I'm going to amplify a DC optical input power. Can AD8015 deal with it?
Acoording to the datasheet of AD8015, there is no illustration about this DC application.
Can this chip amplify a DC optical input power?
AD8015 is in a quantizer design. Its primary application is for optical telecommunication application.
With a DC optical power applied to a PD then to AD8015, AD8015 will output logic 1 or 0 at its differential output, when input power is large enough to saturate the AD8015.
Unless you wants to detect DC optical power in the AD8015 input linear range: < |25uA|, you would only get
logic 1 or 0 at AD8015 differential outputs.
Btw, in the AD8015 input linear current range, AD8015 differential output amplitude linearly correlates to the input current. PD photo-current is also linearly correlated to the optical average power shining on the PD, in general.
Thank you for your reply!
I used the DC optical input power to shining on the PD, the circuit is attached below.
I applied the +Vs to 4.5v and the-Vs to 0v.
And I tested the +output is DC 2.29v, the -output is DC 3.34v and the voltage of pin 2 is DC 3.23v.
No matter whether the optical input power is shining on the PD, these values seem to be no change.
Are there some mistakes in my circuit? Why the voltages of pin6 and pin7 are both positive? Shouldn't the output change with the optical input power?
Btw, I tried the AC optical power with the frequency 2khz-1Mhz, there is no AC output from pin6.
In your tests, what is the optical DC power applied to the PIN photo-diode connected to AD8015 input?
Please note, AD8015 provides differential outputs, OUT+ / OUT-. When you applied an digital optical signal to the photo diode, AD8015 should output a digital output differentially. Datasheet Fig. 7 shows the differential signal formed by the two single ended output pin signals. As you can see, both pin 6&7 would output positive voltages around +Vs - 1.3V typ.
When you read the Fig. 7, please pay attention to the photo-diode current direction. Positive current means input current to AD8015.
If we assume the photo-diode current is always input to Ad8015, then pin 6 output voltage is always larger than +Vs - 1.3V, and pin7 output voltage us always lower than +Vs-1.3V, depends on the input current amplitude.
In your DC optical power input application case, if the produced photo-current Id satisfies: 0 < Id < 40uA, you would see
linear amplification at both pin 6 & 7 (around +Vs - 1.3V typ.).
What is your AC optical signal scheme and power applied to the photo-diode?
Hope these are useful to your AD8015 applications.
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