OPA818: OPA818

Well Giora, 

Would probably need the more complete schematic - source C and load. As first guess, the diode current would simply pass around the disabled device and divide into the load through the feedback R. 

That is a pretty large swing, have you checked slew rate?

Hello Michael,

Thanks.

Here is the schematics and the output at VF1.

You will need to use the little paperclip in the ribbon to insert figures

Thanks.

Here is the schematic + output at VF1.

Well this must be an active normal sim. You are AC coupled at the input, why is the output not centered on 2.4V as the V+ input bias. If you want to reduce the disabled output swing, you might just need to put like a 100ohm to ground on the OPA818 output. 

Thanks.

The output is +2.4V when there is no current from the diode.

When there is a current it is flowing into the 40K feedback resistor - cause to the voltage at the out to drop down towards -2.4V

Giora.

Sorry, unipolar current of course. 

You have a total 8.4pF on the inverting node including those diodes and the OPA818 input Ccm+Cdiff

with a 2.7GHz GBP, the noise gain zero will be 1/(2pi*40kohm*8.4pF) = 470kHz and then the Fo will be sqrt(2.7GHz*470kHz) = 36Mhz. 

To get a closed loop Butterworth with F-3dB = Fo=36MHz, set the feedback pole at .707*36Mhz or 25Mhz, requires a Cf = 0.16pF - pretty close to what you show. My measurements of SMD R's show about 0.18pF, so you will be a little more bandlimited just with parasitic unless you are playing some other tricks.