I have the below circuit to translate a slow digital MCU on/off (3.3V) signal to a higher voltage signal (range from 12 to 24V). I have a specific attention for the Vin=12Vdc case, the voltage drop from Vin to Vsignalout shouldn't be more than 2V (extra safety guard is always good). With the below circuit I have about 1.7V drop that is ok.

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But I have a big problem with the signal fall time is about 200 us that is too much (acceptable level should be about <80 us for my purpose).

This is the signal monitor for the above circuit (R2 is 1k):

enter image description here

The green signal represents the base signal of U$2 (now I see, I could rename it as T2 anyway), and the yellow one represents the signal at the collector.

Below one is for R2 is 100 k: enter image description here

(The green probe is on the collector this time) First, I thought lowering the current helped the fall time, but e.g., I tried with 1 M ohm that had the same signal.

I am not really sure why my fall time is terrible and how to fix the problem. Do I make a mistake by not using an extra base resistor for U$2?

Here are the datasheet of the transistors: U$2: PZT2222AT1G Q1: BCW66GLT1

One interesting thing is that I use MJD31C for the same circuit configuration that gets quite decent signal (but the voltage drop is too much that I am trying to replace it with PZT222A).

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I got stuck on it and would highly appreciate for any suggestion

I was playing with R1 resistor by lowering the value to let it conduct more current, placed 47 ohm, but didn't help (MCU's GPIO has 2mA limitation).

If anyone asks U$1 is a LDO to limit the output voltage to 15V, and the end diode is a CLD to limit the output current, I tested with them by removing and short circuit their paths, but it didn't help the signal ( my circuit should work with those components anyway).

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    \$\begingroup\$ You'd be better off with a PNP e.g. MJD32C with series , Shunt R's req'd \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 24 '18 at 13:39

You've probed around in a good fashion but you neglected to look at the emitter signal. It will fall more rapidly but the diode then becomes reverse biased due to stray (or actual) output capacitance holding the cathode at a positive potential therefore the true output trundles back to 0 volts much more slowly.

Between the base-emitter region shutting down and the reverse bias on the diode you have the problem. Try putting a real load on the circuit or even a 1 kohm from emitter to ground or both.

  • \$\begingroup\$ Do I think correct that placing the diode on U$2's collector would also solve the problem or would it trap the voltage between the collector and diode (when the transistor is off), then it'd damage or wear the transistor?(then I guess I'd need a resistor in parallel to diode, right?) \$\endgroup\$ – johan elm Aug 24 '18 at 14:09
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    \$\begingroup\$ I'm not sure why you need a diode at all? \$\endgroup\$ – Andy aka Aug 24 '18 at 14:17

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