For a project of mine I need to generate UART-like signals with higher voltage (up to 18V) and a baudrate of 115200 bps and later possibly 250000 bps. I tried a basic BJT level shifter which worked as shown below:

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(here and later: yellow in IN, green is OUT, plots are 2V/div and 5us/div)

I decided the delay due to BJT saturation may become problematic at higher baudrates, so I threw in a Schottky diode which helped:

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However, that's not how a traditional Schottky transistor is made, and I have seen advice that it's better to avoid saturation in the first place rather than fighting the consequences. So I implemented the third prototype which didn't work as I expected:

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I understand that the higher LOW level is due to the fact that I now avoid saturation, but the shape of the edges is totally unexpected, and can hardly be called an improvement.

Is this how a Schottky transistor is supposed to work, or can anyone offer an explanation as to why it didn't?

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    \$\begingroup\$ Your second circuit is partly using the schottky as a capacitor to overwhelm the parasitic BJT capacitance (as a kind of "speed up") and also using it to help rapidly discharge built-up charge, too. For an experiment, try replacing it with 150pF of capacitance, instead. You will see some of the benefit, but without the forward biased discharging which is also a good idea to add. Your 3rd schematic spends a LOT of time trying to charge up the schottky capacitance (through your two resistors) when it becomes reverse-biased. That's bad. Keep the 2nd circuit. \$\endgroup\$ – jonk Aug 25 '18 at 17:53
  • \$\begingroup\$ In addition to "transistor problem", your scope shows unphysical waveform, likely due to super-sophisticated digital post-processing (interpolation) inside your scope, with undersampled signal (on edges). \$\endgroup\$ – Ale..chenski Aug 25 '18 at 17:57
  • \$\begingroup\$ @jonk I have tried to replace D2 with a 100pF cap in the second schematic and it worked exactly the same. Perhaps the signals I have are too slow for a Schottky transistor to make any difference. \$\endgroup\$ – Dmitry Grigoryev Aug 25 '18 at 20:09
  • \$\begingroup\$ @DmitryGrigoryev Yes. That's likely, then. At slower speeds, it does a job. I was thinking about rates at 100 kHz or so. At those rates you really want the diode, as it helps quickly discharge stored charge on the rising edge of things. \$\endgroup\$ – jonk Aug 25 '18 at 23:11
  • \$\begingroup\$ Related: electronics.stackexchange.com/a/236216/72179 \$\endgroup\$ – Dmitry Grigoryev Aug 27 '18 at 9:18

Your Schottky diode is in lieu of a Baker clamp, which is like this:


simulate this circuit – Schematic created using CircuitLab

and the diodes are switch diodes; that 1N5819 diode is a rectifier, which has higher current (and stray capactance) than a switch. The 1N5819 is about 110 pF of capacitance, while a 1N4148 switch is 4 pF.

Circuit #2 might be showing the benefit of paralleling the base resistor with a 100 pF capacitor, which speeds up transistor turnon and turnoff, rather than the clamp benefit of reducing base charge storage. Your circuit #3 might be showing the capacitance C-B causing that ramp, so a change of component, from Schottky rectifier 1N5819 to Schottky switch 1N5711, about 2 pF, should help. A small capacitor across R1 is a good bet, too.

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    \$\begingroup\$ Common low-capacitance Schottkys typically used for clamping are SD101 (through hole) or BAS70 (SMD). \$\endgroup\$ – CL. Aug 25 '18 at 23:37
  • \$\begingroup\$ Thanks, that makes sense. I finally decided to go for #2 circuit with a 100pF cap instead of a diode, which works just as well as the original #2. \$\endgroup\$ – Dmitry Grigoryev Aug 25 '18 at 23:42

The schottky diode between B and C would be more useful if the emitter were grounded and one inputs the signal to the top end of your R1. Then it would prevent the saturation like it does now, too, but it would not slow down the turning from ON to OFF. There also should be a small, say 50pF capacitor in parallel with R1 to make some actual speeding.

This is totally different configuration, it's a common base circuit. As commented, your version 2 works well, it discharges the base fast when the emitter is pulled up and the transistor turns from ON to OFF. Turning the transistor from OFF to ON is fast enough without any speeding.

  • \$\begingroup\$ Are you saying that Schottky transistors can only be used in common emitter configuration? \$\endgroup\$ – Dmitry Grigoryev Aug 25 '18 at 18:42
  • \$\begingroup\$ @DmitryGrigoryev not that, but I haven't got it working other way. People who have designed the interiors for ex. of 74S-TTL ICs know something that's needed. The problem in 3 is that the base potential isn't held the same when the collector potential rises. \$\endgroup\$ – user287001 Aug 25 '18 at 19:06

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