Solution for "No drop" LDO

Question

I have a very noisy 5V supply, the noise is in very low FQ around 1kHz.

From all the components In my project there is only one I have to provide with 10Hz - 2kHz clean 5V.

But because I need 5V and I only have 5V. I can not just use LDO. So my current idea is to use low noise LDO with boost converter:

• LT1109
• LT1761

^{simulate this circuit – Schematic created using CircuitLab}

I'm looking for integration solution and right now I have found only this one:

I found this one (http://cds.linear.com/docs/en/datasheet/3537fd.pdf) but Rejection Ration of this LDO is disporting + I would not have a way to hand solder it.

Any suggestion ?

Answer 1

If you absolutely need 5 V, that is all your input is, and passive series filtering isn't enough, then a boost followed by a linear regulator does make sense. However, there is no point boosting to 12 V if you just want a clean 5 V. There are plenty of LDOs that can work with under 1 V headroom. Boosting to 6 V would provide plenty of voltage room to loose a little due to series filtering, then still provide enough for many LDO's to do their job.

A trick I have implemented a bunch of times now is to use the B-E junction of a transistor as the threshold detection above a LDO output. That's usually 600 mV or so. It's not hard to find LDOs that work with half that or less headroom. For example, here is a snippet of a schematic from a real product where I employed this trick:

The thing controlling the switcher pass element (Q1) compares the signal out of the R4/R6 voltage divider to a internal 600 mV reference to decide if the switcher output is above or below the regulation threshold. Q4 is set up so that it turns on when the "5.6V" line is the B-E drop above the LDO (IC2) output. The reason for scaling the collector of Q4 down with the divider is so that the system doesn't get fooled into thinking it is above the regulation threshold during startup when the voltage is rising and the LDO input might be 600 mV above its output even though the output is not yet in regulation. In that case Q4 still turns on, but the voltage divider prevents the feedback signal from reaching the 600 mV reference until the LDO input is close to full voltage. I've done this a bunch of times, and it has always worked very well.

Answer 2

You don't need an LDO if you use a boost converter, but in any case I suggest using two separate devices for this task. There are not many one-device solutions.

For example, you could use a DC-DC converter or boost converter to generate 9V or so and use a TPS7A4901 regulator (which does happen to be an LDO) to get the clean DC.

Answer 3

Many devices which require "5V" can actually work with a lower voltage. This is why there are many "4.85V" or "4.75V" LDOs available with low enough dropout to enable operation from 5V power (such as USB). This is commonly used for audio circuits, as many high end converters require "5V" for analog power (e.g. AKM converters) but in reality, a 4.85V linear-regulated power supply may be used if the 5V is too noisy to use directly. As an example, the AKM AK5552 ADC operates as low as 4.75V and can be paired with a TPS 76248 4.85V LDO. Note that the dropout voltage is < 100mV at 100mA load, so that LDO can easily provide a "cleaned up" 4.85V from 5V USB power.

Answer 4

There are many DC/DC boost chips available these days.

Try something like the Diodes AP3012 would provide 500mA with adjustable output up to 29V:

\$V_{out}=1.25 * 1+\frac{R1}{R2}\$

So tune it just above 5V and use your favorite LDO.