Calculating the smoothing cap. on a AC/DC converter

Hi, this is my first time around.

I've searched and I haven't find another thread discussing this issue, on this particular layout. i'm currently involved on the development of a small power supply, the input is 12/24 Ac or DC, and the outputs are 9VDC @ 0.4A, 5VDC @ 0.2A and 3v3VDC @ 60mA, all the loads are digital loads, with long sleep times and short wake ups. I was trying to use as small as possible smoothing cap.

First I rectifie the Ac using a full wave rectifier, then use a soothing cap., after capacitor I use a step down, to go from 24 or 12 to 9 VDC.

I used the expression, C = I/2fVrpp. Then I checked the application note of the step down, and found this expression used for calculating the input cap, C1 = (Io x ton)/Vrpp, which I understand is calculated for already rectified DC signal. Has anyone experience with this kind of design? I'd like to have the smoothing cap. the smaller the better, cause I have serious size constrains.

Thank you in advance for your time and your advice, I find this forum really interesting and useful.

Best regards.

Here is a method using the RC time constant. You can use it to check your result with the other method.

The capacitor will charge during voltage peaks, then discharge during idle gaps.

Suppose your ripple spec is 5 percent on your 9V supply. This means voltage can go down by .45 V during every 1/100th (or 1/120th second depending on your mains frequency).

Your load will draw 660 mA total. This calculates to 13.6 ohms on a 9V nominal supply.

Your RC time constant is the time it will take your smoothing cap to drop 63 percent.

If your spec says it can drop 5 percent in .01 second, this is equivalent (more or less) to dropping 63 percent in .01 * 63/5...
or .126 sec. This number is your RC time constant.

Your resistance is 13.6 ohms. Therefore your smoothing capacitor should be .126 / 13.6...
or 9,000 uF.

This simulation tells pretty much the same story:



You can reduce the cap value if:
* your mains is 60 Hz rather than 50 Hz
* if you can permit greater ripple V
* if you reduce your max current drain.

BradtheRad said:

Here is a method using the RC time constant. You can use it to check your result with the other method.

The capacitor will charge during voltage peaks, then discharge during idle gaps.

Suppose your ripple spec is 5 percent on your 9V supply. This means voltage can go down by .45 V during every 1/100th (or 1/120th second depending on your mains frequency).

Your load will draw 660 mA total. This calculates to 13.6 ohms on a 9V nominal supply.

Your RC time constant is the time it will take your smoothing cap to drop 63 percent.

If your spec says it can drop 5 percent in .01 second, this is equivalent (more or less) to dropping 63 percent in .01 * 63/5...
or .126 sec. This number is your RC time constant.

Your resistance is 13.6 ohms. Therefore your smoothing capacitor should be .126 / 13.6...
or 9,000 uF.

This simulation tells pretty much the same story:



You can reduce the cap value if:
* your mains is 60 Hz rather than 50 Hz
* if you can permit greater ripple V
* if you reduce your max current drain.

Click to expand...

Hi BardtheRad, thanks for your reply. I have 60Hz mains. The buck is the load of the rectifier, so that changes a little bit the numbers, Doesn't it?
The load from the rectifier perspective, is the buck's input, 12VDC and 0.550A ( calculated for a 90% efficiency and 9VDC @ 0.660A Vout on the buck ) I'd be able to handle bigger ripple V, about 10%.

12 VDC divided by .55 A, means the load is 18.2 ohms average.

You can allow 10% ripple (1.2 V) every 1/120 second. This translates to 63% in .052 sec. Your RC time constant is .052.

.052 / 18.2 ohms...

equals 2,880 uF for your smoothing capacitor.

A simulation bears this out:



(I added the resistor near the 12 VAC supply, to represent some amount of internal resistance. If I omit it, the output voltage approaches the theoretical peak.)

BradtheRad said:

12 VDC divided by .55 A, means the load is 18.2 ohms average.

You can allow 10% ripple (1.2 V) every 1/120 second. This translates to 63% in .052 sec. Your RC time constant is .052.

.052 / 18.2 ohms...

equals 2,880 uF for your smoothing capacitor.

A simulation bears this out:



(I added the resistor near the 12 VAC supply, to represent some amount of internal resistance. If I omit it, the output voltage approaches the theoretical peak.)

Click to expand...

Hi, thank you for replying back. The thing is that I got the same caculations, but I was guessing that as the 9 vdc is regulated by the buck converter,It'd be able to handle more ripple on the input.

I tried today the worst case, no smotthing cap at all, and the step down was not able to work continuously, due to the high ripple, then I tried with 1000uF and evenif the Vin is has arround 15 to 20% of ripple, the buck is able to handle it without problems, Do you think this is to mutch? Any experience with this kind of layout?

Best Regards.

It is commendable that you are running tests with real components. It's as good as theory, if not better.

1.

I suppose your buck converter operates at several kHz (or tens or hundreds of kHz)? 120 Hz is a gradual change in comparison.
Evidently the high speed gives it time to respond to the variations in the supply V.

2.

The real test will be whether operation is smooth even when you add or reduce loads suddenly. The smoothing capacitor may require a few cycles to charge/discharge to your desired volt level. Your converter will need to make quick changes to the frequency and/or duty cycle.

3.

Does your control unit derive a 9V reference from the power supply in any way?

Have you seen the power supply waveform drop below the 9V mark, due to ripple? If it were to drop below 9V at any time, would that stop your converter in its tracks?

BradtheRad said:

It is commendable that you are running tests with real components. It's as good as theory, if not better.

1.

I suppose your buck converter operates at several kHz (or tens or hundreds of kHz)? 120 Hz is a gradual change in comparison.
Evidently the high speed gives it time to respond to the variations in the supply V.

2.

The real test will be whether operation is smooth even when you add or reduce loads suddenly. The smoothing capacitor may require a few cycles to charge/discharge to your desired volt level. Your converter will need to make quick changes to the frequency and/or duty cycle.

3.

Does your control unit derive a 9V reference from the power supply in any way?

Have you seen the power supply waveform drop below the 9V mark, due to ripple? If it were to drop below 9V at any time, would that stop your converter in its tracks?

Click to expand...

Hi,

Yes I'm running the test with real stuff, mounted the whole system, an started to scope the voltage. Answering to the points;

1. Yes, the buck is operating from 260KHz to 380KHz, compensating the variations on the Vin, is a ti LM25010.

2. As far a as I saw yesterday, the output of 9V has 10% ripple, but part of this ripple is high frequency noise from the control boards going backwards.

3. This is the key. On the worst case situation, with no smoothing cap. the voltage has high ripple, that makes the input drop bellow the 9V, so the buck is not able to work continuously, as it never reaches an stable point above 9V.

The 9V is used to power the board an two external boards that hang from this supply. The 9V supply is then regulated to 5V with a linear regulator. That's why I'm guessing that I can use a smaller smoothing cap., as the voltage has two regulation stages

Yesterday I tried with a big cap, double the calculated value, and the Vin is smooth and clean, but after connecting the load, I started to see ripple again. That was making no sense, so I used a DC supply, and the noise was still there!. It was from the control boards hanging on the 9V supply, these boards use high frequency pulses, and the noise was going back through the regulators an the buck, to the input.

Thank you for your replies BradtheRad.

Best regards.