# Discrete Schmitt Trigger: make output logic level at 0V, not 3V

I designed a Schmitt Trigger with discrete components (using an emitter-coupled pair) that operates with a Low Disable Trigger of around 3.5V and a High Enable Trigger of around 4V on circuit simulation. The circuit simulation uses a sine wave function generator to simulate a changing analog waveform.

The desired output of the Schmitt Trigger for high and low are 5V and 0V respectively, but this Schmitt Trigger outputs a logic high of 5V and a logic low of 3V, which is not suitable for the rest of my circuit, as they require 5V-0V logic levels to work.

The question is how can I make the Schmitt Trigger output a logic level of 5V-0V from a logic level of 5V-3V? What circuit can I "bolt on" to this design to make it output that logic level, or even better, an entirely different circuit altogether that achieves similar functionality (a discrete-component comparator that can achieve a 5V-0V logic level is also cool with me).

Keep in mind that I cannot use any ICs due to project requirements, and I also cannot use any other power supply other than 5VDC.

The basic problem is how the positive feedback in your Schmitt trigger robs low output voltage. The feedback voltage is generated across RE, which is always added to the low level output voltage.

A Schmitt trigger is basically a amplifier with a little bit of positive DC feedback. Can you think of other ways of using two transistors to make positive gain? Perhaps some of those configurations allow for a feedback path that does not limit the output voltage swing.

Hint 1: Your exising amplifier topology can be used with a different feedback path.

Hint 2: The two transistors can be a mix of NPN and PNP. That may open your thinking to more amplifier topologies.

Logic levels for 5v CMOS are not necessarily 0v and 5v; 1v-3.5v will usually work fine. You can however get 0v-4.5v with two extra transistors and resistors.

In theory, yes, you could use a negative supply.

Or, since the specs for TTL allow a 400mV output voltage with full noise margin you could use a few hundred mV of that up for the hysteresis and divide the input voltage down and probably get to your goal, but that's a bit ugly.

Even unloaded you can't get to 5V/0V output swing because a transistor will have some 10's of mV across it when saturated, so you need some slightly more relaxed requirements. You can get close to 0 and 4.9x V (unloaded) with a fairly simple addition on the output side.

You could use a negative supply (about -3V) to offset the low output level downwards by 3V, but this will alter the switching thresholds, and by readjusting all the resistances to restore them, you also change the low output level. It's a balancing act.

Much easier is just to add a third transistor stage (blue box below) to obtain 0V and +5V from +5V and +3V inputs:

simulate this circuit – Schematic created using CircuitLab

However, that stage inverts, so now you have an inverting schmitt trigger:

If you require non-inversion, another inverting stage will correct it:

simulate this circuit

RE (plus transistor base) sets the lower trigger point. Reduce this to ~56R for (falling) triggering at ~3.5V with ~0.3V low output.

Adding a small resistor between T1 collector and T2 base allows fine control of the hysteresis; 100R will give (rising) triggering at ~4.5V with 5V output.

R1, RC1 and RC2 can be increased quite a lot to reduce power wasted, but will interact a little with the triggers - simulate and hand tune.

• Note that this thread died almost 8 years ago. Commented May 4, 2023 at 2:44

Try shorting/removing RE and tie the emitters to the negative of the supply (ground). In order to get 0v on the low side, the emitter of T2 must be tied to 0v. Any other components will affect that low side output voltage.

The same goes for T1; if its emitter isn't connected to 0v, then it may not fully turn off T1 when it's in its on state.