Hello,
I'm playing with my SMPS step up (boost) design, where the controller is a =
PIC.

I thus have +12V power input, an inductor, a MOSFET, and a diode that charg=
es a
high voltage capacitor, and a voltage divider before the diode, to monitor =
voltage
(via a first PIC ADC input) of the cap without actually discharging it cont=
inuously.

The gate of the MOSFET is connected to a +5V rail through a 1k resistor and=
 I use
a PIC output in open drain configuration to ~quickly stop the MOSFET from c=
onducting
(the opposite doesn't need to be ultrafast, as the SMPS will be always used=
 in
discontinuos mode, thus current is zero whenever the MOSFET starts to condu=
ct).
Switching frequency is pretty low.

While it's certainly not the most performant boost circuit, I like its simp=
licity
and low components count.

Now, I also need to monitor current (for which I have devoted a second ADC =
input),
and thus I have placed a shunt resistor between the source of the MOSFET an=
d the
0V rail. I can be 100% sure that the voltage out of it will never exceed th=
e PIC
ADC max input voltage, also due to insufficient input current from the powe=
r supply.
The shunt has also pretty low resistance so the voltage never reachs 0.5V (=
not much
resolution from the ADC, but enough to not want to add an OpAmp, as space o=
n board
is very very limited).

Question is: should I put a series resistor between the PIC ADC input and t=
he
shunt? Theoretically the voltage will never go below 0V and will never go a=
bove
0.5V (as explained above), but I fear some insidious aspects that maybe I h=
aven't
been able to consider, maybe for some hard to grasp parasitic capacitance, =
etc..
could the shunt voltage go below 0V and thus fry my PIC ADC input pin unles=
s I put
a series resistor? Is this series resistor necessary?

If it's totally useless, I don't want to put it, it's a matter of principle=
.. :D

Thank you very much for augmenting my design knowledge and experience.

Kind regards,
Mario

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