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electrolytic cap ripple

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explorick

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Hi,
A capacitor is used before a Linear regulator. The input voltage to the capacitor is 12VDc. There is a sinusoidal ripple of 0.5Vrms and 13KHz frequency upon it. I calculated the ripple current using the formual Vrms/Impedance. Impedance is metioned as 0.15ohms at 100Khz,25degC. The ripple current value is 0.5/1.5 = 3.3333A, which none of the capacitors are rated for. But this is a noraml ripple that could occur in several applications. Could you please let me know what could go wrong. Is these type of calculations needed?
 

Is the 0.15 ohms the capacitor resistance? What is your load current?

Keith
 

The load current is 200mA (LDO input current). What is the need for load current? The input to LDO will be a clean value and capacitor will filter the ripple upon 12V.
 

It is a while since I have calculated ripple currents, but it is the RMS current in the capacitor which depends on load current. You have calculated the peak current. If the load current is zero, the ripple current is zero.

Keith.
 

I calculated Rms ripple current as Vrms/Z. Also how can be RMS current through cap will be zero if there is no load current?
 

I was assuming a rectifier - is that not correct? If there is a rectifier then when there is no load the capacitor will charge to its peak value and stay there - there will be no capacitor current.

If there is no capacitor then you are charging and discharging the capacitor from your power source. Ripple currents can be very high if that is the case. Is this a capacitor in a power supply? Is the ripply sinusoidal or pulsed? A bit more information would help.

Keith.
 

Is the voltage ripple measured with the cap in the circuit?

If not, the ripple should go way down with a large cap out there.

Does your Z account for the ESR?

If not, the series resistance should be added to your z to get your actual current.

One thing you are definitely missing is the fact that your capacitor impedance is not going to be the same @ 13kHz as it is at 100kHz.
 

If you want to know the ripple current, you either can calculate it from known SMPS circuit parameters, or measure it with
a suitable instrument and respective setup. 3.3 A ripple current sounds absurd with 200 mA load, I rather assume a
measurement artefact.
 

The input to the capacitor is as shown in the picture. Its a sinusoidal ripple upon 12V DC. It is fed to the capacitor followed by an LDO. In such case let me know your feedback
 

I didn't doubt, that you have seen the ripple in a measurement, but if it really represents a current of 3.3 A. I understood, that a
switching power supply would be involved in the circuit, that causes the ripple. If this isn't the case, only an oscillation either on the
load or the source side can explain the ripple. But can it create a high current amplitude? The LDO input current e.g. is uni-directional.
With 200 mA load, it won't generate more than 400 mA pk-to-peak at worst case. But you didn't tell about other circuit parts connected
to the capacitor.
 

Keith:
Even if rectifier exists if the output of rectifier is low compared to the capacitor voltage then the capacitor voltage discharges and charging it again causes current.
In my case the 12V input is from a battery and the ripple noise is caused by the other devices that are connected to the same battery line.
stefannm:
The voltage ripple is not a measured value. It is the system requirment given to desing my circuit. This ripple could occur on the voltage input line of the LDO. So to avoid it I am intending to use 220uF cap. If that large ripple is going away for that large 220uF then cpaacitor is bypassing the ripple voltage and is causing the ripple current which is too high than the rated value.
As of my knowledge, the Z value takes ESR into account. Please refer to the datasheet link and justify if wrong.
I agree that the value specified at 100KHz cannot be taken for 13KHz application. But the impedance will not vary much between 10KHz and 100KHz. It could be 0.17ohms at 10KHz. If I use this value the ripple current is 2.94A which is considerably high compared to rated value.
FvM:
I understand that the 3.3A rippple current is too high. Whatever I am doing is only theorotical calculation only. I didnt complete the circuit. I wanted to know what is going wrong in my calculation, as this kind of noise ripple could be common in several designs.
 

You still haven't shown how the ripple occurs. The circuit causing the ripple is important in determining the real ripple current. It would save a lot of time if you would post you circuit and explain what you are doing instead of being continually vague. The are lots of people here trying to help you and you won't tell them anything.

Keith.
 

Keith:
As of now I dont have the circuit schematic of the entire system, as it is in the design stage only. This is used in an automotive environment. So this ripple can be from the alternator circuit that is generating the voltage to charge the car battery. The ripple could due to the variation in the speed of the car, etc..
 

In which case the answer is simple - add some series resistance between your power source and the capacitor.

Keith
 

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    explorick

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Yes, I think thats what I missed :D. If I consider the wiring resistance to be 1 ohm. Then the current would become 0.5/1.15=0.434 which is well below the rating of capacitor. Is this correct?
 

Correct. However, for an automotive application I would add as much resistance as you can tolerate. Another option is a series diode (Schottky maybe) and also a series resistor. With only 200mA I would have thought you could add 10 ohms.

Keith

Added after 50 minutes:

Here are the results of a few simulations to help you compare different options. I haven't included large resistors such as 10 ohms, but that would be my preferred option unless you cannot afford the voltage drop (but in an automotive application you need to be able to withstand total dropouts usually).

Keith.
 

If a schotkky diode is introduced it is not helping in reducing the current. The input voltage is 0.5V ripple upon 12V. So the diode will be conducting and will not cause any difference. In simulation 4 the current might exceed the spec of the shottky diode an it could damage the device. If the ripple is 0.5V without any DC content, then the shottky diode will not conduct till certain voltage(say 0.3) and will cause a reduction in current 0.3/0.15 for positive cycle and will not allow the negative ripple.
Please rectify if my understanding is wrong.

Happy learning
 

What the diode does do is help with power dropouts. When there is a sudden heavy load on the car battery the voltage can drop considerably. If there is no diode then that will discharge your capacitor, your regulator will dropout and the circuit will stop working. If the diode is there it stops that happening and the capacitor can supply some current for a short period.

Do you have a brownout specification?

The waveform with the high peak current is the one with no series resistor or diode, so it cannot damage the diode!

Yes, you lose 0.3V across the diode, but you have 12V. What is the regulated voltage you are trying to achieve?

Keith.
 

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