I am an electronics newbie looking to broaden my horizons, so I'm going through The Great Courses's Understanding Modern Electronics course. In Lecture 4, they give a project in which the goal is to simulate a DC power supply.

The supply should step-down a 120-volt rms, 60 hertz AC signal to a 5-volt DC signal. I got the correct schematic, but I came up with a different turns ratio than the one given in the project solution. The solution gives a turns ratio of 28:1, but I calculate a turns ratio of 21:1 by solving: $$5.7V=\frac{N_2}{N_1}(120V)$$

where 5.7 volts is the voltage before the voltage drop across the diode.

I simulated the circuit in LTspice XVIII, and it confirms that the turns ratio should be approximately 28:1, not 21:1. The closest guess I have as to why this would be would be that I ought to use a value of 4.3 volts instead of 5.7 volts, which corresponds to 5 volts dropping across the diode. However, this conclusion is contradicted by my simulation below, which has the expected value of 5V, not 4.3V, across the load.

How can I correctly calculate the turns ratio for my transformer?

My schematic and output


The answer is in your simulation voltage source. The 120VAC is quoted as RMS which is \$1\over\sqrt{2}\$ times the peak value, so the simulation uses 170V as the peak voltage (340Vp-p) 60Hz.

That's what comes out of the wall in North America.

So use the peak voltage in your calculation (169.7 or 170V) rather than 120V.

In general, if you see a power source given in volts AC it will usually mean RMS unless otherwise specified. For a sine wave the ratio of peak to RMS is \$\sqrt{2}\$. For other waveforms it generally differs.

  • \$\begingroup\$ When I did that, I got 29.8:1 for the ratio, which I thought may have been too far off. Do you think the difference is negligible? \$\endgroup\$ Jun 14 '20 at 5:18
  • \$\begingroup\$ I guess 28:1 in the solutions seemed oddly specific. \$\endgroup\$ Jun 14 '20 at 5:19
  • \$\begingroup\$ I don't know your course material scope, but if you adjust it for half the ripple you will come up with a slightly lower ratio. \$\endgroup\$ Jun 14 '20 at 5:19
  • \$\begingroup\$ That doesn't mean anything to me yet, but that probably is enough to make a meaningful difference given the results of the simulations. Thanks! \$\endgroup\$ Jun 14 '20 at 5:20
  • \$\begingroup\$ If you zoom in on a precise interval containing an integral number of cycles (eg. 0.1 second to 0.2 second, which contains 6 cycles of 60Hz mains) and ctrl-click the trace label LTspice will display the average value of the waveform- the output voltage for example (also RMS, which you don't need here). You can also delete the input voltage trace and just look at the output voltage, which will show the ripple waveform more clearly. \$\endgroup\$ Jun 14 '20 at 5:34

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