Sizing Decoupling Capacitors for 240AC to 5VDC Conversion

[Studiot]

It's a pretty basic calculation, that doesn't need a simulator.

The reactance of your capacitor = 2\pifC in ohms. Note this is 2 time pi, not 2 raised to the power pi. D__mmed Latex

This value is slapped across the rectifier output at switch on and so draws an initial current of

V_rect/2\pifC

Until the voltage builds up on the capacitor.

To protect the bridge, you only need a current limiting resistor. In addition to protection, the pi filter also reduces input ripple to the regulator considerably.

go well

 

[Bob S]

I note Bob is using single discrete transistors in his simulations, rather than a 78xx regulator.

The regulator's performance will be far better than this at 1 amp output.

Perhaps Bob is doing this for emphasis?

My LTSpice simulator doesn't have 78xx regulators in its repertoire, so I had to substitute a simple NPN regulator to provide a constant current sink. Bob S

 

[Phrak]

My LTSpice simulator doesn't have 78xx regulators in its repertoire, so I had to substitute a simple NPN regulator to provide a constant current sink. Bob S

So add a stack of 5 diodes to simulate the overhead requirements and see what you get.

(incredibly, this has gone on to three pages) It's like a McDonald's shake consumed through a cocktail straw.

 

[Pagedown]

using the calculation Current drawn by cap=Vrect/2pifC.

I get 12v/(2 pi x 120hz x 2000uF)= 8 amps of current!

So how do I size my resistor?
why do you need 1000u + 1000u rather than a single 2000u cap?

 

[Bob S]

I plot the currents at thee places in the regulator circuit in the attached thumbnail. (I have added two very small resistors to measure the current.) The current into the first resistor (green) is about 5 amps peak repetitive. The current into the second resistor (pink curve) is about 2 amps peak. The current into the regulator transistor (red curve) is about 1 amp dc.

Using two 1,000 uF caps separated by a 2-ohm resistor reduces the power dissipation in the regulator transistor and simultaneously increases the inter-cycle minimum voltage, because the pi-filter configuration suggested by Studiot integrates over the 120-Hz cycle (RC = 2 milliseconds). See thumbnail plot in post #27.

Bob S

 

[Pagedown]

I researched about pi-filter, and its two cap separated by an inductor.

By using a 2 ohm resistor can reduce the current taken? I still don't really understand it.

So for my situation, 2 ohm resistor can do the trick?

 

[Studiot]

You're getting there and learning stuff as well.

 

[Pagedown]

ya, I've been learning a lot from you guys...studiot & bob..

 

[Pagedown]

any kick start for my pi-filter understanding?
i thought a bigger resistor would do more current limiting.

 

[Bob S]

A pi filter can use either a resistor or an inductor between capacitors. In either case, the purpose is to reduce the repetitive peak surge currents. In my post #35, the 2 ohm resistor reduces the peak repetitive surge current from 5 amps (green trace) to 2 amps (pink trace) at 1 amp output. Note that the minimum current in the pink trace is higher than in the green trace, due to the RC time constant (R=2 ohms, C = 1000 uF, so RC = 2 milliseconds). In my post #27, the voltage fluctuation on the second capacitor (C2, green trace) is from about 9 volts to 11 volts at 1 amp output. Because the recommended minimum voltage into the 7805 is about 8 volts, there is about 1 volt margin. This means that you could increase R from 2 ohms to 3 ohms max. So a 3 ohm resistor is better than a 2 ohm resistor, but a 5 ohm resistor would be a disaster.

Bob S

 

[Studiot]

You should note the wattage requirements for the filter resistor.

At 1 amp a 2 ohm resistor dissipates 2 watts etc.

If you are using a circuit board, mount the resistor well clear of the board, to allow free airflow. A 5 watt resistor would do nicely.

Using a choke (inductor) instead substantially reduces this diddipation, but is not normally done in modern circuitry because of physical size of the necessary choke.

 

[Pagedown]

whats the difference if I use just one capacitor of 2000u with 2ohm resistor?

 

[Bob S]

whats the difference if I use just one capacitor of 2000u with 2ohm resistor?

The voltage on the 2000 uF capacitor is roughly sinusoidal (120 Hz), with a minimum of 8 volts and a maximum of 10 volts. The peak repetitive surge current in the 2-ohm resistor is about 3 amps.

Bob S

 

[Phrak]

I don't see the advantage of a pi filter over an R-C-R-C filter, where the effective resistance is evenly distributed between 1) the source and first capacitor, and 2) the two capacitors. The dynamic impedance of the bridge should supply some of this to the first, but without a simulator I could only guess how much.

 

[Pagedown]

1. What's the transfomer rating? should it be 1Ax12V=12VA?

2. But in the configuration from the bridge, current flowing is 8 amps? thus bridge must be able to handle 8 amps of current?

