KSA992 vs ZTX795A

Does anybody have experience with both Fairchild KSA992 and Zetex (now Diodes Inc.) ZTX795A as input dif pairs? Any experience comparing the two is appreciated.

The ZTX supposedly have a little more bandwidth — but I'm not fluent enough in data sheets to discern if there are other benefits from one or other other. ZTX seem to be easier to match—but I don't know about their linearity or noise.

FWIW, I have ZTX pairs with Vbe identical to within ∆Vbe=2±2µV and beta identical to within 0.002% (my measurement precision). Best I can do on the Fairchilds is about ∆Vbe~100µV.

I'm not sure it's a meaningful parameter, but I've also computed ∂(∆Vbe)/∂Ic at my operating point of Ic=1.2mA, and the ZTX have a factor of 10 smaller slope.

For people looking to match their own in the future: Distribution of Betas (hFE) on my ZTX sample has mean of 380 and standard deviation of 19. Distribution on the Fairchild's (FB gain grade) has mean of 429 and standard deviation of 17.
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The ZTX795 is designed for higher currents and lower impedances - for instance the Miller capacitance is 15pF rather than 2pF for the 2SA992. The high input capacitance could affect compensation requirements too.

The noise parameters can be important for input pairs, the ZTX would likely be a good performer at 50mA and 10 ohms source impedance(*), but at the current and impedance levels of a standard input pair I don't know - perhaps someone has measured this?

(*) Due to the very low saturation voltages implying very small base- and emitter-spreading resistances.
Thanks Mark!

Here's my circuit (Pioneer M22 rebuild), in case someone has thoughts as they pertain to this application....

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I found that Vbe was incredibly sensitive to temperature, and have to take special precautions on the setup with both air currents and sunlight. But I found the Betas to be pretty stable by comparison. The Betas did change a lot as a function of Ic, so I measure it at my idle current. I'm measuring voltage ratios using an HP3456A over Vishay z-foil resistors.


Paid Member
2003-06-12 7:04 pm
Maine USA
...computed ∂(∆Vbe)/∂Ic at my operating point of Ic=1.2mA, and the ZTX have a factor of 10 smaller slope....

Shockley showed that ∂(∆Vbe)/∂Ic is inherently 26mV per doubling of current, plus parasitic resistance which should not be significant at amplifier current.

So a 10:1 difference of slope suggests something wrong with the parts or the math.
To a first approximation the dependence of current with forward voltage for a junction is set by the physics of thermal distribution of charge-carrier energies and depends only on temperature and the charge per carrier, not the semiconductor device (unless tunneling is involved as in a back-diode).

In other words there's no way to damage the parts to give a 10:1 ratio unless somehow a tunneling junction has been made!
PRR & Mark,

I am still learning about transistors, so your responses make me wonder if I've done something wrong.

In the spirit of making sure we're all on the same page, let me clarify that I did not measure Vbe directly. I measured ∆Vbe between pairs using the differential setup shown here. I then computed ∂(∆Vbe)/∂Ic, not ∂(Vbe)/∂Ic.

I did this by adjusting the base bias-voltage to achieve a current 0.2mA above and below my operating point, and re-did the ∆Vbe measurements at each point. From that pair of ∆Vbe measurements, I compute the slope ∂(∆Vbe)/∂Ic. I have just reviewed the math in the spreadsheet and it appears correct to me.

With that clarification, does this now seem plausible, or do you think I've erred elsewhere?

If I parse PRR's message correctly, I think he's suggesting ∂(Vbe)/∂Ic≈26mV/mA, which means ∂(∆Vbe)/∂Ic should be zero. But I could believe that there might be imperfections from the ideal model, perhaps related to heating of the substrate, that would lead to a nonzero ∂(∆Vbe)/∂Ic. Perhaps that's what I've measured.
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