered VL voltage with the specified ripple. \$\endgroup\$ Dec 12 '17 at 19:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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