Capacitor Calculations... Correct? - Page 1
Forum section: electronics / beginners

author: t1d
author_rank: Super Contributor
timestamp: October 02, 2018, 08:50:30 am
content: This is a new area, of learning, for me... So, think noob...This is not my homework. I need this information, for my current project… A voltage rectifier/regulator, for a computer-style fan.I have a 12vac voltage source. It is half-wave rectified, to dc. I want to use a capacitor, to fill the 50% duty cycle, to run a 12vdc computer fan.Have I calculated this, correctly?Calculate voltage available. How does the 50% duty cycle effect this calculation? 0.4 x ½ = 0.2; therefore the factor should be 1.2, not 1.4?Xvdc = 12vac x 1.4 = 16.8vdcXvdc = 12vac x 1.2 = 14.4vdcOr, maybe use 12/2 = 6 volts, times one of the above?Calculate watts needed. Assume 1/2a draw, for the fan.12vdc x 0.5a = 6wTime – Half Wave Rectifier60Hz/2 = 30Hz = 0.0333 secondsCalculate Joules6w x 0.0333s = 0.1998jYou should only use the first 1/3 of a capacitor’s charge? Is this accounted for, in the needed joules calculation?3 x 0.1998j = 0.5994jCalculate capacitance, for the needed joules, and resistance, for the needed time. Are either of the below calculations correct? If so, which one? If not, what needs attention?Because the zener is in the circuit, the cap will never see higher than 12vdc? So, the calculation should be based on a 12v starting point. Correct? Is the clamp schematic correct?Thank you, for your help...

author: IanB
author_rank: Super Contributor
timestamp: October 02, 2018, 08:58:16 am
content: If you start with 12 V AC and half wave rectify it you will get 8.5 V DC, because you have kept only half the available power from the original AC supply.You now will have an uphill task getting back to 12 V for the fan. It would be much better to start with full wave rectification. Is there a particular reason for using half wave rectification?

author: t1d
author_rank: Super Contributor
timestamp: October 02, 2018, 10:45:48 am
content: Hi, Ian, thank you, for your reply... I already have a half wave rectifier for driving op amps with a positive and negative supply. I thought trying to combine a full wave rectifier with it might cross up the negative, with the ground. I posted a separate question, about that, and only received information as how to use the half wave supply. Do you know of a way to use both types of rectification, on the same board?If only 8.5VDC is available, that is no problem; I will just use a 5v fan. But, even for that, all of my questions stand, just at 5v, instead of 12v.Thank you, for your help.

author: Ian.M
author_rank: Super Contributor
timestamp: October 02, 2018, 11:46:13 am
content: Your other topic:https://www.eevblog.com/forum/beginners/how-to-use-a-half-and-full-bridge-rectifier-in-the-same-circuit/had a 12V AC secondary grounded at one end, which when rectified will provide 17V peak less diode Vf drop.You don't *NEED* a common ground for your fan and your OPAMP circuit. Use a bridge rectifier, with its AC terminals connected across the 12V AC supply, and do *NOT* connect its - terminal (or any other part of the fan circuit after the bridge rectifier) to ground.For a full wave rectified supply, in the USA there's 8.33ms between peaks. You'll loose about 1.5V to diode drops in the bridge rectifier, so expect about 15.5V peak, under load. You need to keep the regulator input voltage above 12V, so you can tolerate 3V ripple. From Q=CV, and substituting It for Q, and numbers for I, V and T, solving for C:C= 0.5 * 8.33e-3 / 3 = 1388uFDue to component tolerances, I wouldn't recommend less than 2000uF.Its *STUPID* using a linear regulator for a cooling fan - it just makes more heat to get rid of, and the fan's BLDC motor makes switching noise anyway, so you might as well use a switching regulator. For a one-off or low volume prototyping I would suggest a generic 3A buck converter module, commonly available from EBAY, other Chinese suppliers and a number of western hobby robotics and maker suppliers.A 1nf capacitor between the bridge - and ground may be advisable to suppress EMI from the fan and regulator.

author: t1d
author_rank: Super Contributor
timestamp: October 02, 2018, 12:47:42 pm
content: Thanks, Ian,Is this what you intend? Please look closely at the connection to the 12vac supply.I appreciate the help.

author: Ian.M
author_rank: Super Contributor
timestamp: October 02, 2018, 12:59:22 pm
content: Pin 2 of J5 still needs to be grounded for your existing PSU for the OPAMPs to work, and obviously J4 is redundant. Otherwise its close to being O.K. Add 2200uF 25V between bridge + and bridge - and 1nf between bridge - and ground and it will be exactly what described earlier

author: t1d
author_rank: Super Contributor
timestamp: October 02, 2018, 01:55:13 pm
content: Thank you, Ian, so very much!

author: Ian.M
author_rank: Super Contributor
timestamp: October 02, 2018, 02:21:10 pm
content: Looking good. If you need to run any sort of temperature sensing and fan control circuit, its easiest if you use the 12V buck converter output, but remember the negative side of it is *NOT* ground, so if you need the control circuit ground referenced, you'll have to optoisolate.

author: ArthurDent
author_rank: Super Contributor
timestamp: October 02, 2018, 09:16:51 pm
content: Having multiple threads on the same basic subject like you have decreases your chances of getting an answer and it clogs the forums.

author: t1d
author_rank: Super Contributor
timestamp: October 02, 2018, 10:45:47 pm
content: AD, agreed. It did not happen, by intent. Thank you for your help, with the circuit solution.