Am I calculating/considering linear regulator voltage ripple correctly?

I need a clean and power efficient -1.22V supply for a ADC driver fully-differential op-amp, the main power source will be a Lipo battery cell.

I'm considering ADM8828 followed by a LT1964. The LT1964 has been performing well supplied by my bipolar bench supply, but I'm not sure I know what I'm doing with the ADM8828.

From the charge-pump datasheet:

I need < 5mA output @-1.22V so from Fig 5 using a 3.3uF capacitance it looks like my ripple would be ~10mVpp however I'm not sure how that relates to Fig 6 which shows values 10x higher.

From fig 10 it looks like the ripple will have a ~80kHz to 180kHz fundamental. From the regulator datasheet:

Looks like I can only expect 25-30dB ripple rejection using 10uF output cap, so 100mVpp input ripple becomes ~5mVpp? That's not very good relative to the 30uV RMS noise spec. The ripple frequency is outside of my ADC range of operation (audio frequencies) so maybe I don't need to worry about it? It could alias though. The figure above is for load current of 200mA, is rejection likely much better at <5mA?

Am I approaching this correctly? What would be the normal design approach for something like this? Could I practically add my own filtering to improve ripple rejection?

• You can filter out the ripple by putting a low-pass filter between the charge pump and LDO. Since it is only 5 mA, you can use a 10 Ohm resistor in series and a cap in shunt. You can also use two stages of filtering (4.7 Ohms each). The 10 Ohm resistor will only cost you 50 mV. Just make sure the LDO won't drop out. Feb 26 at 17:45
• You've picked a regulator with poor high frequency rejection. A better one would give you less noise. Feb 26 at 18:23

I need < 5mA output @-1.22V

If you are only supplying up to 5 mA then there are better options than a negative voltage regulator; consider using a precision shunt regulator. Firstly, it doesn't care whether you are generating +1.22 volts or -1.22 volts; it only cares about regulating the voltage across it.

So, if the ADM8825 is supplying -5 volts$$\^1\$$ you'll need a series resistance of about 680 Ω to drop the voltage to the shunt regulator. And, if you look into what the dynamic impedance of a shunt regulator might be you'll probably find it can be as low as about 3 Ω in the frequency range 80 kHz to 180 kHz. See the data sheet for the LM4041 for example using a 1 μF output capacitor: -

So now, you can work out what the ripple rejection will be by taking the ratio of 3 Ω to 680 Ω and converting to decibels. More precisely it's this: -

$$20\log_{10}\left(\dfrac{3}{680+3}\right) = 47.14\text{ dB}$$

And, you can improve this by spitting the 680 Ω feed resistor in two and putting a 1 uF capacitor at the junction. This lowers the perturbations of ripple current being handled by the shunt regulator. You might find that at 80 kHz and using a 1 uF capacitor, the ripple is reduced by another 30 dB.

Am I approaching this correctly? What would be the normal design approach for something like this? Could I practically add my own filtering to improve ripple rejection?

Well you need to consider shunt references for sure - they are suitable for positive and negative voltage control and are more commonly available than negative voltage regulators. I feel confident this is a better approach and will give you a more repeatable initial tolerance on voltage, lower drift with temperature, lower noise and better long term drift.

The LM4041 is available with an output voltage of 1.225 volts and an initial tolerance of up to 0.1%. The basic device is 1% initial tolerance and this means it will produce a voltage of between 1.2128 volts to 1.2373 volts.

Look at the LT1964 - it has an initial voltage variation of between 1.202 volts and 1.238 with a 1 mA load hence, the shunt reference is better. Looking at the LT1964 it looks like it will drift more with temperature than the LM4041 too.

I'm not trying to press you into using the LM4041; I'm just saying that there are better devices with more options to consider if you think about the shunt reference device.

$$\^1\$$ I'm assuming you are generating -5 volts because it roughly corresponds with this graph: -

• If I understand the ADM8828 it will generate the negative voltage of whatever is supplied to it. A Lipo cell will have vary from 4.2 to 3V. If my load current will be static at 1.8mA, I'd need a 946Ohm Rs. Worst case LM4041 power draw will be 9.3mW versus the LT1964 which yields 5.3mW. I'll need to decide if that's a worthwhile tradeoff for the benefits identified. Feb 26 at 21:36
• The power draw will be virtually the same actually. Feb 26 at 21:46