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I was looking at the schematic for Dave Jones' uCurrent, and I can't seem to make sense of why there are 270 ohm resistors on the inputs and outputs of the op amps in his circuit. The resistance is low enough that it wouldn't affect the filter in the feedback loop, and is negligible compared to the input resistance of an op amp.

Any ideas why these resistors are put in series with the input/output of these op amps?

Here's a snippet of the main section I was looking at, with resistors R12 and R8. The full schematic can be found here: Full uCurrent Schematic. Here's a link to full description of the project: Full uCurrent Design Article. The schematic is on page 16, and a short sentence on the resistor in question can be found on page 8 (search for "R10").

Dave mentions that he uses the resistors to provide "output stability". How exactly does this small resistor provide stability, and why?

enter image description here

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  • \$\begingroup\$ Side note: The application of the TPS3908 here almost seems as if he had that part lying around and included it in the design. It really seems like an overkill to use it as an LED driver. Any comments? \$\endgroup\$ – sherrellbc Aug 4 '14 at 20:26
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    \$\begingroup\$ Well, I think the TPS3908 serves as a nifty little battery life indicator. It's essentially a comparator with an internal reference voltage - it drives the LED at full power until the battery voltage drops below 2.64v, at which point it does not drive the LED. It's a cheap part, and simplifies the battery life indicator. Better than attempting to read the "dimness" of an LED! \$\endgroup\$ – ccpmark Aug 5 '14 at 14:35
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    \$\begingroup\$ Dave is a cool (but busy) guy. I emailed him a link to this question and he replied: "Yes, the resistor is for capacitive load stabilisation and also some series current limit protection." \$\endgroup\$ – JYelton Aug 5 '14 at 18:30
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Regarding R10, the virtual ground is driving stray capacitance. If you loaded the output of U2 directly with that capacitance it would probably oscillate.

R8 actually hurts stability of U4. It would improve it only if C4 was connected to pin 1, however in most cases users will not load the output capacitively (eg. by attaching a long cable) so it's probably meant to limit the short-circuit current from the amplifier, and keep it from driving the virtual ground too far.

R12 limits current in case of a fault such as a large voltage applied to the input- he uses 270R to reduce the number of different part values- but a larger value might be appropriate too. Also perhaps it acts as a jumper for his single-sided layout. ;-)

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  • \$\begingroup\$ About R10, as far as I can tell it isn't directly bypassing C1 or C2, correct? The power supply (battery) is directly bypassing C1 and C2 - the output of U2 isn't directly loading any capacitors. \$\endgroup\$ – ccpmark Aug 4 '14 at 15:11
  • \$\begingroup\$ How might R8 impact the stability of U4? what exactly do you mean by the capacitors are bypassing the virtual ground? How was the virtual ground generated? \$\endgroup\$ – sherrellbc Aug 4 '14 at 20:32
  • \$\begingroup\$ @sherrellbc The virtual ground sees stray capacitance since C1 and C2 are across the battery. More than 100pF or so can cause oscillation on a low power op-amp. The virtual ground is a voltage follower from the two 100K resistors across +V to -V. Op-amps set up as a voltage follower have more tendency to oscillate than op-amps with resistors to give higher gain. \$\endgroup\$ – Spehro Pefhany Aug 4 '14 at 20:53
  • \$\begingroup\$ @ccpmark Yes, you're correct. It's not directly driving the capacitors only stray capacitance. Edited my answer. \$\endgroup\$ – Spehro Pefhany Aug 4 '14 at 20:53
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Yes,the resistor R8 - in conjunction with the capacitor C4 - can stabilize the amplifier against capacitive loading, which however is not shown in the diagram. This can be explained similar to the function of a frequency-compensated voltage divider (oscilloscope probe). In principle, it is a compensation of an additional pole (caused by the cap. load) by introducing an additional zero at app. the same frequency.

EDIT: Here are the equations:

Zero at: wz=1/[(rout+R13+R14)C4]

Pole at: wp=1/[(rout+R8)CL]

rout: opamp output resistance (open-loop); CL: cap. load.

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  • \$\begingroup\$ Ohh yes - I just have seen the answer from S. Pefhany. And he is right - my explanation above holds only if C4 would be connected directly to the opamp output \$\endgroup\$ – LvW Aug 4 '14 at 15:03
  • \$\begingroup\$ Thank you, that makes perfect sense. I had forgotten about the effects of low output impedance op-amps and capacitive loading. However, why would R12 be of use? There's no capacitive loading on the input of the op-amp. \$\endgroup\$ – ccpmark Aug 4 '14 at 15:03

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