1
\$\begingroup\$

In the book Learning the Art of Electronics, one of the labs is to build a differential amplifier with a CA3096 transistor array. The author admits that the chip may be difficult to find and gives the HFA3096 as a suitable replacement. However, the datasheet for the HFA3096 gives the following breakdown voltages:

enter image description here

The circuit is a standard differential to single ended amplifier:

schematic

simulate this circuit – Schematic created using CircuitLab

Is it just me or does this circuit violate the transistor breakdown voltages? The author doesn't mention anything about using reduced supply voltages with the alternate chip. Am I interpreting the breakdown voltages correctly?

\$\endgroup\$
  • 3
    \$\begingroup\$ I would interpret this as a mistake on the author's part. It does indeed seem that the HFA3096 is being used outside of its specifications here. The HFA3096 appears to be designed as a replacement for and/or high-speed variant of the obsolete CA3096, but it does not have all the same specifications. \$\endgroup\$ – Hearth Mar 20 at 12:12
  • 1
    \$\begingroup\$ The NPN transistors in the CA3096 and HFA3096 are very different in my opinion, for example the GainBandwith product fT is more than 20x larger in the HFA3096. The HFA3096's NPNs are much faster but can handle less voltage as well. So I do not consider the HFA3096 a "drop in" replacement for the CA3096. Both chips are NPN transistor arrays but that is where the similarity ends. \$\endgroup\$ – Bimpelrekkie Mar 20 at 12:33
  • \$\begingroup\$ @Bimpelrekkie To be fair, they appear to be the only 3xNPN+2xPNP matched transistor arrays on the market. \$\endgroup\$ – Hearth Mar 20 at 12:51
  • \$\begingroup\$ @Hearth I agree that there isn't much choice. But I do think the author could have mentioned that the HFA3096 contains more modern transistors which cannot handle the same voltages that the CA3096 can. So the circuits can be the same but have to be re-dimensioned for smaller voltages. \$\endgroup\$ – Bimpelrekkie Mar 20 at 13:05
  • 1
    \$\begingroup\$ Also respect the collector-to-substrate breakdown. \$\endgroup\$ – analogsystemsrf Mar 20 at 13:45
2
\$\begingroup\$

Your actual question has been answered in the comments, but I'm going to go ahead and do so here, and list some suggestions for fixes.

Answer: that circuit violates the transistor's breakdown voltages.

Suggestions:

  • You could fix that by dropping the supply voltages to \$\pm\$6V or so.
  • The 100\$\Omega\$ resistors force an awful lot of matching on the transistors. That input stage would probably be mostly OK with randomly-chosen small-signal transistors, even ones that came from some surplus house's grab-bag.
  • That circuit doesn't need a bunch of matched transistors -- it just needs a matched pair. I checked on DigiKey, and got back a bunch of hits. There appear to be at least four different part numbers for NPN matched pairs, and at least a hint of PNP matched pairs. If you understand which pairs transistors need to come in matched pairs and which don't, then you can do your discrete op-amp design without using that specific chip (which was getting hard to find when I was the IEEE store manager for Portland State University back in the mid 1980's).
\$\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.