PSRR - Power Supply Ripple Rejection  - and dB calculations from datasheet - Page 1
Forum section: electronics / beginners

author: ez24
author_rank: Super Contributor
timestamp: August 25, 2015, 07:54:54 pm
content: HiI am looking for a LDO with low dropout voltage and high ripple rejection to be used after a buck convertor just to reduce ripple from the buck. It will be 3.3v at 2aI came across this datasheet with F on the x axis and PSRR on the Y axis.The buck I have has a spec freq of 300kI watched a YT video on PSRR and I thought he said "low" PSRR is better (I cannot find the video now) so I started to get myself confused. And his example looked just like this chart.I wonder if the Y axis could be in -dB or +dB. Since high freqs usually cause problems I figure as the graph goes to the right and down the rejection gets worse. ie the mV ripple voltage will be higherQuestions1. Am I reading the chart correctly - as the line goes down the rejection gets worse ?2. The dB at 300k and 3A is about -15dB or 15dB ? ie would I say this LDO has 15dB of rejection or -15dB of rejection ? (there is a "-" on the Y axis)3. Assuming the ripple going into the LDO is 50mV what would be the output ripple be (at 300k and +15 or -15 dB rejection)? How do I multiply 50mv by 15 dB ? And since we say "rejection" I assume a "-" sign is used in the math.4. Simple question - does this LDO reject more ripple at 100 Hz or 100 kHz ?thanks to anyone for answering such stupid questions

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: August 25, 2015, 08:20:42 pm
content: Hi,If you look at the graph:There are two curves, one is for Iout = 1mA, the other is for Iout=3A.The way to read the graph is pick the switching frequency, in your case 300 kHz, find 300 kHz on the x -axis. find the intersection with the curve and then move over to the y-axis. In this case the answer is 15 dB.15dB can be converted into a number byattenuation = 1015/20= 5.6So the ripple on the input will be attenuated 5.6 times.At 1mA the attenuation is 40 dBattenuation = 1040/20= 100The ripple will be attenuated by 100x at 1mA of load current.Now you have to consider that the ripple is not sinusoidal, it will have some harmonics, these harmonic will be attenuated less.Most (if not all) switching regulators have 'spikes' these have very HF noise.Suggested reading:http://cds.linear.com/docs/en/application-note/an101f.pdfRegards,Jay_Diddy_B

author: Chris C
author_rank: Frequent Contributor
timestamp: August 25, 2015, 08:29:20 pm
content: This was written in response to a prior thread, before I realized you'd already posted this one. I'm moving it here, though  already provided a good answer. And with a few quick additions to make sure all questions have been answered:I looked up a sample LDO, a LT1763.http://cds.linear.com/docs/en/datasheet/1763fh.pdfThis one does give detailed specs on input ripple rejection vs. frequency, at the bottom of page 10. Some others may not, in which case I would assume nothing, and use a Pi filter on the input unless I have experimentally verified it is not necessary.For this part, assuming Cbyp is present at any value (and which I would always use regardless), the worst represented case (at 1Mhz) looks to be 23dB. Yes, lower on the graph is worse. Converting that into something that makes sense isn't too hard, so long as you know the formula:10^(23/20) = 14.125x reductionIf you're running Windows, put the calculator in scientific mode. Do '23 / 20 =', then click the 10xbutton. Voila. Now assuming 50mV input ripple at 1Mhz:50/14.125 = 3.540mV output rippleI'm not sure exactly how to calculate the response of a Pi filter. There's not only the normal capacitances/inductance to figure, but the parasitic ones too. I won't attempt such maths unless absolutely necessary!But from experience, I know even a non-optimal Pi will reduce that to about 1mV or less. 0.5mV or less if you pick the ferrite bead to have good performance up to 1Mhz, as I earlier mentioned. Put a Pi filter on the output too, and double the ripple reduction, getting you down to less than 0.25mV or 0.125mV.One more thing. If your switching regulator runs at 300Mhz, that's the frequency at which you'll find ripple, in the traditional sinusoidal sense. However, they can also emit some spikes too, which occur when the switch turns off, and which can go up to 1Mhz or more. Normally these spikes are of lesser amplitude than the ripple, but some cheap Chinese power supplies are an exception - I've seen documented cases of the spike exceeding 100mV in counterfeit iPhone chargers!

author: ez24
author_rank: Super Contributor
timestamp: August 25, 2015, 10:03:01 pm
content: Thanks Chris C and Jay_Diddy_BMy phone can do these calcs. So with the info, links and datasheets I have enough to keep me busy today.I will also go through Digitkey and order some LDOs to play with.QuoteGoing from 50mV to 3.5 mV looks good to me BUT someday I would like to use the buck to power a PIC32 soWhat is an acceptable ripple to power a PIC32? (I know it probably is in the 1,000 page datasheet) but I am asking someone with real world experience.(maybe this could be a lesson topic - the effect of ripple on electronic components)I am making an assumption that the 3.3 volt rail needs lower % ripple than the 5 v rail because of more sensitive components. For the 12v I do not plan on using a LDO to reduce ripple.I am making a bench PSU with 3.3, 5, 12 and adjustable volts (whatever I can do) using reasonable good bucks from China like this one.http://www.aliexpress.com/item/Wholesale-Price-C-D-C-CC-CV-Buck-Converter-Step-down-Power-Module-7-32V-to/32339838120.htmlThe max current for the 3.3, 5 will be 2 amps based on suggestions from members to be the max current used with breadboards.The source is a Toshiba like power supply that furnishes 19v at 4.74 amps like this onehttp://www.ebay.com/itm/New-19V-4-74A-90W-AC-Adapter-Charger-For-Toshiba-Laptop-Power-Supply-5-5mm-2-5mm-/262019218190?hash=item3d018ffb0eSo I will be using a lot of linears and bucksthanks again

author: Chris C
author_rank: Frequent Contributor
timestamp: August 26, 2015, 03:48:23 am
content: Quote from: ez24 on August 25, 2015, 10:03:01 pmFrom direct experience, it handles 50mV 340Khz ripple with no major issues.Most of the PIC32 actually runs at a lower voltage, generated by a built-in regulator. That provides a layer of protection from ripple.Powered directly by 3.3V are the circuits that directly interface with pins. Doesn't affect digital pins, but analog is another story. For example, the analog-to-digital converter (ADC). With a resolution of 10 bits (1,024 levels), it has 3.3/1024=3.2mV accuracy. 50mV of ripple can therefore alter the reading by 50/3.2=7, in either direction, and that's very close to what I saw in real-world tests. Reducing the ripple helps, but isn't necessary unless you actually need/want the accuracy.But I did want some more accuracy.Connecting my Pi filter to the switcher output drops the ripple from 50mV to 5mV. At that point the inaccuracies of the ADC itself, and stray signals picked up from unshielded analog traces and connections, seem to exceed errors caused by ripple. That was good enough for me. (Also good enough for my counterfeit nRF24L01+ RF transceiver ICs, which lose lots of data when exposed to 50mV ripple; the genuine item does not.)Also analog are the comparators, and charge-time management unit (CTMU, which is typically used for capacitive touch switches but can be used in other creative ways). I'd expect them to have similar losses of accuracy from ripple, though I haven't used them yet on PIC32 and can't speak from experience. Again though, you may not require utmost accuracy.

author: ez24
author_rank: Super Contributor
timestamp: August 26, 2015, 09:49:53 pm
content: QuoteThanks I have been looking for a real number.