8
\$\begingroup\$

I found this application in AD's OP77 datasheet:

enter image description here

Apparently they (ab)use a 2N930 transistor's B-C's junction as a low-leakage diode. Is there any reason why you would choose a transistor over a real low-leakage diode for this?

\$\endgroup\$
8
  • \$\begingroup\$ Seems odd to me too. \$\endgroup\$ – Olin Lathrop Jul 17 '11 at 11:47
  • 2
    \$\begingroup\$ Wow. Normally I'd be worried about noise pickup in that floating emitter. I like how they threw FET in there as well (and drew the symbol wrong; D and S shouldn't be connected like that). \$\endgroup\$ – Mike DeSimone Jul 17 '11 at 12:56
  • 4
    \$\begingroup\$ @OrigamiRobot - No. A base-emitter junction has the same polarity as a base-collector junction. This is an NPN, so the base is the P, or the diode's anode. \$\endgroup\$ – stevenvh Jul 17 '11 at 15:16
  • 2
    \$\begingroup\$ I guess it might be cheaper- if your buying in bulk and you're using that tranny in another part of your design. \$\endgroup\$ – Jim Jul 17 '11 at 16:41
  • 4
    \$\begingroup\$ @Jim - I don't think that's the reason. Application examples in datasheets are usually not cost-optimized. They're given so that the design engineer can get started. Cost optimization is part of the design, which only starts there. Thanks for your thoughts, though. \$\endgroup\$ – stevenvh Jul 17 '11 at 16:45
9
\$\begingroup\$

The reason is a recovery time. Diode can have 3 microseconds, BJT is just a 5pF * N ohm ~ dozen of nanoseconds. The diode is better than BJT in slower range, it has pA current, when BJT has nA.

Physics of fast recovery for BJT can be explained by very low volumetric capacitance (low charge) of base, because BJT base is very thin by design.

I guess low leakage diodes have much longer distance across junction (falling doping gradients or even gaps), when carriers must approach region by thermal diffusion, tonneling, which takes time. The diode gap has very large volume (comparing to BJT base volume) to fill with carriers during recovery time.

\$\endgroup\$
2
  • \$\begingroup\$ +1 for making me think .Does this mean that a CB junction could be a good diode for a SMPS ? Assuming that the transister has a reasonable base current rating .Would performance be the same if BE were shorted on this CB diode? \$\endgroup\$ – Autistic Jan 19 '16 at 19:26
  • \$\begingroup\$ @Autistic your question in the comment was interesting. Have you found anything in the last 6 months to answer your questions ? \$\endgroup\$ – efox29 Jun 24 '16 at 1:47
8
\$\begingroup\$

The reason is all about leakage current. The B-C junction outperforms all low leakage diodes by more than a factor of 10. A highly used low leakage diode, the BAV116, is rated to 5nA of rev leakage current. The B-C junction of a 2N3904 is well below 30pA at room temperature. Using the B-E junction is even lower leakage, usually around 5pA. You can't touch this with standard low leakage diodes. There are very pricey FET based diodes that are lower, but they are rare. The B-C junction advantage is that its reverse voltage is that of the transistor. Using B-E the junction will "Zener" at 6-7 volts. Still quite useful in low voltage circuits or even as an asymmetric clamp.

I have used 2N3904s for low leakage diodes for years with excellent results. Just pay attention to maximum forward current through your selected junction.

In the example circuit the B-C junction was used because of the range of voltage. It increased the holdup time of the sample capacitor by reducing rev leakage back to the op-amp output. I would think the leakage of the reset mosfet will dominate this circuit as the input bias of the AD820 is typically 2pA.

\$\endgroup\$
1
  • \$\begingroup\$ Note that B-E junction is subject to breakdown typically at several V only. But e.g. at -1 V, an ordinary transistor C945 was observed to leak 230 fA through B-E junction and 300 fA through B-C junction. So it B-E is marginally better. \$\endgroup\$ – dominecf Jan 6 at 8:16
0
\$\begingroup\$

Thermal characteristics? BJT can be heat sinked quite easily with a clip-on, or case mounted for mechanical thermal coupling. If the op amp OP77 are also chosen as circular case, then a dual-mounted heatsink is simple to mechanically fit.

\$\endgroup\$
2
  • \$\begingroup\$ This is not exactly a power application, the dissipation in the diode would be in the mW range. I don't think that's the reason. Thanks for your thoughts, though. \$\endgroup\$ – stevenvh Jul 21 '11 at 13:19
  • \$\begingroup\$ Thermal matching will match beta's even at cool temps. Not necessarily intended to dissipate heat. Just guessing at the original purpose however. \$\endgroup\$ – Jonathan Cline Jul 22 '11 at 23:09

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.