# Op Amp Output having oscillation

In my signal processing chain, the first stage is an inverting gain stage using Op AMP LT6200CS8-10.

My output DC operating point is 1.24 V. I am providing an exponential signal (Amplitude 1mV) as the input. This is my schematic:

C43 is not populated. The layout is shown below:

This stage is driving another op amp gain stage using AC coupling cap C38 shown in layout above.

When I don't apply a signal, this is how the output (net op_amp_1) looks:

When I apply an exponential signal with 1 mV amplitude high and 0 mV amplitude low, the output (net op_amp_1) looks like this

The flat portion shows oscillation:

How I can reduce this oscillatory behavior of the output?

• What do you mean "when I apply a signal"? How do you apply or not apply, by disconnecting a cable, or how? Aug 10, 2017 at 18:11
• I have the coax cable from signal generator to my board. When there is no signal, I mean the signal generator's output is off.
– Ash
Aug 10, 2017 at 19:27
• What do you mean "Off"? Is the 50-Ohm source disconnected? Then you effectively change the feedback network parameters (R23-R24-siggen), which changes the gain, and cable delay also changes the feedback transfer function. The gain is about 50X-100X, it is easy to cross the stability threshold. Aug 10, 2017 at 20:23
• I mean output off, the cable is still connected. So the source impedance should be still there.
– Ash
Aug 10, 2017 at 20:42
• It is not certain that the impedance is still there. If yes, then did you consider that the oscillations could be coming from the source? Aug 10, 2017 at 20:45

1) Your most likely culprit is the ground connection on pin 4 of the LT6200. That skinny trace is just asking for trouble. Make a via to ground right at the pin.

2) Your decoupling caps likewise. A via for each. And R29/C40, too.

3) Your R23/R24 combination will produce an effective input impedance of 20 ohms, assuming the LT6200 is working correctly. This will not cause oscillation, but it does represent a gross impedance mismatch to your presumed 50 ohm signal.

ETA - Point 3 applies to AC components only. DC doesn't count since it is blocked by C35.

• Thank you for the answer. Could you suggest any rework which I can do to improve this on the current revision of board?
– Ash
Aug 10, 2017 at 1:59
• R24 is connect to the (-) pin which is a high impedance path. The low impedance path is at the opamp output. 50 ohm is actually parallel to (33 + 2k7) ohm. Aug 10, 2017 at 2:58
• @Ash - Sure. Drill a small (.050 or less) hole just below pin 4. Insert a fine wire and solder on both sides. Do the same for R29 and C40. Aug 10, 2017 at 21:24
• @JasonHan - Sorry, no. Negative feedback will keep the - input at 1.24 volts regardless of signal - it's called a virtual ground. All AC components of the signal will therefor see a 33 ohm to ground. And there is no way 33 adds to 2.7k. Aug 10, 2017 at 21:27
• @JasonHan - exactly. No (appreciable) current flows into the op amp. However, the pin is forced to remain at a DC level of 1.24, which is indistinguishable (in AC terms) from ground. So AC flows into the node and is compensated for by the output through the feedback network. The node is kept at a DC level and is indistinguishable from ground from an AC point of view, so the 33 ohms is effectively feeding ground from the point of view of the input signal. This is in parallel with the 50 ohm terminator. Aug 11, 2017 at 0:17

Use a star ground topology at C40. Use a trace from J3 Gnd and connect directly to C40, then punch a via to the ground plane. Pin 4 of U10 should also have it's own via to the ground plane.

You may also consider the new value for R15, R29 and C1 (new) electrolytic cap. Which gives better low pass filter from the 5V supply. Also consider using a new C40 value that can lower down the ripple in your scope. It's in the capacitor datasheet that shows ceramic cap impedance vs frequency curve.

simulate this circuit – Schematic created using CircuitLab