Several implmentation questions

Hi Infraredx ,

Good day. We'll look into this and get back to you. Thank you.

Regards,
Gilbeys

OK, looking forward to what is posible here.

Hello?

Hello Infraredx ,

Apologies for the delayed response. Please refer to the following:

1) The first design iteration was to use the AD8338 in a normal way providing 0-80dB gain. We unextpectly will be getting more gain from the APD itself, requiring the gain of the AD8338 be modified to be -20 dB -> +60 dB. There are instructions on how to do this on p.17 of the datasheet after equation (12), also pointing to Figure 43. The recommendation is to drive the INPD & INMD pins with 5K external resistances, and then short the INPR & INMR pins. Any reason why you couldn't just add resistors to the INPR & INMR pins and leave the INPD & INMD pins unconnected? Drawbacks to doing that?

Answer:

 

The INPR and INMR pins have internal 500-ohm resistors that's why connecting additional resistors externally will shift the gain range below 0db. You can connect external resistors to the INPN and INPR pins and leave the INPD and INMD pins unconnected, however, to get the Gain range that you want, you will have to add the values of the external resistors to the values of the internal resistors to get the new RP and RN, then substitute them in the formulas from page 17.

In order to get a gain range of -20dB -> +60 dB using the INPR and INMR pins, the total Rp and Rn should be 5kohm each. This would mean that you would need a 4.5k ohm external resistor to be in series on the internal 500-ohm resistor within the said pins.

To confirm this, I've run a simulation in LTSpice. As you can see below (5th plot), when the voltage at the Vgain pin is 0.1V, the gain is about 0.1 or -20dB. Alternatively, when Vgain = 1.1V, the gain is about 1000 or +60dB
 

 
2) On one of the other threads on AD8338 in EngineerZone someone asks what the transimpedance gain is if you use the AD8338 as a photodiode TIA directly. Advantages/disadvantages of doing that over using the MAX40213 first before the AD8338? 

Answer:

When a photodiode is directly connected to AD8338, the overall transimpedance gain would just be equal to:

V_{out_diff} / I_IN,

where:
V_{out_diff} = voltage difference between the OUTP and OUTM pins (see equations 9 and 10)

I_IN = generated current from the photodiode
 


 

Since the VGA core inside AD8338 only accepts current, one advantage is that there will be no need for resistance values in the input pins (INPD, INMD, INPR, INMR), since the generated photodiode current will be directly connected to the VGA core via the INPD and INMD pins. This can reduce voltage-to-current conversion errors in the input due to resistor tolerances.
 

3) Just a nit - I think there is a typo in Equation (11). Shouldn't there be a log function after the 20x operating on the resistor calc in parentheses?

Answer:

Yes, you are correct. The correct formula should be:
G(dB) = 80 x (Vgain) + 20log((2xRfbk)/(Rp+Rn)) - 34

Regards,
Paul

Thank you, Paul, for the detailed reply. In the end I may drive the INPD & INMD directly through 4.99K resistors to have more control of the gain. But in the text around Figure 43 it says in driving it that way you need to short INPR to INMR to prevent stability issues. Is this true? What stability issues are expected? Also, do these external resistors need to be highly matched, or will 1% ones do?

Regards,

Mark

Hello Infraredx,

Yes, that is correct. The INPR and INMR pins should be shorted to one another to avoid stability issues. If they are not shorted together, unwanted oscillations or ringing can occur on the output signal. You can check out this KWIK Video series for more information regarding Op Amp Stability: 

https://ez.analog.com/precision-technology-signal-chains/w/kwik-circuits/32237/op-amp-stability-basics-part-1

Regarding your 2nd question, if you want the gain range to be as accurate as possible, you should choose external resistance values with lower tolerances. But it's also okay to use 1% tolerance.

Regards, 
Paul