Level Shifting – How to Shift Levels of a 3 State Pin

The circuit you propose is roughly OK. The schematic not so much.

This is a common base amplifier. One use of such things is to handle a large voltage with a small voltage. The first resistor at left (use component designators next time) is probably better omitted, depending on what you are really trying to do. Just tie the base directly to the 3.3 V supply. Put a resistor in series with the emitter to limit the current to a safe level for the digital output and the transistor.

One disadvantage of this approach is that you don't get any current gain. All the collector current goes out the input. Make sure the digital output can handle whatever current you want to sink.

There is no danger to the digital circuit driving the input as long as the transistor meets the appropriate specs. Obviously it must be able to withstand 50 V C-E when off. You don't show resistor values so I can't comment on the current requirements.

Here is the circuit I described above:

With IN fully down to 0 V, there will be about 5.5 mA thru R2. This same current, minus a little bit for the base, will also be the collector current. That is still enough to pull the bottom end of R1 as low as the transistor can, which will be a little below 3.3 V. As long as the digital output driving IN can sink 5.5 mA, this will work fine, and you get about 47 V of swing from a 3.3 V input.

From what I read (but I never used those ICs) if the open circuit voltage is 6.2V you can omit the 6V zener. In fact, the voltage at its output decreases with the load, and the maximum rating of the IC is 7V, so up to 7V it should not break. You can test it, or wait for someone more expert than me to come and say that I am completely wrong (but I hope I'm not).

As for the microcontroller and LED, your circuit is ok for the LED (it will light only when STAT is tied low), but I don't think the uC will like it. When STAT is tied to VCC the voltage can reach 6V, while usually the uC can bear Vcc+0.3V on their pins (so 5.3V if it is 5V powered or 3.6V if it is 3.3V powered). I suggest you modify your circuit in

^{simulate this circuit – Schematic created using CircuitLab}

I refer to the graph in figure 13 and 14 in the TLMS1000 datasheet. The scheme, with a VDD of 6V, provides you these data:

• STAT = VDD:
  • voltage at STAT PIN = 6V
  • voltage on the whole LED branch = 0, current in LED = 0, luminosity = 0
  • voltage on ADC is 47/57*6V = 4.95V
• STAT = GND:
  • voltage at STAT PIN = 0V
  • voltage on the whole LED branch = 6V, current in LED = 2mA, luminosity = 100% of specified luminosity
  • voltage on ADC is 0V
• STAT = HI-Z:
  • if we estimate a current around 50uA-100uA, the voltage on the LED is not in the graph (the graph goes as low as 100uA). In any case, I expect it to be roughly around 1.5V-1.6V)
  • if the voltage on the LED is 1.5V, the current in the branch is (6-1.5)/(2.2+10+47) = 75uA. The luminosity of the LED is around 2-3% of the nominal luminosity (cannot be seen)
  • voltage on ADC is 47*75uA = 3.53V

If you have a 3.3V microcontroller, to keep almost the same values you can substitute the 10k and 47k resistors with, respectively, a 27k and 33k one. This way the LED will see the same circuit, while on the ADC

• STAT = VDD: voltage on ADC is 33/60*6V = 3.3V
• STAT = GND: voltage on ADC is 0V
• STAT = HI-Z: voltage on ADC is 33*75uA = 2.48V