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I'm designing a power supply block and need to measure the current flowing trough a LDO. It is clear that the current in front of the LDO and after the LDO are the same (apart from some tiny bit used by the LDO). But does it matter if I mount the shunt in front of the LDO or after the LDO? Are there any differences in noise or voltage precision? Shunt after LDO

Shunt in front of LDO

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  • \$\begingroup\$ With the shunt before the LDO, you'll be measuring not just the load current, but also the current the regulator itself takes to operate. This is likely to be a very tiny amount, but it can be significant depending on your application. \$\endgroup\$ – Hearth Apr 9 '20 at 14:21
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    \$\begingroup\$ I mentioned this difference already in the question. But your absolutely right. \$\endgroup\$ – Botnic Apr 9 '20 at 14:26
  • \$\begingroup\$ Ah, so you did. In any case, see my answer for further consideration on the benefits and drawbacks of each position. \$\endgroup\$ – Hearth Apr 9 '20 at 14:28
  • \$\begingroup\$ Just be careful you can still maintain stability with the pole you are creating with Rs and Cout. That part is a linear regulator with a charge pump to achieve a low-dropout voltage. Not an actual LDO running a current-mode control loop. \$\endgroup\$ – sstobbe Apr 9 '20 at 15:52
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With the shunt before the LDO, you'll be measuring not just the load current, but also the current the regulator itself takes to operate. This is likely to be a very tiny amount, but it can be significant depending on your application.

A counterpoint, though, is that having the shunt after the regulator reduces your load regulation. Higher currents out of the regulator mean a lower voltage after the sense resistor, because that's what resistors do. If the shunt is before the regulator, its effect will be mostly regulated out by the regulator doing its job.

So it's a tradeoff. Do you care more about accurate current measurement, or good voltage regulation? The answer to that determines where you should put your sensor. If you care significantly about both, then you need to start looking into other methods of current sensing.

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But does it matter if I mount the shunt in front of the LDO or after the LDO?

I don't see a problem with the shunt being in either positions given that: -

  • You appear to have plenty enough input voltage headroom for a 1.2 volt regulated output as per this: -

enter image description here

  • Or, if implemented in the output stage then, your voltage feedback components are after the proposed output shunt position (as shown in your diagram).

However, I do have a concern about the BAT60 diode - it will have significant leakage current and, if your load is very light (just the 36 kohm of the FB network) then potentially 100 uA of leakage current could produce 3.6 volts on the output at moderate ambient temperatures: -

enter image description here

At elevated temperatures (85 degC) the leakage current might rise to about 10 mA and this I see as a distinct and serious problem.

I also see you may have countered this with the 120 ohm resistor (ref 120E) on the output and, by the looks of it, you are aiming to have an output voltage of 1.2 volts but, all the same, at 85 degC and maybe 4 volts across the BAT60A you still might exceed 1.2 volts a fair bit.

Additionally, wouldn't you want the current draw from the 120 ohm resistor to be "measured" by the current monitoring? That current would not be "measured" if the shunt is in the output as shown.

And, one more thing - the 10 nF feed-forward capacitor (across the 12k1 feedback resistor) should ideally be implemented before any current shunt on the output: -

enter image description here

The \$\color{red}{\text{devil}}\$, as [nearly] always, is in the detail.

It is clear that the current in front of the LDO and after the LDO are the same (apart from some tiny bit used by the LDO).

If you want that to be largely true then you have to consider moving the current shunt to before the 120 ohm resistor: -

enter image description here

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  • \$\begingroup\$ Oh, that is indeed an incredibly leaky diode, I didn't catch that at all! \$\endgroup\$ – Hearth Apr 9 '20 at 20:08

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