OPA855: OPA855

Discussed here, 

Michael,

Thanks.

So connecting the PD pin  to 0V ( Ground) will put the amplifier in the ON state!?

Hi Giora, as discussed in Michael's link, the PD pin is referenced to Vee. in Figure 23, Vee=-2.5V, so Vpd= -2.5V is OFF, and Vpd= +2.5V is ON. These are the same supplies as your circuit, so these Vpd voltages will also disable and enable your circuit respectively. The Vpd voltage that will trigger a transition from low to high is Vee+1.5V. Once the OPA855 is enabled, Vpd will have to fall below Vee +1V for the OPA855 to transition back to powered down. This hysteresis improves the noise immunity of this feature.

Sean,

Thanks for your answer.

We tried the connection to +2.5V and it does not work  - The amplifier stayed at "state" off.

As I said only connection to +5V through a 1K resistor "did the trick".

Regards,

Giora

Hi Giora, So you needed a voltage above the supply to enable the OPA855? That's not how it's supposed to work. Maybe one of the input ESD diodes got damaged and is starting to leak? What is the input current when you connect 2.5V straight to the PD pin?

or perhaps the device is not actually an OPA855 but one of the +Vcc enable devices. 

Michael,

I am really do not know.

Will look on the marking on the component.

Thanks.

Hi Sean,

The amp is connecting as a transimpedance amp - an APD is connectes to the minus pin , the feedback resistor is 30K and a 30k connects from the positive input to ground. Only the ADP dark current ( several Na ). The APD is biased with  -40V.

connecting the PD pin straight to +2.5V (regulator) drops the +2.5V to 0.6V. The regulator can supply 200Ma.

Hi MIchael,

It is OPA855. We tried both.

Glad you figured it out, good luck with this