My intuitive understanding is that the BJT designed for higher currents will always have higher transconductance simply because of the physically larger junction, almost like a number of smaller transistors in parallel.

From this perspective it would be beneficial to use bigger transistors even in low-power amplification circuits to get a higher open-loop gain. Does it work this way? Is it the lower β of the higher-power transistor that prevents the gain increase?

Edit: I just realised that for a BJT at least at the fixed \$I_c\$ the transconductance \$g_m\$ doesn't depend on a structure size. It's defined purely by the \$I_c\$ itself

  • \$\begingroup\$ I would think doing is more important than than size when it comes to beta. \$\endgroup\$ Jul 24, 2022 at 12:05
  • \$\begingroup\$ There are too many variables to simply say β * Pmax is constant or some equation, but we can say diode capacitance increases with current rating which reduces the maximum transition frequency. Higher base-emitter doping also increases β and reduces GBW and special Diodes Inc superbeta have lower Rce thus can switch more current but size affects power dissipation. Also higher current ratings tend to produce greater gm which is not specified in datasheets with low β yet gm is more constant over a wider range above 0.65V at 25'C. Then geometry has a big effect with planar transistors. \$\endgroup\$ Jul 24, 2022 at 13:53
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    \$\begingroup\$ I looked at 500 power NPN's and could not find any simple correlation or constants with 6 key factors \$β, g_m, P_{max}, f_T, R_{ce}(sat), V_{ceo}, I_{max}\$ yet for Pmax >> 5W , β tends to be much lower than small signal packages \$\endgroup\$ Jul 24, 2022 at 13:58

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I just realised that for a BJT at least at the fixed Ic the transconductance gm doesn't depend on a structure size. It's defined purely by the Ic itself

Indeed. It only depends on current. The power BJT can have higher transconductance if it passes a higher current, which it can do without burning. For example MJL3281 above a few amps of current... the few tens of milliohms resistance in PCB traces, bondwires, internal resistance in the chip, etc, set the limit on transconductance.

To elaborate on your question, a transistor can have a high current rating without necessarily having a large chip.

Big chip benefits:

  • More thermal mass -> larger safe operating area.

  • More contact area with leadframe and then heatsink -> lower RthJC -> higher power dissipation.

  • Higher current ratings, also due to more bondwires.

Big chip drawbacks:

  • More capacitance and charge storage, lower transition frequency (ie, slower).

  • More bulky expensive, bigger package, etc.

Along with that, power transistors usually have lowish hFe along with other undesirable characteristics that you have to live with, like lead inductance and collector capacitance to hetasink.

You can compare a high current small transistor like 2SC5706, with a big power transistor like MJL3281 or MJL21193, and pay attention to parameters influenced by chip size, like SOA, power, capacitance, fT, etc. Another thing is weight. The TO264 weighs 10 grams, and most of it is in the thick slab of copper just below the chip. The DPAK transistor weighs 0.3 grams. Thermal mass is important if you need high peak power dissipation.

So if you're thinking about amplification, a bigger die with the associated huge capacitance would be a drawback at medium to not so high frequencies.

The datasheets usually don't talk much about charge storage or other parameters that influence switching speed, unless the BJT is specifically sold for that use, but then why not use a MOSFET instead...

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    \$\begingroup\$ One more difference: noise figure is often better for larger devices. LT1115 uses input transistors bigger than the outputs. Some have found excellent noise figure from ZTX851 and even MJE15028 (Art of Electronics, 3rd ed.) Another way to put it: a big die is lots of smaller transistors in parallel, so their noise averages out as 1/sqrt(N). \$\endgroup\$ Jul 24, 2022 at 22:11

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