Electronics > Beginners
Capacitor Calculations... Correct?
t1d:
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 = 16.8vdc
Xvdc = 12vac x 1.2 = 14.4vdc
Or, 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 = 6w
Time – Half Wave Rectifier
60Hz/2 = 30Hz = 0.0333 seconds
Calculate Joules
6w x 0.0333s = 0.1998j
You 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 = 0.5994j
Calculate 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...
IanB:
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?
t1d:
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.
Ian.M:
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 = 1388uF
Due 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.
t1d:
Thanks, Ian,
Is this what you intend? Please look closely at the connection to the 12vac supply.
I appreciate the help.
Navigation
[0] Message Index
[#] Next page
Go to full version