OPA340: OPA340 driving TLV2548 oscillation

Chris Boustred

Part Number: OPA340
Other Parts Discussed in Thread: TLV2548,

Hi,

I have a OP340 in non-inverting configuration driving into a TLV2548 ADC. At lower steady state amp input voltages (in the region of 0.65V) I'm see an oscillation at the output of the amp/input to the ADC as shown below, as I increase the input voltage the oscillation will disappear and the circuit will behave as expected. Increasing the values of R1 and R2 (R9 and R42 in my attachment) 10 fold eradicates the noise. I've not been able to replicate this via simulation and am looking for advice as to the cause of the oscillation (value of R9 too low?) or any alternative recommendations to remove the noise. I've also tried an additional low value series R between the amp and ADC to no effect.

Thanks

over 6 years ago

0 kai klaas69 over 6 years ago

Guru 140200 points

Hi Chris,

is the oscillation dissappearing when you disconnect the ADC from the output?

The ADC has an input capacitance of 50pF. So, I would add a series resistance of 22...47R to the output of OPA340. And as the OPA340 is rather fast I would decrease R9 and R42 a bit, to make it more immune to stray capacitances. Then I would put a small cap in parallel to R9. This and the 22...47R output resistance should recover the phase margin enough to make the OPA340 to work stable. The input of OPA340 should also have some low pass filtering. Have a cap directly from the +input to GND.

The supply voltage is 0V and 5V?

Kai

0 Marek Lis over 6 years ago

TI__Guru 64850 points

To learnChris,

The AC gain of your configuration is defined by: Gain(f) = 1+R9/[R42||Zcin(f)] where Cin is a total input capacitance, Cin_tot = Cin_diff+Cin_cm=3pF+6pF=9pF.

For this reason, 1M||Cin creates a zero in the close-loop transfer function, at frequency f=1/(2Pi*1M*9pF) = ~17.7kHz, leading to instability of your circuit. Therefore, in order to stabilize the circuit, you must add a capacitor, CF, in parrelel with the feedback resistor, RF, to create a cancelling pole.

For optimal solution, Rin*Cin_tot = RF*CF. Thus, CF= Rin*Cin_tot/Rf = 1e6*9e-12/33e3 = 272pF - see below.