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Regulator Circuit

I am trying to understand the attached circuit. It's an ADP3333ARMZ-3.3-R7 regulator to provide 3.3V to an analog system. My goals is to have a very low noise regulator and power circuitry for a 16 bit resolution Analog Circuit that is attached to it, but am concerned this design is not optimal for low noise. Does the inductor indicate that it is a noisy regulator? Next to the inductor there is a surge resistor which also sounds like it is expecting noise, is there are better regulator to use for low noise applications? Also, can I replace the polarized capacitor with a non-polarized capacitor? The data sheet for the ADP3333 regulator suggests it can have a capacitor as low as 1uF - why is the capacitor at the input 10UF? Love any insights on how to optimize this circuit.

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Pretty much all LDOs are very quiet, the ferrite bead (inductor with huge AC losses) is not there to suppress noise from the LDO itself but to filter out any VHF noise that comes from the main power supply, ferrite beads work best for attenuating noise from 10MHz-1GHz (they work at higher and lower frequencies, but that's the most common range).

Page 8, Figure 19 of the datasheet has the LDO's noise attenuation characteristics, at 1MHz the ripple rejection ratio is around 40dB for a 500mA load, 40dB is a 100x reduction in the amplitude of the voltage ripple. 40dB at 1MHz is quite good for an LDO, if you've got 10mVpp coming in (which is reasonably noisy), then you'll have only 100uVpp of noise on the output (on page three it says "Output noise 10Hz-100kHz: 45uVrms, so it's got a pretty clean output to begin with, and a line regulation of 40uV per 1V of input change looks pretty good to me).

The ADC most likely adds at least another 20-40dB of PSRR (power supply rejection ratio) at 1MHz so already you're looking at ~1-10uV worth of noise which is 5-50 times smaller than the LSB of a 16bit ADC operating at 3.3V. Now the ADC and the LDO's PSRR's will degrade at higher frequencies, but so does the input attenuation of your ADC (they have internal bandwidth limits so noise much more than 10x your sample frequency is unlikely to be an issue unless it's really big in amplitude)

Now in reality, things like coupling and parasitic inductances cause far more noise problems, so I'd say that that LDO is most likely just fine for your application.

Now, as for your other questions: Yes, you can swap out the polarised cap for a non-polarised one, but as they're usually more expensive and have higher ESR, I'm not sure what the advantage would be.

There is no reason why you couldn't change up to 100uF or 1000uF for the input capacitor, it's just a bulk filter cap, it's value isn't too important. The datasheet only refers to how much output capacitance you need, this is because some (but not many) LDOs can become unstable if there either isn't enough or there's too much output capacitance, its internal control loop gets all confused. They're just saying that it'll still work with as little as 1uF of output capacitance so you could use just one really tiny ceramic capacitor as your output filter cap - it's mainly a space and money saving thing.

Have a look at Figure 2 on Page 1 of the datasheet "Typical Application Circuit", that is the best place to start as it tells you what the minimum recommended setup is, there's just a 1uF input cap and a 1uF output cap and that's it.

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  • \$\begingroup\$ I would take issue with the statement that non-polarised caps are more expensive and have higher ESR; a 10uF general purpose 10V ceramic has an ESR of around 3 to 5 milliohm and can be had for a few pennies. uk.farnell.com/murata/grm188r61a106ke69d/… \$\endgroup\$ – Peter Smith Jun 23 '16 at 7:01
  • \$\begingroup\$ So I can replace the 10uF on the input with a ceramic non-polarized, although the output(the picture is the original design) I should add a 1uF cap between the diode and ground to match the suggested use in the datasheet? (Right now there is just the 560ohm resistor and LED). Would you still think the surge resistor - R31 is necessary, or the bead inductor? \$\endgroup\$ – jtoul Jun 23 '16 at 18:40
  • \$\begingroup\$ @jtoul Ceramic capacitors make excellent filter caps (I thought you meant non-polarised electrolytic), ceramic caps have such low ESR and ESL that they will block most of the electrical noise on their own. What do you mean by "cap between the diode and ground"? Figure 2 on page one of the ADP3333ARMZ datasheet is all you need to make this regulator work quite well, anything more than that is just icing on the cake, you can add the 560ohm resistor and the ferrite bead if you want a super clean output, adding more ceramic caps will have a similar effect too \$\endgroup\$ – Sam Jun 23 '16 at 23:52
  • \$\begingroup\$ @PeterSmith I took "non-polarised capacitor" to mean an NP electro, usually ceramics are just called "ceramics" to distinguish them from non-polarised electrolytics as there's no such thing as a polarised ceramic so the prequalifier in that instance is unnecessary, but there are polarised electros (and tants and EDLCs) so the distinction is chemistry dependent. \$\endgroup\$ – Sam Jun 24 '16 at 0:00
  • \$\begingroup\$ Thanks @Tom I will use a ceramic 10uF on the input instead of the electrolytic. I've also added a 1uF capacitor on the output(with LED and resistor in series with it to show when it is powered on or not). I have left the bead inductor and surge resistor in place, hopefully this is the most stable low noise implementation? You can see my updated circuit diagram in the original post above. \$\endgroup\$ – jtoul Jun 24 '16 at 21:23
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enter image description here

I've posted my updated circuit design proposed answer from the info in the comments from @Tom's post. Changed the input capacitor to a ceramic instead of an electrolytic, and added a filter cap on the output(1uF) in line with the power LED. I've removed the surge resistor and the bead inductor too as it was mentioned that's likely overkill. If it isn't overkill and you think it would cut down on noise I will add it back in!

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  • \$\begingroup\$ C5 needs to go from the output of the LDO straight to ground (just like the 10uF), it should be in parallel with the resistor/led string. You can't pass DC power through a cap so the LED isn't going to light up. Other than that, it looks good, you may get better performance by having another small capacitor really close to the adc so the adc has a clean and low impedance supply. The output is probably going to be clean enough for a 16bit ADC with or without the ferrite bead and the surge resistor. \$\endgroup\$ – Sam Jun 24 '16 at 23:31
  • \$\begingroup\$ Great thank you. So you would add a cap close to the adc on the main mcu? I just took out the bead inductor and the surge resistor, changed the output cap to go straight to ground. Now it is a fairly cut down design. \$\endgroup\$ – jtoul Jun 25 '16 at 2:47
  • \$\begingroup\$ Yep, that sounds pretty good to me, that should be good even to 18bits. \$\endgroup\$ – Sam Jun 26 '16 at 0:09
  • \$\begingroup\$ Thanks! Will let you know what happens when I go put it together . \$\endgroup\$ – jtoul Jun 27 '16 at 1:44
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Just a couple of observations. If the aim of the application is as low a noise floor as possible I would reconsider the use of ceramic capacitors if the circuit is likely to encounter mechanical vibration as ceramic capacitors are inherently microphonic in nature due to the piezo electric effect. If the purpose of the LED is to indicate the presence of power, perhaps consider moving it to the input side of the regulator and adjusting the series resistor value to suit the input supply voltage. LEDs generate noise albeit perhaps not much, but once again if the intention is to keep noise as low as possible I'd eliminate it completely unless it was necessary. Actual physical layout of the components will have an impact on the noise performance of the circuit. The best way IME to discover which designs work best is to try different component types as has been suggested by others and if you have access to appropriate test equipment, take measurements and document your findings for future reference.

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  • \$\begingroup\$ Please answer the question, yeah you have some good comments, these go in the comment section. When you get some points for answering you will earn the privilege of commenting. \$\endgroup\$ – Voltage Spike Jun 14 '18 at 5:56

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