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I am a building a circuit which has a differential amplifier part. To keep things simple I have simplified the circuit to only show the relevant part below.

One thing from op amps which is still difficult for me is the selection of the feedback elements, that is, in what range should you choose the resistors. I have seen answers such as: high value resistors will cause issues due to bias current, while low value resistors require higher currents which lead to dissipation, etc.

In the simulation below all the resistors have the same value and are swept from 1k to 100k in steps of 20k. The op-amp becomes unstable as R increases. Even at 10k! You already see some oscillations.

I expect that this is related to the input capacitance of the op-amp. Is this the case? Or are there other factors here? What op-amp parameter should I look for in the datasheet that could give a stable output with high resistance values? And how can one predict based on the input signal what value might lead to instability?

The input of the system: source with 3 µs rise and fall time.

LT1214 non-inverting amplifier with gain of 2

sweep 1k - 100k with 20k steps

where x = swept from 1k - 100k with 20k steps

10k

with x = 10k

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    \$\begingroup\$ Opamps are limited in the output current; I'd not go above 10 milliAmps. Thus a 10 volt output requires Feedback of MORE than 1Kohms. Resistors (as do opamps) produce a random (thermal agitation) voltage noise; Thus I'd keep your resistors below 100Kohm, which produces 40 nanoVolts RMS/rtHz and in a 10,000Hz bandwith will become 40nV * swrt(10,000) = 40nV * 100 = 4 microVolts random noise spread uniformly over the 10,000Hz bandwidth. Having be beaten around head and shoulders by these limits (Ishort_circuit_output, and thermal noise) for decades, my default opamp resistor is 10K ohms. \$\endgroup\$ – analogsystemsrf Nov 21 at 21:08
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This particular op amp has a note in its datasheet on page 14 about the feedback resistor and how to choose components to ensure there is no oscillation:

Feedback Components

Because the input currents of the LT1213/LT1214 are less than 200nA, it is possible to use high value feedback resistors to set the gain. However, care must be taken to insure [sic] that the pole that is formed by the feedback resistors and the input capacitance does not degrade the stability of the amplifier. For example, if a single supply, noninverting gain of two is set with two 10k resistors, the LT1213/LT1214 will probably oscillate. This is because the amplifier goes open-loop at 6MHz (6dB of gain) and has 45° of phase margin. The feedback resistors and the 10pF input capacitance generate a pole at 3MHz that introduces 63° of phase shift at 6MHz! The solution is simple, lower the values of the resistors or add a feedback capacitor of 10pF or more.

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  • \$\begingroup\$ spelling error in the datasheet *ensure. \$\endgroup\$ – DKNguyen Nov 21 at 20:08
  • \$\begingroup\$ I usually put the option for a capacitor across the feedback resistor and keep the summing node traces short to minimize capacitance. \$\endgroup\$ – Kevin White Nov 21 at 20:17
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    \$\begingroup\$ aha, can you give some background info about the 10pF, does it introduce a zero to add phase shift in order to cancel the phase shift introduced by the pole \$\endgroup\$ – Navaro Nov 21 at 20:17
  • \$\begingroup\$ @Navaro yes, it does. \$\endgroup\$ – TimWescott Nov 21 at 20:29
  • \$\begingroup\$ @Navaro The purpose of the feedback capacitor is explained, e.g., here. You can think of it as making the circuit more like an integrator, which has very low bandwidth. \$\endgroup\$ – Null Nov 21 at 20:31

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