Math Formula to calculate DC power supply capacitor value?

Pictures are worth a 1000 words. There is nothing about solid state circuits in my old text book copyright 1960. I'm not sure transistors were invented yet. The whole chapter on power supplies is 6 pages. I'm sure technology has advanced some since I was in college. LOL My 1973 Radio Amateur Handbook has better information on power supplies.

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Transformerless PSU for a valve (tube) radio
A selenium rectifier,

That is not electronics, it is industrial archaeology.

JimB

ChrisP58 said:

I should have said, "..... low voltage supplies for solid state linear audio amplifiers.

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Modern solid state linear audio amplifiers have such a good Power Supply Rejection Ratio (-120dB which is one millionth) that they do not need very low ripple nor a voltage regulator.

Gary seems quite good at asking questions and then refusing to believe anyone who answers. A number of threads have demonstrated this.

To select a suitable bleeder resistor, you can consider the relationship between the momentary voltage across the capacitor Vt, the bleeder resistance R for discharge, the initial voltage Vu. t is the momentary period and the total capacitor capacitance is C. Then, you can use this equation to estimate the required value of your bleeder:

• Vt=Vu x e^(−t/RC)

There is always a trade off between the speed the bleeder works at and the amount of power waste in the bleeder. Lower values of bleeder resistance will give you a faster time to reach safe voltages when the device is powered down, but they will also waste more power during operation.

Vt must be 22 volts meter said PS volts is 22 vdc.

R=?

e=?

Vu must be 22 volts again?

t= what is monentary period?

C=3300 uf x 12 = 39,600. uf

22 = 22 x e ( -t/?39,000.)

I don't know how to do this formula. I usually put a resistor across the power supply then time how long it takes to discharge. Then use that information to cross multiply to find resistor value for 4 minutes.

gary350 said:

I found this formula in my college text book. Teacher skipped over the power supply chapter for some reason.

%Eripple = Eripple / Edc x 100

Book says, 1% ripple is exceptable

Shunt capacitor between plus & minus filters the voltage ripples.

Choke coil in series filters the current ripples.

filter.

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What is often overlooked is that the Va requirement of the supply transformer goes up as ripple is decreased with increased capacitance at the required load.
Max.

JimB said:

Define "gone".
From a purely theoretical point of view, you will never remove all the ripple with a simple (inductor) capacitor smoothed PSU.

This is why most PSUs use electronic regulation of some kind.

Consider a simple 12v PSU which uses a 7812 regulator IC.
As long as the input voltage to the 7812 is above 14v, the output will be a nice smooth 12v DC with the noise and ripple attenuated to a very low level, regardless of how much ripple is on the input of the IC, as long as the troughs of the ripple do not go below 14v.

JimB

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I understand this. I have some 7812 regulators I experimented with a 15 vdc power supply 7812 gives me 12 vdc but it over heats. My first heat sink was too small and second larger heat sink was better but it still over heats. When its to hot to touch its too hot. I don't think it should get too hot to touch?

MaxHeadRoom78 said:

What is often overlooked is that the Va requirement of the supply transformer goes up as ripple is decreased with increased capacitance at the required load.
Max.

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My transformer is 1500 watts. Output voltage is 22 volts. 1500 / 22 = 68 amps

Now the circuit works better it idles at 1 amp and never pulls more than 25 amps under load.

This is one type of simple "boosted" regulator:

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The 7800 series regulator acts as a voltage controller and driver for the PNP power transistor, just drawing enough directly to give the required base drive current.

If you need a voltage that's not a standard for the 7800 series, you can use the nearest below plus a resistor divider across the PSU output, with the 7800 common / ground connected to the divider junction - or am LM317 adjustable regulator, which is in effect a "780(1.25) and is usually used with the resistor divider setup to give the desired voltage.

[Edit - sorry, it won't post links to the images I was trying to use].
Try again: https://i.ytimg.com/vi/nB2a93gpKfg/maxresdefault.jpg


For the negative rail, mirror the circuit with a 7900 (or LM337) and NPN power transistor.

The transistors will need good heatsinks for high current operation.

gary350 said:

I don't know how to do this formula. I usually put a resistor across the power supply then time how long it takes to discharge. Then use that information to cross multiply to find resistor value for 4 minutes.

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Are you making a valve amplifier? - NO! - then stop asking questions that relate solely to valve amplifiers.

You DON'T need a choke, and you DON'T need a bleed resistor - these are specific to valve circuits.

This is a bit difficult to follow. In one place, you talk about drawing 25 amps and in another, you talk about 7812 linear voltage regulators.

To add some value to this thread....

The power dissipation of a linear regulator (any and all) is

P = ΔV × I, where

P = power dissipated in watts

ΔV = voltage drop across the regulator

I = current in watts


If the supply voltage to a 7812 is 16 volts, and you're drawing 1 amp,

P = (16 - 12) × 1 = 4 watts

To put that in perspective, 4 watts is a standard night light bulb.

JonSea said:

To put that in perspective, 4 watts is a standard night light bulb.

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To put that in perspective, WAS a standard night light bulb

Incandescent bulbs are history.

Nigel Goodwin said:

To put that in perspective, WAS a standard night light bulb

Incandescent bulbs are history.

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Fair enough Not entirely history yet, but just about.

Nigel Goodwin said:

Are you making a valve amplifier? - NO! - then stop asking questions that relate solely to valve amplifiers.

You DON'T need a choke, and you DON'T need a bleed resistor - these are specific to valve circuits.

