Power supply filtering and bleeder resistor calcs

[seware]

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.

Steve

 

[John Popelish]

seware wrote:

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.

Steve

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.
--
John Popelish