2
\$\begingroup\$

The following circuit is nearly exactly what I need for the output stage of my function generator:

enter image description here

datasheet: https://www.renesas.com/us/en/www/doc/datasheet/ca3140-a.pdf

What I don't understand is how an op-amp with GBP = 4.5 MHz and slew rate = 9 V/us, followed by a unity gain buffer, could possibly have an overall nominal bandwidth of 10 MHz and SR = 28 V/us, especially in a gain of 10 configuration. What am I missing here?

\$\endgroup\$
  • \$\begingroup\$ Opamp gain is open loop, then it is further amplified with transistor pair, then feed back with an attenuator 1:9. It is not clear what is the input voltage range VS. output voltage range 18Vpp. How do you know that overall gain is G=10? \$\endgroup\$ – Marko Buršič Aug 23 at 11:18
  • \$\begingroup\$ @MarkoBuršič Transistor pair doesn't amplify, just buffers. Gain of 10 comes from noninverting configuration with 1:9 attenuator, as you said. \$\endgroup\$ – pr871 Aug 23 at 11:28
  • \$\begingroup\$ Therefore the opamp gain is 20dB and the BW is only 0.3MHz. Probably the 10MHz BW is referring at transistor buffer stage. \$\endgroup\$ – Marko Buršič Aug 23 at 11:41

Your Answer

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

Browse other questions tagged or ask your own question.