3. Whats the choke size if I were to really use an inductor instead of a power resistor?

 

[Pagedown]

4. the surge current is around 5A-8A starting. Should i buy a current rating of full wave bridge rectifier sized to 8A then??

 

[Phrak]

Now you're in a bit of trouble because you have been modeling using an ideal 12VAC source. Select a transformer with a VA rating greater than the time averaged power output. This is the integral of V and I out of the transformer. Now you have to do some guessing. The output voltage is somewhere around 85 to 90% of the unloaded voltage at the rated VA with resistive load. This converts to an effective series resistance with either the primary or secondary.

You no longer have a 12VAC supply and some redesign is in order to get 1A and 5V. Ever heard of an LM317?

 

[Studiot]

Diodes/bridges are rated for RMS current.

The peak current is therefore \sqrt{2} RMS rating

ie a 1 amp diode expects to receive 1.4 amps twice per cycle.

The twice per cycle is also important because rectifiers also have a 'non repetitive surge rating' to handle the inrush current.

For a 1 amp diode this will be about 3 amps.

So to handle 1 amp RMS continuous comfortably I would look for a 3 amp rectifier.

 

[Pagedown]

Yea, I have heard of LM317. Is it better than 7805.But I think will stick to the 7805 for now.

I have attached my final(hopefully) designed circuit. The filtered output to 7805 is 9-11V at around 1 A.

The inrush current from rectifier is around 6A and repetitive 4A peak cycles after that.

Is really 3A rectifier enough to do this? or a 5A is better?

 

[Studiot]

Much of the world's electronic kit runs happily on
1N4001 diodes for "currents up to 1 amp" either as 4 discrete diodes or in monlithic bridge format and used with 500 to 1000 microfarad reservoir capacitors.

Your design requirements are pushing the upper edge of this so the logical step is to go to the next series up ie

1N5401 diodes. These are the 3 amp diodes I was thinking of.

Most designers would simply choose these without further ado, from experience. In this thread we have tried to share the underlying reasons for these choices.

go well

 

[Pagedown]

but it should be handling 4amps from the simulation.

 

[Pagedown]

we designing backwards, thus now the transfomer and rectifier is giving some headaches.

 

[Phrak]

Yea, I have heard of LM317. Is it better than 7805.But I think will stick to the 7805 for now.

I have attached my final(hopefully) designed circuit. The filtered output to 7805 is 9-11V at around 1 A.

The inrush current from rectifier is around 6A and repetitive 4A peak cycles after that.

Is really 3A rectifier enough to do this? or a 5A is better?

It's your design, but it won't handle 1A without dropout unless you are willing to put up with a 25VA+ overkill transformer. You've used an ideal transformer in your spice model. However, there's nothing magical about 1A, and you probably won't use all of it anyway.

 

[Pagedown]

So if I use an underated transfomer or bridge, at full load, the voltage will drop?

It won't effect heating or maybe damage??

 

[Pagedown]

I think there is a mistake of using Vrect/ 2 pifC.

Impedance of capacitor is : 1/ 2pifC

I=V/R = Vrect / (1/2pifC) = Vrect x 2 pi x f x C

 

[Studiot]

I do apologise you are quite correct.



Well done for spotting that.

 

[Pagedown]

So if I use an underated transfomer or bridge, at full load, the voltage will drop?

Will it cause heating or maybe damage??

Anyone can answer me pls?

 

[vk6kro]

This may be an English problem.

If a transformer is used at full load then it is being used properly, but at the limit of its ratings.

So, a 10 amp transformer can be used to give currents up to 10 amps although it is considered bad practice to operate components at the current they are rated at.
Normally a safety margin is allowed. So, you might design this power supply so that the load never exceeds 7 amps if the transformer was rated at 10 amps.

If the load were 20 amps and you used a 10 amp transformer, then the transformer is said to be under-rated for that load.

What actually happens depends on the transformer. It may get very hot due to heat dissipated in the transformer not being able to escape fast enough. This can cause the transformer to fail eventually.

It may fail instantly, although this is less likely.

The voltage out of the transformer would probably be lower than if the load was normal and this is due to the extra voltage drop across the resistance of the wire in the windings as well as saturation effects in the iron of the transformer.

Semiconductors which are used beyond their ratings are very likely to fail instantly or in a very short time.

I like to use 30 amp bridges for power supplies, even if they are only going to deliver a few amps. These are very cheap and good insurance.

 

[Pagedown]

I am suspecting the simulation. How correct is it? cause even with lower loads the current in the secondary before the bridge is at least 1.5A.
for full load, it goes until 3-4 A.
even with lower capactitance.
Is it so in reality?

 

[Pagedown]

Due to my earlier designing of 1A, the transformer requirement at least 60VA(tested in multisim) is too bulky and expensive.

I have been trying now to design towards 0.2A since my microncontroller and accessories use a lot of less current. But from simulation, the secondary side still takes in around 1A. Is this true??the 7805 will at least take 1A??