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I have inherited a tachometer signal conditioning circuit.

It has an NPN transistor (2N3904) with a Schottky diode (BAT54) between the emitter and the base. It is preceded in the circuit by an LT1017 comparator - so input should be a 5 V square wave. It is followed by a low pass filter.

I am trying to understand its purpose in the circuit.

enter image description here

enter image description here

Null
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Michael Nunan
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  • If the base could go lower than the emitter, then the diode would protect the \$be\$ junction from a large reverse bias – Aaron Oct 06 '21 at 13:50
  • Normally, you would add such a diode if you had any reason to expect that the base might be pulled far enough below the emitter (or the emitter driven high enough) to force the BE junction into reverse breakdown. The diode protects the transistor. – Dave Tweed Oct 06 '21 at 13:50
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    Where does that other line go that is going down? Assuming that to the left is the comparator. – Aaron Oct 06 '21 at 13:52
  • The line going down is to a capacitor 220pf (part of RC Filter) then it continues to a connector that will go to input on an Autopilot (very power low load). – Michael Nunan Oct 06 '21 at 15:49

1 Answers1

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Q4 provides current gain when the LT1017 output is high. When the LT1017 output is low D11 conducts, bypassing Q4.

This arrangement is needed because the LT1017's current source capability (high output) is many times lower than its current sink capability (low output).

JYelton
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GeBJT
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    So it's basically turning an (almost) open collector output into a push/pull output. From the datasheet: Output Current sink: 70mA, source 200uA. – Aaron Oct 06 '21 at 13:59