Inverting Schmitt trigger design with subtle changes to the hysteresis loop and tripping voltages

Question

Currently, I'm trying to analyze a Schmitt trigger circuit using the 741 IC op-amp, and how it overcomes the problems faced by an open loop comparator circuit.

This is a typical positive feedback induced inverting Schmitt trigger:

The design specifications are as follows -

1. \$ V_{CC} = +-8V \$
2. \$ V_{threshold} = 2V \$
3. Noise band tolerance \$= +- 0.3V \$

In this case, \$ V_{out} = +- V_{sat}\$ , and \$ V_{sat} = +- V_{CC} \$.

The noise tolerance refers to the hysteresis band, in my opinion, and that of width 0.6V. With the equations relating the design of the Schmitt trigger and its resistors R1,R2 , I can find R1 and R2 with the help of the lower and upper tripping voltages \$ V_{LTP} = -V_{sat}\frac{R2}{R1+R2}\$ and similarly for \$V_{UTP}\$, but all I've read about hysteresis loops and tripping voltages are symmetrically situated about 0V. Here, the threshold voltage is 2V, so as a result, \$V_{UTP} = 1.7V\$ and \$V_{LTP} = 2.3V\$.

I don't know how to set the reference around 2V, as I have only read about tripping voltages being of the same magnitude but different signs. Is there any possibility that I'm interpreting wrongly? Can someone help me out in this?

EDIT 1:

PS: Owing to helpful answers, I have figured out that I need to set up a reference voltage. The problem now is that I've only the dual power supply, and 3 resistors to make it happen, and I can't seem to think of any divider bias as such to set the reference voltage to 2V. It would be helpful if someone could atleast throw some hints so that I can proceed.

EDIT 2: Here are the equations -

Connecting the resistor R1 to +Vcc, R2 to ground and R3 as the feedback resistor, I get $$ \frac{V_{CC} - V_{+}}{R1} - \frac{V_{+}}{R2} - \frac{V_{out} - V_{+}}{R3} = 0$$

Therefore, I finally get - $$ V_{+} = (\frac{1}{R1+R2+R3})(\frac{V_{CC}}{R1} + \frac{V_{out}}{R3}) $$

So, to get 2.3V and 1.7 V, I can plug in the saturation values in high state and low state to get the respective resistor values. But, I don't understand how does this help me to create a reference of 2V. It would be helpful if someone could confirm if I'm proceeding in the right way.

Answer 1

I don't think it is possible to select resistor values that will achieve your desired result. You need a different circuit. Since this looks like homework I will stop with that hint.

EDIT: Yes, you need to create some kind of reference voltage. Since you asked for hints, think about voltage dividers and Thevenin equivalent circuits.

Answer 2

If you rely on the 741 datasheet for output nominal voltages and the Vcc Vee to be stable you can compute the Vref from Vcc but the accuracy is poor due to changes in Vout with saturation.

Assuming Vout = -6.9, +7.5 V for supply = +/-8V on a 741 style Op Amp.

The results may be computed or simulated as you have done but with pullup to 8V.

The noise was increased to 800 mVpp 10kHz over a 7Vpp signal with 2Vdc

Answer 3

While this may not be the best approach, what I explain here should demonstrate how you might achieve the 2V "offset" to the switching thresholds. See below for a summary.

Start with an open-loop comparator, with a switching threshold set by a resistor divider:

^{simulate this circuit – Schematic created using CircuitLab}

This isn't a schmitt trigger yet, because there's no feedback, but it does permit us to derive some voltage \$V_X\$ that we can modulate with feedback from the output of the opamp, via another resistor. Before we do that though, to facilitate the calculations of exactly what feedback resistor is needed, I recommend converting the system of R1 and R2, and the potentials -8V and +8V into its Thevenin equivalent:

^{simulate this circuit}

Thevenin resistance \$R_X\$ is easy to calculate; it's the combined resistance of R1 and R2 as if they were connected in parallel:

$$ R_X = R_1 \parallel R_2 = \frac{R_1R_2}{R_1+R_2} $$

The Thevenin voltage \$V_X\$ is:

$$ \begin{aligned} V_X &= (-8V) + (+8V - (-8V)) \frac{R_2}{R_1+R_2} \\ \\ &= 16 \frac{R_2}{R_1+R_2} - 8 \end{aligned} $$

By reducing the network down to a single resistor and potential, when we add the feedback resistor it will form another two-resistor divider, which is much easier to solve for than with superpositions or nodal analysis:

^{simulate this circuit}

Calculate the potential \$V_P\$ as a function of \$V_{OUT}\$:

$$ \begin{aligned} V_p &= V_X + (V_{OUT} - V_X) \frac{R_X}{R_X+R_3} \\ \\ \end{aligned} $$

By setting \$V_{OUT}\$ to the two extremes of opamp output potential (±8V for a rail-to-rail output device; more like ±6V for a 741), the values you obtain for \$V_P\$ are the switching thresholds, and the difference between them will be the hysteresis.

The final circuit you have is:

^{simulate this circuit}

Think of R1 and R2 as providing a potential close to the 2V threshold you require, and R3 as modulating that potential slightly, depending on the output.

Unfortunately you cannot make the assumption that R1 and R2 must provide exactly 2V, because relative to this non-zero offset, the opamp output is no longer symmetrical. A high output of +8V is only 6V higher than 2V, but a low output of -8V is 10V lower! This asymmetry renders the problem far less trivial to solve.

It may be easier just to solve for these two conditions:

^{simulate this circuit}

This of course is the same resistor configuration, but with conditions set explicitly rather than with the opamp. You could set the potential \$V_P\$ at node P to the switching threshold you desire in each case (for eample \$V_P = +2V \pm 1V\$), and find equations that yield values for R1, R2 and R3 that would produce those potentials, by whatever means you choose.