Sizing Decoupling Capacitors for 240AC to 5VDC Conversion

[Pagedown]

I am using a step down transformer 20:1, 240Vac to 12Vac
then to full wave bridge rectifier, then to the 7805 voltage regulator to get +5VDC.

Are my decoupling capacitors sized correctly? I tried to read a lot about those capacitors choosing, but i still don't really get it... thanks for anyone whos helping...

 

[Studiot]

C2 needs to be 50 to 100 times larger and C1 needs to be 10 times larger.

 

[davenn]

Just keep in mind, ...
on the DC output of a bridge rectifier, the "rule of thumb" value of the smoothing capacitor is 1000uF / Amp of current.
So in your case the 7805 is capable of ~ 750mA to 1A if properly heatsunk. Therefore you should have a minimum of 1000uF Electrolytic at 25V rating for C2 on the input of the 7805 regulator.

Dave

 

[vk6kro]

You could also allow for the negative output to be ground.

To do this, connect the bottom of the transformer secondary to the bottom of the bridge and remove the ground connections to these points. Then just ground the left side of the bridge.

This way, the negative side of the power supply would be connected to ground. Or, you could leave the ground off and have a floating supply.

 

[Phrak]

C2 needs to be sized according to load. You haven't supplied a load, whether resistive or capacitive or inductive. What is the maximum expected load?

Also, if you plan to build this, 78xx's don't like to be reverse biased, and can fry is they are. Put a reverse biased 1N400x diode across the 7805 from IN to OUT if there will be other loads across the bridge.

 

[Pagedown]

The load is just a small microcontroller. The datasheet has its properties right? Then how can I calculate when i get the properties? Maximum I think should be 1A.

Okay, I will find out about that diode, must it be that particular diode? I dun really understand how the diode can protect the 7805 from being reverse biased?
Reverse biased means the 7805 has negative on its IN/OUT right?

 

[vk6kro]

No. You need a diode because the output capacitor can hold a charge when the input voltage is removed.

Then you have more voltage on the output than the input, so the chip is reverse biased

The diode just shorts out this voltage.

It is mainly a problem if the output capacitance is greater than the input capacitance, but it is cheap insurance anyway.

Usually the output capacitance is much smaller than the input capacitance.

 

[Pagedown]

Thanks, I finally understand the diode now.

But I still don't really know how to estimate the best value for both input and output caps.

 

[Bob S]

Two things:

1) You have a lot of voltage headroom between the peak dc output of the rectifier and the minimum voltage input for the 7805, so you could allow for a repetitive 4 volt drop in the capacitor voltage between the 120-Hz recharging cycles at 1 amp average current output. Using the relation Q = CV, ΔQ = C ΔV. At 1 amp and 120 Hz, ΔQ = 1/120 Coulombs. Using ΔV = 4 volts max droop between recharge cycles, The minimum capacitance for C2 is ΔQ/ΔV = 1/480 = 2,000 uF.

2) The MDA3551 bridge rectifier choice is inappropriate for this application. If you expect to use this at 1 amp average current output, the rectifier should be rated at at least 3 amp peak repetitive surge current.

Bob S

 

[Phrak]

The load is just a small microcontroller. The datasheet has its properties right? Then how can I calculate when i get the properties? Maximum I think should be 1A.

Search the datasheet for "electrical properties". As an example, a small 8 bit micros like the Microchip PIC16F15xx runs a typical 1.3 mA supply current at 5V and 20 MHertz. But still, 1 amp. is good to have available if you want to run a few LEDs or something, but to utilize the full 1 amp. you will need to heat sink the 7805.

An LM317 could be a better choice for a simple regulated 5V supply using lower input voltage. But BobS has given you some solid design values already.

 

[Pagedown]

the datasheet suggest a 0.1uF output cap and 0.22uF input cap. Any idea why?

 

[Studiot]

These capacitors are for very high frequency (transients). This is because the larger value (electrolytic) capacitors also used do not respond at very high frequencies and might as well not be there. Luckily the current in these transients is quite small so the capacity need is small.

These should be in parallel with the 500 - 1000 microfarad input capacitor I have already suggested and with the 100microfarad output capacitor.

You should also incorporate a 10 -100 microfarad capacitor at the other end of the power supply line near to each integrated circuit you are supplying. This is very important and your system will be unreliable without them.

go well

 

[Bob S]

Hello Pagedown-

I show a 12-volt 60Hz sinewave driving a full-wave rectifier and a voltage regulator in the attached thumbnail. The green waveform is the voltage across the filter capacitor, and the red waveform is the regulator output. With the 500 uF filter capacitor and a 1 amp output, the voltage droop between charging cycles is about 9 volts, and the regulator is dropping out of regulation. Very clearly, the 500 uF capacitor used in the simulation is too small for this application.

You should learn how to do "back of the envelope" estimations of the required capacitor size like shown in post #9, or use actual Spice simulations like in the thumbnail.

Bob S

[Sorry. LTSpice did not have a 5.6 volt zener, so I had to use a 6.2 volt one. ]

 

[Pagedown]

Bob S, my input to rectifier is 12V 60Hz, why do you use 120Hz in ur calculation in post 9?
And why 4V headroom?

 

[Pagedown]

I used the specifications suggested,simulated and found out the output results were great.

But the input voltage seems to be distorted? Is it because of the additional diode?

 

[Bob S]

Bob S, my input to rectifier is 12V 60Hz, why do you use 120Hz in ur calculation in post 9?
And why 4V headroom?

