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Had a question about a grounded base transistor configuration. I was repairing a music synthesizer which uses CA3080 transconductance opamps. One of the later revisions to the schematics placed a grounded base transistor between a TTL logic output and the bias current input of one of the CA3080s. When I looked at one of the datasheets for the CA3080, I found a circuit example (attached) that called for exactly that configuration, with the note that it was “to minimize capacitive feedthrough” (see text in attached). However, in other CA3080 datasheets, there is no transistor... bias current is provided to the CA3080 simply with a current limiting resistor. My question—where is the source of the capacitance they’re referring to... is it capacitance in the output of the driving TTL logic that the bias input of the CA3080 would “see” if being driven only through a current limiting resistor, that it doesn’t see when coupled through a grounded base transistor? I’m not sure why some circuits would need the transistor in addition to the current limiting resistor to provide bias current, and others wouldn’t. enter image description here

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  • \$\begingroup\$ if it is being used as a voltage controlled attenuator/amplifier, most likely the feedthrough is leakage of input signal to output. The usual reason to program the current with a transistor is to avoid the voltage drop that you would get with a resistor. Possibly if the voltage on pin 5 (I think that is the pin that sets gain/attenuation) is not near 0V, there is more leakage? Would need to study the datasheet to be sure. \$\endgroup\$
    – danmcb
    Apr 5, 2022 at 6:53

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Elsewhere in that Application Note 6668 they refer to "The grounded-base configuration is used to minimize capacitive feed-through coupling via the base-collector junction of the PNP transistor".

So they want to minimize the digital signal coupling through the C-B junction during switching. If you compare the oscilloscope captures from each version the difference is not large.

In this case the V- is -15V so pin 5 will be at about -14.3V when the OTA is active.

To my mind, the more important function of the PNP transistor is to shift the voltage/current so a 0/5V signal becomes a switched current source to a fixed ~-14.3V input terminal

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  • \$\begingroup\$ Thanks very helpful-- let me make sure I understand why the transistor is necessary... If I omitted the transistor and just used a resistor to limit the current assuming a 19.5V differential between the TTL high and the -14.5V fixed voltage, the TTL output would see a high negative voltage, and possibly current would flow backwards from the 3080 to the TTL output. By putting the transistor in, I get isolation from the -14.5V. If the 3080 was designed such that Pin 5 was at 0 volts, I would be able to set the current simply with a resistor. Am I thinking about it the right way? \$\endgroup\$ Apr 7, 2022 at 16:33
  • \$\begingroup\$ Yes, that's basically correct., \$\endgroup\$ Apr 7, 2022 at 16:46

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