-1
\$\begingroup\$

I'm using ADA4350 as a transimpedance amplifier with various gains for feedback paths (30 ohms,3k,300k and 3 mega ohms respectively). When I give a constant current source of about 23mA as input to it, the Inverting terminal of the op amp at the first stage gets an offset voltage of 0.4 volts (ideally, it should remain at 0 volts, thus following the non Inverting terminal). However, when I give a constant current source of 15mA and 19mA , there is no offset voltage at the Inverting terminal. I saw in the datasheet that the absolute maximum rating for INN pin is 20mA. But I am confused about what this means. Does it mean that I can not use a constant current source of more that 20mA if I use an arrangement as shown in figure 61 in the datasheet?

\$\endgroup\$
9
  • \$\begingroup\$ You’ll probably find that the combination of feedback resistor and power supply rail are not enough to take 20 mA into the opamp output. \$\endgroup\$ – Andy aka Jan 20 '20 at 18:01
  • \$\begingroup\$ Any time the differential inputs are non-zero (within offset range) the outputs must be current limiting or saturated to rails. \$\endgroup\$ – Tony Stewart EE75 Jan 20 '20 at 18:04
  • \$\begingroup\$ @Andyaka the power supply to my op- amp is +-14 volts and I'm using the 30 ohms feedback path, hence I should be getting about 0.6 volts at the output for a current for about 23mA. \$\endgroup\$ – blazingcannon Jan 20 '20 at 18:13
  • \$\begingroup\$ What are you actually getting? \$\endgroup\$ – Andy aka Jan 20 '20 at 18:14
  • \$\begingroup\$ @Andyaka about 0.25 volts \$\endgroup\$ – blazingcannon Jan 20 '20 at 18:16
0
\$\begingroup\$

Datasheet indicates for +/-5V supply :

Linear Output Current P1 or M1, V OUT = 2 V p-p, 60 dB SFDR 18 mA rms (Typical)

Only 9mA for single 5V supply.

So 20mA is pushing the design limit. If you need more current than specified, add complementary emitter followers inside the feedback loop. This also avoids the self-heating of the Op Amp under such loads of 100mW range.

This has nothing to do with the Absolute Maximums
Input Current (IN-N, IN-P, VIN1, RF1, and REF) 20 mA

Any time the differential inputs are non-zero (within offset range) the outputs must be current limiting or saturated to rails.

warning* I just noticed in comments you are using +/-14 or 28 V when the max is 14V or +/-7V.

ABSOLUTE MAXIMUM RATINGS Table 10. Parameter Rating Analog Supply Voltage 14 V

\$\endgroup\$
0
\$\begingroup\$

As Tony pointed out, the design is way beyond +/-5V typical power rail spec. It is better to lower it.

Typical output capability of 18mArms corresponds to 25.452mApk, and this might mean the part can handle 25.452mAdc. However, if the load is 100Ω per datasheet example, under 23mA input, the output is -690mV, and needs to sink ~30mA(23mA+6.9mA), which is higher then 25.452mAdc. For 19mA input, the output actually sinks 24.7mA(19mA + 19mA*30/100), which is lower than 25.452mAdc. Hopefully, this explains what you have experienced.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.