0
\$\begingroup\$

Ignoring the voltage drop on the diodes, for a full wave rectified DC,can I say that:

1) Vrms of the original AC = Vrms of the rectified DC. ( I have just flipped one polarity. Power is same)

2) Vdc = V average = Vrms / 1.11072 ( the ratio between Vrms and V average)

( 0.7071 / 0.6366 = 1.11072)

(0.6366 / 0.7071 = 0.9003)

If I view the rectified DC as a mixture of pure DC, and AC component, the V average is the pure DC component.

3) Since I know the Vrms and the Vdc, can I calculate the AC component voltage this way: enter image description here enter image description here

The Vrms of the AC component would then be Vrms of the original AC sine wave / 0.435. enter image description here

If this is all wrong, what is the correct ratio? I am looking for an answer without using functions. Should the AC component be referred to as Vrms or V effective or both are ok?

\$\endgroup\$
  • \$\begingroup\$ 0.6366 / 0.7071 = 0.9003 \$\endgroup\$ – nidhin Jan 18 '15 at 18:40
1
\$\begingroup\$

The ac component is usually referred by \$V_{ac}\$. And it is given by

$$V_{ac} = \sqrt{V_{rms}^2 - V_{dc}^2}\tag1$$ $$(1)\Rightarrow V_{ac} = V_{dc}\sqrt{1.11072^2 - 1} = 0.483V_{dc}$$ $$(1)\Rightarrow V_{ac} = V_{rms}\sqrt{1^2 - 0.9003^2} = 0.435V_{rms}$$ $$V_{ac} = 0.483V_{dc} = 0.435V_{rms}$$

The ripple factor, $$\gamma = \frac{V_{ac}}{V_{dc}} = 0.483$$ where The actual value is \$0.482\$. The variation in answer is because of the round off errors.

\$\endgroup\$
  • \$\begingroup\$ Thank you for the answer. If I am after the Vac and not the ripple factor, is the 0.435 the correct value? \$\endgroup\$ – sparky Al Jan 18 '15 at 19:50
  • \$\begingroup\$ I think, that when calculating Vac, 0.482 is the multiplier from Vdc, and 0.435 from Vrms \$\endgroup\$ – sparky Al Jan 19 '15 at 1:41
  • \$\begingroup\$ Yes. Vac = 0.435Vrms = 0.482Vdc. \$\endgroup\$ – nidhin Jan 19 '15 at 2:55
0
\$\begingroup\$

The wave is asymmetric, so you need to clarify what you mean by Vac. The Vpp of the rectified wave will obviously equal Vac (ignoring the voltage drop) of the input, or Vpp-in/2.

In the conventional sense, the Vac of the rectified signal will be the peak amplitude, i.e. the point furthest from the average. In this case this peak is at zero DC. So (again in this sense) the amplitude will be 2/pi = 0.6366.

In any case (i.e. if we choose to define Vac in some other way), the formula in 3) doesn't seem to make sense.

\$\endgroup\$

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.