Ripple Voltage Equation Confusion - Page 1
Forum section: electronics / general

author: beaker353
author_rank: Unknown
timestamp: September 19, 2013, 04:15:08 am
content: I'm taking Solid State Devices/Circuits I at the moment and running into some confusion with two equations that I thought were for the same calculation, but I'm not sure now. I'm trying to make sense of come lecture notes that don't jive with what the book has. Here are the two equations:Vripple = (1-e^(-1/(fRC)))Vp(rect)Vripple = (1/(fRC))Vp(rect)My notes say the first equation for half-wave rectifiers and the second equation for full-wave rectifiers, but the book doesn't even mention the first equation which leads me to believe there is something else at play I must of missed. Is one more accurate than the other and both can be used for half or full-wave, or is there proper situations for one verses the other? Thanks!-EM

author: Unknown
author_rank: Unknown
timestamp: September 19, 2013, 08:11:22 am
content: The equation for full wave or half wave will probably be similar, it's just that f = 120 Hz for full wave and 60 Hz for half wave. 1-exp(-x) can be approximated by x for small values of x (try doing a Taylor expansion), so the second equation is just an approximation of the first.If this is a college level course, then I think you may be focusing on the wrong details. The point of the exercise was most likely to show how to do steady state analysis of an RC circuit, not to memorize the equations. An exam might feature an exercise where you have to calculate the ripple for an LR circuit, for example. Memorizing the equation for an RC circuit won't be of any help for that. Rather focus on the process of deriving this equation.

author: Wytnucls
author_rank: Super Contributor
timestamp: September 19, 2013, 09:04:11 am
content: As Alm says.These approximate equations can be used to determine the capacitor value required for a set ripple.Full wave rectifier: Vr=(1/2fRC)VpHalf-wave rectifier: Vr=(1/fRC)VpThe higher the load (lower resistance), the larger the capacitor has to be, to maintain the same ripple.

author: IanB
author_rank: Super Contributor
timestamp: September 19, 2013, 09:11:09 am
content: Also as alm says, try to understand where that equation comes form and how to derive it. Draw the circuit, sketch out the expected graph of voltage over time, and figure out how to calculate the ripple voltage from first principles.

author: Bertho
author_rank: Regular Contributor
timestamp: September 19, 2013, 10:16:16 am
content: Quote from: Wytnucls on September 19, 2013, 09:04:11 amYou should mention that the actual equation comes from the relation (charge equation):Q = I * t = C * VThe time t is 1/f for half-bridge and 1/2f for full-bridge. The current I can be (approx.) expressed as the load resistance and (average) output voltage in relation to the ripple voltage and the peak input voltage..Vout = Vp - 1/2 * VrippleI = (Vpeak - 1/2 * Vripple) / RloadQuoteI guess you mean "the higher the load", which implies alowerresistance. I know, splitting hairs, but is quite a reciprocal difference.

author: Wytnucls
author_rank: Super Contributor
timestamp: September 19, 2013, 10:30:25 am
content: Of course, thanks, writing too fast, post corrected.

author: beaker353
author_rank: Unknown
timestamp: September 19, 2013, 09:12:53 pm
content: Thanks for the information guys (and possibly gals). After pounding Google for a few hours (aka, sorting through useless results), it appears that both equations are just approximations of a much more complicated calculus based analysis. The first one is better for approximating half-wave and the second better for full-wave. Being that the EET (electrical engineering technology) classes I'm taking are non-calculus based, our analysis are all algebraic based which is why I believe I am a bit lost how these equations were derived.If I understand this correctly, the actual ripple waveform is made up of a bit of a rising sine wave (during the cap charge) followed by a RC time constant discharge (during the cap discharge), then the sine wave takes back over once its voltage is dominate over the cap voltage. Ripple Vpp would be equal to the top of the sine wave portion to the bottom of the RC tc portion right before the sine wave takes back over. Because the cap discharge portion is such a short period in comparison to the time constant for that RC combination, I have a feeling that we can approximate it to being linear even though it really isn't. Am I on the right path even though the math is a bit beyond me?-EM

author: Bertho
author_rank: Regular Contributor
timestamp: September 19, 2013, 09:22:08 pm
content: Quote from: beaker353 on September 19, 2013, 09:12:53 pmYou hit it spot on.The approximation goes because RC >> 1/f and the charge_time << 1/f. So you ignore the charge-time and assume the discharge linear.The approximation becomes less effective when the ripple rises. But, we normally don't care because you already need at least a 10..20% margin in the design (most caps are +/-20%) and even more depending input variability. The approximation is usually within a few percent.