linear-regulatorvoltage-regulator
I am looking at an older circuit, and I was wondering why the designer added these two resistors in series with the input voltage of this 5 V regulator. The regulator would be powering a system that pulls 30 mA maximum.
I think I had seen that some people add resistors to try to dissipate heat, but the specified resistors are 1/4 W 0805 resistors. Is that really what they are for?
The regulator is MCP1703-5002E/DB (datasheet).
Good question.
Almost always, what damages things is heat, not current. The answer to your question is in the difference between local and global, or between sub-parts and the whole thing. The current may go through several sub-parts, and each one of them may have different power ratings. A bonding wire may have, by itself, a dissipation limit that is way below that of the whole device. A series sub-part, that needs to be there for whatever reason, may have a dissipation limit lower than that of the main sub-part. For instance, because it has a small volume, and a small surface in contact with the rest of the device, and cannot put out heat at the same rate as the main sub-part. In cases like these ones, it makes sense to specify a limit for the current, even after having specified a limit for the total dissipation, because those little, "secondary" sub-parts may not be able to dissipate heat at the same rate as the device as a whole.
Just an example. Imagine the situation in the following figure. The power rating for the whole device is 5 W. It would look like operating the device at \$V_d\$=2 V and \$I_d\$=1.5 A would be within safe limits, because \$P_d=V_d·I_d=2·1.5=3\$ W < 5 W. The global heat dissipation would be below its corresponding limit. However, \$I_d^2·0.1=1.5^2·0.1=0.225\$ W > 0.1 W, so that small sub-part would be exceeding its local dissipation limit, and hence the need to specify \$I_{dmax}\$ in addition to \$P_{max}\$. But the key is that, even for that current limit, the real reason behind it is heat, not current by itself. Except for a few cases (like electromigration), current by itself does not directly damage anything.
If they are signals to the pi then use a resistor potential divider. A 1k ohm across the input and 0v of the pi and a 9k resistor from the pi input to a high voltage logic signal will reduce that voltage by ten to one. You should be looking for slightly less than a 10:1 reduction. Can you work this bit out yourself? See the link.
Best Answer
There are obvious possible (and interconnected) reasons: -
Drop-out voltage from data sheet: -
So, at a load current of 120 mA the drop-out voltage is about 0.16 volts and the minimum voltage needed to run the regulator is 5.16 volts hence, the voltage across the resistors will be around 6.9 volts at a current of 115 mA.
There could be another reason too: -
If the load is fairly stable at around 100 mA there would be a guaranteed volt drop across the resistors of 6 volts and this means that if the input rail (called 12v_ISO) rose too high, the regulator would be somewhat better protected. The regulator has a maximum input voltage of 16 volts but the designer may be aware that this voltage might spike up to over 16 volts in certain situations.
A trick that the designer may have missed is not applying more input capacitance. With 100 nF in series with 60 ohms, this acts as a 27 kHz low pass filter but the main improvement to the device's PSRR is to be made at frequencies starting at a few hundred Hz: -
So, based on that I'd be making the input capacitor more like 10 uF resulting in a cut-off of 265 Hz and this will improve PSRR quite a bit. The data sheet tends to favour an input capacitor of 1 uF for most of its performance graphs so this might be another little thing that the designer missed.
Data sheet quote: -