0
\$\begingroup\$

enter image description here

Assuming the peak voltage at \$V_1\$, \$V_{1p}=9.25V\$ and ideal silicon diodes (\$0.7V\$ drop in forward bias) and the DC output in \$V_0\$ is \$5V\$. I want to find the \$V_{op}\$, peak voltage at \$V_0\$. Given that the ripple voltage, \$V_r=0.2V\$ Then:

\$V_{DC}=V_{op}-\frac{V_r}{2}\$

And we have \$V_{op}=5.1V\$

I can understand why I couldn't apply KVL and write \$V_{op}=V_{1p}-1.4\$. Shouldn't this be the \$V_{op}\$?

If I apply KVL with \$V_{op}=5.1V\$ the equation doesn't hold true. What's going on? Excuse me because I this question may look similiar to another one I posted a while ago. This exercise is confusing me.

\$\endgroup\$
3
  • \$\begingroup\$ Could you rewrite your question? I can't follow your thinking process. You have \$V_{DC}\$ in an equation, but I have no idea where it really comes from. It seems like you conflate it with \$V_o\$, too. But is it just made up? And normally "peak" doesn't mean "peak-to-peak", so I'm not sure whether \$V_{1_p}\$ is peak or peak-to-peak. Write more and spend some more time shaping your question more precisely. \$\endgroup\$
    – jonk
    Oct 15, 2016 at 23:19
  • \$\begingroup\$ It is just made up and it is not peak to peak. Is given by the exercise. \$\endgroup\$ Oct 16, 2016 at 0:19
  • \$\begingroup\$ So this is an exercise, then. Something seems missing. I'm not sure I'm following it. \$\endgroup\$
    – jonk
    Oct 16, 2016 at 0:41

1 Answer 1

0
\$\begingroup\$

I can't tell if this is homework, or not. But if you actually are trying to figure out how to design a bridge rectifier system that yields a desired output and ripple specification, and you have complete freedom to build a transformer for it rather than being forced to buy one from a list of readily available units, then the process is something like this:

Specifications:

  • \$V_{OUT_{REGULATED}}=5\:\textrm{V}\$ (the final regulated voltage)
  • \$V_{DROP_{REGULATOR}}=1.5\:\textrm{V}\$ (the minimum overhead used by the linear regulator)
  • \$I_{MAX}=1\:\textrm{A}\$ (the maximum load current to be supported)
  • \$V_{RIPPLE}= 500\:\textrm{mV}\$ (the ripple voltage at the input of the linear regulator)
  • \$f=60\:\textrm{Hz}\$ (the operating frequency of the mains supply to the transformer)

From that you can work out the rest. For example, let's use the above values to get:

$$\begin{align*} t_{charge} = \frac{\textrm{sin}^{-1}\left(\frac{V_{OUT_{REGULATED}}+V_{DROP_{REGULATOR}}}{V_{OUT_{REGULATED}}+V_{DROP_{REGULATOR}}+V_{RIPPLE}}\right)}{4 \pi f}&\approx 1.6\:\textrm{ms} \\ \\ C = \frac{I_{MAX}}{V_{RIPPLE}}\left(\frac{1}{2 f}-t_{charge}\right)&= 13.5\:\textrm{mF}\\ \\ I_{DIODE_{MAX}} = \frac{\sqrt{2}\cdot V_{RIPPLE}\cdot C}{t_{charge}}&\approx 6\:\textrm{A} \end{align*}$$

At this point, even though the capacitor isn't a standard value, you can make an estimate for the voltage drop of your rectifier diode. I generally take this to be about \$800\:\textrm{mV}\$ when supplying \$1\:\textrm{A}\$, so in this case I'd guess about:

$$V_{DIODE}=800\:\textrm{mV}+120\:\textrm{mV}\cdot \textrm{ln}\frac{6\:\textrm{A}}{1\:\textrm{A}}\approx 1\:\textrm{V}$$

Now, you can estimate the transformer's RMS value as:

$$V_{RMS} = \frac{V_{OUT_{REGULATED}}+V_{DROP_{REGULATOR}}+V_{RIPPLE}+2 V_{DIODE}}{\sqrt{2}}\approx 6.3\:\textrm{V}$$

Luckily, that happens to be a standard transformer value.

But note that this is under load. The capacitor chosen matters here.


I'm going to stop at this point and let you clarify your question some more. Perhaps the above process isn't what you wanted, at all. But it gives you an idea of some of the factors involved. Perhaps someone else understands you better and can provide a better response. But for now, I'm still not sure what you are doing.

\$\endgroup\$

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.