The way these things work is that the power line gives a pulse every
1/60 seconds (or 1/120 seconds, if you are using a full bridge with 4 diodes). So, the caps will be 'pumped up' to full voltage that often, and then power your supply between pulses.
You will probably also use some kind of voltage regulator. If you were building a 5V supply, for example, you might use a 7805. These regulators will always require some 'headroom', meaning that the input voltage must not drop below some minimum for the regulator to maintain the output voltage. For a 7805, for example, this is usually stated as
3V. Thus, you don't want the input voltage to drop below 8V.
So, the caps get pumped up to 12.6V every 1/120 second, meaning that the drop from 12.6 to 8V cannot be faster than 1/120 seconds.
Now, you need to decide how much current your maximum is. For a 7805, for example, the maximum current is about 1A. So,
I = C * (dV/dt)
I = 1A
dV = (12.6-8)
dt = (1/120)
so C = 1A / (4.8*120) = 1812uF
Usually, some fudge is thrown in, so if you use 2,000uF, you'll be safe.
For a bleeder resistor, say you want the thing to discharge in 5 seconds. The rule of thumb is that a resistor will discharge a cap in 5
RC. So RC in this case is 1. Thus, R = 1/2000uF = 500 ohms. Thus, 470 ohm is a standard value which is near enough.
--
Regards,
Robert Monsen
"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.
There are no hard and fast rules for these things. More capacitance produces less ripple voltage at full load current but a bigger start up surge and a longer bleed down at turn off. Personally, I hate to wait for a lab supply to fade out after I turn it off, because I may do that dozens of times a day when working on circuit variations. So I like serious bleeders on a lab supply. The supply for an audio amplifier, I am not in such a hurry about, and there is some minimum load there that pulls the voltage down, anyway.
But here are some formulas starting with the transformer. A capacitor input filter (cap directly to the bridge rectifier) and load heats the transformer more than a resistor that draws the same average current, connected directly to the transformer.
The bigger the cap, the worse this gets, because the cap voltage tends to stay near the peak of the rectifier output wave, so all the current has to occur in a small blast right at the top of the wave, when the rectifiers forward bias. So don't expect more than about 75% of the transformer's current rating coming out as DC. Your transformer will supply about 4 amps continuously from a capacitor input filter.
That peak voltage is almost 1.414 times the RMS AC voltage of the transformer, so your 15 volt transformer will pump an unloaded cap up to about 15*1.414=21.2 volts. How low the cap voltage sags to before another peak comes along depends on the ratio of capacitance to load current. If you want the output to be regulated to no more than about
12 or 15 volts, you can afford a lot of ripple on the capacitor, but if you want to regulate something like 18 volts or use the supply unregulated, you will probably want to keep the full load ripple less than a volt or so.
To simplify the math a bit, lets assume that the 60 hertz line frequency (if that is your line frequency) charges the cap instantaneously every half cycle or every 8.3 milliseconds. That means that the 4 amp output current is running entirely from the cap all that time. The formula that relates current to rate of change of voltage is I=C*(dv/dt) or current (in amperes) equals capacitance (in farads) times the rate of change of voltage (in volts per second). So to have a 1 volt sag before the next recharge, the 4 amps has to cause a rate of change of voltage of 1 volt per .0083 seconds or 120 volts per second. So the required C is 4/120 farads = 0.033 farads = 33,000 microfarads. If you can tolerate 2 volts of ripple, half that capacitance will do. Or you can substitute your available capacitance and calculate the ripple voltage.
Bleeders are really only essential for safety reasons (to dump the voltage to a safe level before you can get the case open). But you can decide what is a reasonable wait if you decide you want the supply to fade out on its own, even if it isn't a shock hazard. The time it takes to fade it to 37% of full voltage is R*C seconds where R is in ohms and C in in farads (or R in meg ohms and C in microfarads, etc.)
After you get that all working to your satisfaction and you have load tested it with some pairs of 12 volt bulbs in series (or other 24 volt capable loads) to see if the ripple voltage is as expected and the transformer doesn't overheat, etc. you can start thinking about regulation.
I am wanting to build a power supply and need a bit o' help. I have an "Electronics for the hobbiest" type book that goes into fair detail on the subject, but where it talks of filtering and bleeder resistors it speaks of their use and suggests values for an example power supply but not how to calculate the values for your own. Their example uses a transformer with a
12.6V 3A secondary wth a full-bridge rectifier. The bridge outputs have 3
1000uF 35V caps in parallel and then a 1K 1/2W resistor in parallel with the caps for the bleeder. How do I calculate my needs for a transformer with a
15V 5A secondary? A worked example would be nice but I can plug in numbers if someone will explain the reasoning. I understand that the size of the bleeder depends on how fast I want to discharge the caps, but I don't know what kind of time is reasonable. Should I reverse engineer the RC constant in the example and work forward from that or are their some better rules to follow? Thanks to all you professionals who sustain the questions of all of us wannabees.
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