The 120Hz frequency arises because the full-wave bridge rectifier changes the negative half cycle of the 60-Hz sinewave to a positive half cycle, thus creating 120 positive half cycles per second.
If you look at my LTSpice simulation in post #13, the voltage on the capacitor drops from +15 volts to +6 volts every half cycle (for a 1-amp load). This 9-volt droop is too large with the 500 uF capacitor. The 7805 I believe requires at least 8 volts input to meet all regulation specs, so the input voltage droop can be a maximum of 15 volts minus 8 volts = 7 volts "headroom" at full current (1 amp). You should select the filter capacitor to maintain a minimum of 8 volts out under worst case conditions (max dc current output, minimum input line voltage, tolerances on components, etc.).

In your simulation, you have no resistive load. You should add a 5 ohm resistor to test the circuit at 1 amp (or another resistor for a lesser load). Be sure to graph the filter capacitor voltage in your simulation to see the voltage droop. Also, you are using one side of the transformer secondary as a ground reference. This then requires that both sides of the 5-volt regulated output be floating. You should consider floating the transformer secondary in order to be able to ground the negative side of the 5 volts out.

Bob S

 

[Pagedown]

bob, that really clears everything up, thanks alot.

About the grounding, is there a disadvantage grounding the secondary side while have others floating? Or merely for easier implementation that I should reverse this configuration?

 

[Pagedown]

1.now that I know how to size my input cap, how about my output cap?
how about my powerline cap mentioned?(near the load)

I should not have any headroom like my input cap, as I want to get constant DC voltage of +5V. and max load should be 1A.

2.Micro-controller/LEDs = resistive loads?

3.components are without datasheet can I determine their properties whether resistive/inductive/capacitive?

4.what if I have a capacitive/inductive load? I have to size output cap differently?

 

[Studiot]

Since none of my posts seem to have been acknowledged I don't know why I bother, but all the information you require is in them.

 

[Bob S]

The recommended output cap is 0.22 uF. See page 5 of datasheet http://focus.ti.com/lit/ds/symlink/tl780-05.pdf
The reverse diode across the 7805 is a good idea. You might be able to get away with an input filter capacitor of 1000 uF at 1 amp output (5 ohm load), even with worst case conditions (capacitor voltage dropping below 8 volts). You might want to consider using 2,000 uF for a little safety factor. Microcontroller and LEDs are a very light load. Don't worry about capacitive/inductive loads unless you are using a large capacitor or inductor (relay, solenoid, etc.) on the output. Too large a capacitor on the output may force the 7805 into a current foldback (overcurrent) mode at turn-on.

On grounding, it is possible that the negative side of the 5 volts becomes a common ground on a printed circuit board, or connected to other circuits via I/O signals, so don't ground either side of the transformer secondary. (The only configuration where the transformer secondary is grounded is in a full-wave CT (center-tap) configuration, which you do not have).

Bob S

 

[Studiot]

One other thing occurs to me.

Many microcontrollers are fussy about their rail voltage and the spread of the 78XX series is pretty wide, so your 7805 output might be to far off 5 volts.

You should either incorporate a trimpot to cope with this or use a better reg IC.

 

[Pagedown]

I read all of the contributors, and I appreciate all of their contributions and help.
Thanks, studiot.

Studiot recommended a 100microfarad for output cap. But in the datasheet Bob sent me, it is not needed? 0.1uF/0.22uF for transient?

 

[Studiot]

In your original circuit you had C2 & C3 at the input with C1 and C4 at the output.

I agree C1 is not always used, but you asked about your circuit, which was nearly there.
C3, C4 are the transient suppressors.

Incidentally is this only a simulation or are you building it for real?
Some of my comments only apply to real hardware - a simulator will not give you 4.85 volts for a 7805, but a real one might.

 

[Pagedown]

I will build it for real =)

I have ommited the output cap,C1.

I have used input cap, C2 as 2000uF.
and C3,C4 as transient suppressors - Bob, my 7805 datasheet recommended this. Is it ok?

 

[Studiot]

I don't have specs for your bridge. The only MDA3551 I can find is a diode voltage multiplier assembly for televisions.

However assuming there is indeed such a bridge rectifier check the peak current that a 2000 microfarad capacitor will draw. This would be too much for some bridges.
You might consider a pie filter instead. That is two x 1000 microfarad caps connected by a low value resistor of a few ohms. This would need to be a power resistor.

 

[Bob S]

Looks good. The 7805 output voltage specs at 1 amp show 4.95 volts min, 5.05 volts max at 25 degrees C; 4.90 volts min, 5.10 volts max up to 125 degrees C. Use the 7805 TO-220 package and mount it to heat sink for operation at 1 amp. It will get very hot.

Why don't you substitute the Radio Shack 4 amp bridge ($2.59) for the one in the circuit diagram.

Please put the oscilloscope directly across the 2000 uF capacitor in your simulation and look at the voltage droop. If the voltage droop is less than 3 volts, you can decrease the input filter cap to 1000 uF, but make sure the minimum input voltage to the 7805 is always over 8 volts. Use Studiot's idea of using two 1000 uF caps separated by a 1 or 2 ohm resistor.

Bob S

 

[Bob S]

Here is the circuit with Studiot's pi filter with a 2-ohm (5 watt) resistor. Green trace is the cap filter output, and red is the regulator output. Looks good.

Bob S

 

[Pagedown]

Actually, I am using an rectron wo2m bridge rectifier (1.5A)
Multisim doesn't have this model.
http://www.datasheetarchive.com/WO2M-datasheet.html

 

[Studiot]

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?

go well

 

[Pagedown]

my rectifier is rated at 1.5Amps, should i need a pi filter?

I can't simulate my rectifier, as the component is not there in Multisim. So I don't know whether the capacitor is drawing too much current? any calculations? Help!

 

[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 ALOT 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