ldoripplevoltage-regulator
I would like to understand how to calculate the ripple voltage of an LDO.
Suppose I don't have an electrolytic capacitor.
I have a 3.3V output loaded with 50mA max with a ceramic output capacitor of 10uF. How to find the LDO output ripple voltage?
If I have an electrolytic capacitor, I will multiply the ESR of the electrolytic capacitor with my maximum load current. Is that correct?
The best way to deal with a LDO that requires a minimum ESR output cap is to use a different LDO.
These LDOs were designed when anything more than a few 100 nF was too big for ceramic or any other technology that has very little ESR. Tantalum was the usual choice for a few µF to a few 10s of µF. These had some inherent ESR, so LDOs were designed to be stable with them.
Somewhere around 2000 multi-layer ceramic caps became small, cheap, and available enough to replace tantalums in these applications. Newer LDOs were designed with these in mind, and are stable even if the output cap has 0 ESR. This latter arrangement is better, because the cap does it's job better the lower the ESR is. You want the LDO plus cap to look like as a low a impedance as possible to the rest of the circuit, and having some resistance in series with the cap defeats that to some extent.
If you really really need to use one of the ancient LDOs that require a minimum ESR, you can use a ceramic cap and add a resistor in series with it. You can try to find a tantalum cap that specifies the minimum ESR required by the LDO, but those are getting harder to find in the range ceramic can nowadays cover well.
It looks to me like a netlist/layout error:
If you are placing your fixed resistor on the dig pot footprint, it would also be messed up. Looks like you accidentally wired the CO+ and NO2- together and the NO2+ and CO- together so the LDO(Vout-Vadj) is not across a resistance as you believe.
Best Answer
Linear regulators (LDO regulators are a sub-class of linear regulators) don't have ripple in the same way that switching power supplies have ripple. You cannot calculate the noise on the output of the regulator just from its output voltage, the current, and the value of the output capacitor.
A linear regulator will have some noise generated by the resistances inside it.
A linear regulator will also allow some amount of any noise or ripple present on its input through to the output.
The noise the regulator produces will be specified in the datasheet.
How much noise and ripple the regulator lets through from its input to its output will also be specified in its datasheet. This will depend on the frequency of the ripple and noise present at the input.
You can't find the noise parameter for the chip you want to use because the datasheet doesn't specify it. That chip has a linear regulator built in to it, but the chip itself is not primarily a regulator.
Compare your datasheet to the datasheet for the LM1117.
The LM1117 has a lot of charts about all kinds of performance characteristics that your datasheet doesn't mention at all.
The LM1117 datasheet has a specific mention of ripple rejection in the electrical chatacteristics. It also tells you how much noise from the chip itself you can expect in the output:
That's from page 6 of the LM1117 datasheet.
It tells you you can expect at least 60dB of ripple rejection at 20Hz. That is, any ripple at 20Hz present at the input will be reduced by 60dB on the output.
Now turn to page 9 of the LM1117 datasheet. This chart shows you the ripple rejection as a function of frequency:
Note that this is again ripple rejection. That is, suppression of garbage coming in to the regulator, not ripple that the regulator generates. Linear regulators don't produce ripple - although they might oscillate and destroy themselves if you don't follow the manufacturers recommendations on input and output capacitors.
The "RMS output noise" is all the noise from the regulator that you can expect.
If the noise performance of the regulator is critical to your circuit then the TJA1128 is the wrong regulator to use because it tells you very little about the performance of the regulator it has built in to it.