OPA857EVM: OPA857EVM

Hi Gerard,

Samir is out of the office right now but will get back to you ASAP.

Regards,

Hi Gerard,

The output of the OPA857 is pseudo differential...i.e one side (OUTN) stays fixed at 1.83V (on 3.3V supplies) and OUTP will change depending on the input current. For a photodiode that sources current it implies that OUTP will get smaller than 1.83V as the current increases and is equal to 1.83V (0V differential between OUTP and OUTN) when there is 0 input current.

In test mode, what I would do to get the correct bias voltage at Test_In is apply a DC voltage to first get OUTP = OUTN = 1.83V. On a 3.5V supply that voltage will be 5/9*3.5 = 1.94V. Once I find the voltage for 0V differential output I would then change the voltage at Test_in and measure the resulting differential output voltage..this will give you the test_in voltage to test_out current transfer function. Hope this makes sense.

samir

Hi Samir Cherian,

Thanks for the clarification, I eventually realized that as well. After some debugging, it turned out that the evaluation board has two differences/mistakes:

- At the output of the TIA, in the resistor network preceding the transformer, the 56 ohms resistor has a 0 ohm resistor mounted instead (measured with multimeter). 

- In the schematic, and at the output of the transformer, VOUTN is connected to ground through a 0 ohms resistor, and VOUTP is open circuit (to get the pseudo-differential topology). However, in the evaluation board, they are flipped (VOUTP is connected to ground). I am not sure if this affects the read out (whether I can still pick up the same AC signal from VOUTN instead). It would be great if you can provide an explanation of why this would/would not be possible. 

After some measurements, I was able to get the expected signal from the output of the TIA (probing the node OUT and OUTN). However, the TIA output is clearly not getting buffered properly to the board output. 

Please let me know what would you recommend doing in this case. I am constrained with time, and it would be great if there is any modification that I can do in the lab (substitute resistors, etc ...) that can make it work. My assumption is substituting the 0 ohms with 56 ohms, and flipping VOUTP and VOUTN should make it work.

Thanks,

Gerard 

Hi Samir Cherian,

Thanks for the comments. The DC bias point of my circuit looks correct. And the output ac signal directly probed (1 Mohm) from the TIA looks good as well (OUTP carries most of the signal, and OUTN has a small ac signal -- but for now, let's assume that in the ideal case OUTN does't change, i.e. voutn = 0 (AC), and voutp = - i x Rf). 

However, the output of the TIA is not being buffered properly still to the output: VOUT_DIFF is not equal to the ac component in (OUT+ - OUT-). Can you please describe how the output stage (resistors + transformer) is supposed to work? 

I am confused. My understanding is that, with the current topology, the voltage across R7 will be 56 / (56 + 237 + 237) x (OUT+ - OUT-), and this voltage is being transformed to the output using the RF transformer (1:1 ratio), to extract the differential AC signal to the output port VOUT_DIFF. However, that voltage is << (OUT+ - OUT-), the original voltage we are interested in. Any explanations?

I eventually will terminate the transformer with a 30 dB 50 ohm voltage amplifier. I will take the loading effect into account: originally, the TIA has 25 ohms, so I expect a reduction of 2/3. However, with the output resistors/transformer circuitry, I should expect that output resistance to change. Correct? 

Thanks,
Gerard 

Gerard,
If I assume you have a 50 ohm resistor on the transformer output then that 50 ohm in parallel with R7 (56 ohm)...lets call that 25 ohm. Now internal to the OPA857 on OutP and OutN are two 25 ohm resistors so the attenuation should be 25/(25+237+25+237+25) = 0.045 V/V.

This is the attenuation you should be seeing. You can simulate this using TINA spice. The test mode itself cannot be simulated, however in this case you are looking simply at the output attenuation network which is correctly modeled.

Samir

Hi Samir Cherian,

Thanks for the clarification, this also matches my calculations. However, why does the design implement a 0.045 V/V attenuation? (or around this value, even for high impedance termination).

This would decrease the gain by a factor of 20. In my application, the current is very small, and few dBs in gain would be important before amplifying the signal. 

Would it be a good idea to bypass the transformer, remove the 237 ohms resistors, and directly connect the output of the TIA to the SMA? (OUTP). 

Best,
Gerard