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I did not know that. I thought all power supplies need choke & bleed resistor. I would like to build a Dynaco 60 tube amplifier but I have no use for it so I probably never well. I built several tube amps many years ago. Now I learn none of my power supplies need chokes or bleeder resistors. Next I learn there are regulators other than 78xx. YOU guys are light years ahead of me. I looked up power supply circuit drawings google search shows lots of circuit drawings most have chokes, caps on 1 or both sides of the choke, some 2 chokes with 3 caps, all have bleeder resistors, not many PS drawings with regulators. I already know, half wave, full wave, voltage double, voltage triplers, now i learn there other types of voltage doubles & triplers circuits. YOU wonder why I keep playing with toy induction heater circuit, I am learning new things. Maybe I need to learn on a different circuit one that makes less smoke and explodes less mosfets. I think there is a limit to what I can learn with this toy induction heater circuit. I see my 68 watt stereo amplifier has built in power supply all I need is 24-0-24 volt transformer. Not much to learn with it connect speakers and plug transformer into 120 vac. I built a mosfet motor speed controller that was fun project but I copy online circuit did not learn very much. I have book called Malcoms 101 electronic projects it should be re named Malcoms 101 electronic projects that don't work. LOL. This is good learning projects for me to study each circuit to learn how to make each circuit work.

gary350 said:

I looked up power supply circuit drawings google search shows lots of circuit drawings most have chokes, caps on 1 or both sides of the choke, some 2 chokes with 3 caps, all have bleeder resistors, not many PS drawings with regulators.

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Where did you find ANY that have chokes?, the first half dozen pages full I lloked at had only two - and they were specifically valve power supplies.

gary350 said:

I found this circuit drawing online I can link it here as an example.

Using a 22 VDC power supply I already have as an example it can produce 75 amps but the circuit never loads the power supply more than 50 amps.

50 amps x 1000 uf = 50,000. uf

How to calculate 10% load resistor often circuit only pulls 2 to 20 amps some times as much as 35 amps.

I don't know how to calculate choke coil value if one is used? What is L with 2 dots Rx mean?

If I use choke does 1/2 the capacitors need to be on each side of choke?

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Hi,

The peak current in the diodes is always high with capacitor filters, so with high current supplies like this a choke is usually added to help smooth out the peaks. The choke usually used is a swinging choke though not a linear choke and the reason is for efficiency. You have to be able to wind your own choke for this usually but i guess you can look around. The idea is to get reasonable filtering at low load but at full load get some filtering but less loss.
If this is just for a power supply for testing something else, then you may not want to bother and just use a regular choke.
The choke is usually connected to the input or the diodes though so that it takes care of that peak diode current problem. A pi filter is not as good at doing that because in fact it is still a capacitor input filter which still shows high peak diode current.

I can show you the best formula for a rectifier circuit of any kind but i have a feeling you wont want to be bothered with the complexity. You can estimate the cap size by using the approximate discharge time and load current and calculate the voltage drop over that time and adjust for your maximum ripple spec. As an example, unregulated 12v, 1 amp wall warts usually use around 3300uf.
The simpler formula is:
C=i*dt/dv

and for 50Hz dt=0.01 which brings us to:
C=i/(100*dv)

where 'i' is the load current and 'dv' is the max ripple you want to see.

So for i=1 and dv=2.5 we get:
C=4000uf

and for 60 Hz multiply by 50/60 and we get:
C=3333uf

which is about 3300uf.

For the choke input filter we'd have to get a little more complicated here with some more theory.

gary350 said:

Is there a math formula to calculate the correct size of a DC power supply capacitor?

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Yes. It comes from the basic formula relating current, time, capacitance and voltage change:

C = I * dT / dV

where C is the needed capacitance in farads, I is the load current in amps, dT is the time during which the capacitor is discharging in between being recharged through the rectifier in seconds, and dV is the voltage droop during the discharge period, i.e., the allowable peak-to-peak ripple, in volts.

EXAMPLE: I have a supply which must provide up to 2 amps of load current, using a full-wave bridge rectifier running off 60 Hz mains. How much capacitance do I need across the bridge rectifier output to keep ripple to 1.5 Vp-p or less?

C = I * dT / dV = 2 * 0.00833 / 1.5 = 0.0111 F = 11,100 uF

Or is it easier to keep making the capacitor larger until AC ripple is gone?

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It will never be "gone": it will merely get smaller and smaller the larger you make the capacitor.

What is best circuit drawing some have capacitors only & some have a choke coil too?

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NO ONE uses chokes anymore to smooth out power supply ripple; those were common back in the valve era, when voltages were very high and load currents very low. For modern low-voltage, high-current supplies, they just add unnecessary weight and cost.

I read bleeder resistor should put 10% load on the power supply? This seems like a lot of wasted power.

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Bleeder resistors, too, are a relic of the valve era, and were used largely for safety reasons. On low voltage supplies, they're not necessary. I haven't used a bleeder resistor or a smoothing choke in a power supply design in more than fifty years.[/QUOTE]

OBW0549 said:

Bleeder resistors, too, are a relic of the valve era, and were used largely for safety reasons.

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And were fairly pointless back then anyway, as they commonly went O/C so ceased to perform their intended function.

As a service engineer I certainly NEVER trusted them to have done their job, and would always short the caps out (usually with a screwdriver) to ensure they were discharged. The screwdriver technique was also quite entertaining

There are probably a few members on here who did repairs back in those days?, and it still applies now, high value resistors tend to go higher in value, and to go completely O/C.

Nigel
The only resistor types which did not go open circuit without fighting the good fight were wirewounds.
And you could tell they had open because the ceramic casing would break from the intense heat.

WHAT is O/C ?