Q
I have a question concerning the AD846, regarding the simplified schematic ofthe AD846 from datasheet. The amplifier has an open loop transimpedance of
200MOhm. I would like to decrease this value to a choosen fixed value, 5k for
example. Is it possible doing this by adding a resistor parallel with the
compensation capacitor on pin5 to ground? Looking to the simplified schematic
it should be possible, but maybe not in the real internal circuit. Is it
possible to get an answer on that?
A
I would strongly suggest that the customer look for a newer alternative partfor new designs. The 846 is very old and expensive. We can suggest an
alternative if you can tell me what the customer requires. If he's looking for
a high speed voltage feedback amplifier for building a filter, take a look at
the AD818 and AD817 for +/-15V operation and the AD805X, AD804X for +/-5V
supplies. (in year 2000)
It is unusual for a customer to want to adjust the open loop tranimpedance (or
open loop gain) of an opamp. Typically a user will use a very high open loop
transimpedance in combination with external feedback elements to achieve the
desired gain.
Note from page 8, right-hand column of the specs that high transimpedance gives
you high dc precision. Lower transimpedance gives you lowered dc precision. If
you would like worse dc precision you do not really need to pay a premium for
that; you could use a worse op-amp (that is probably cheaper) and no additional
resistor.
The AD846 is a current feedback amplifier. Pin 5 is really the high-Z version
of the output. A resistor connected between pin 5 and ground would not do much
more than to draw current from the collector of Q6 or Q8 on the simplified
schematics on page 8 of the datasheet.
If the customer wants to make a voltage-in-current-out amplifier he can adjust
the ratio by selecting the appropriate values for Rs and Rc of the circuit of
Figure 48.
If the customer wants to use the AD846 as a sort of a run-of-the-mill op-amp,
the output voltage is equal to Iin x Z. A lowered Z will result in a lower
gain. If that is all he wants to do then he should just reduce the ratio of the
feedback-to-gain resistor.