Electrical – Schmitt Trigger not reaching full saturation

Design will generate dual frequency signal, but due to asymmetric nature it will change duty cycle more and it would be hard to obtain high hi/lo frequency ratio (precise parts needed).

There are other ways.

1. Use additional capacitor switched by mosfet. It may me parallel or serial to main (parallel shown).

^{simulate this circuit – Schematic created using CircuitLab} 2. Use alternating oscillator topology.

^{simulate this circuit}

Here on schematic second 'slave' oscillator work in two modes: when 'master' is high it is two invertor oscillator (NAND is working as second invertor); when 'master' is low NAND gate is blocked and is is acting as single schmitt oscillator.

Second schematic was discovered while playing with CD4093 quad NAND schmitt. Drawback of it that its frequency ratio hardcoded by schmitt thresholds.

When you solve positive feedback circuits like this, you need some initial values.

We can say that \$V_{sat+}\$ as the upper limit to what the opamp can drive to and \$V_{sat-}\$ as the lower limit.

If we make an initial assumption that \$V_{out} = V_{sat+}\$ then you will get $$ V_+ = V_{sat+}\dfrac {R_1}{R_1+R_2} $$ $$ V_{out} = A_v(\dfrac{R_1}{R_1+R_2} V_{sat+} − V_{in})$$

When \$V_{in} < \dfrac{R_1}{R_1+R_2} V_{sat+}\$ the output will be \$V_{sat+}\$ When \$V_{in} > \dfrac{R_1}{R_1+R_2} V_{sat+}\$ the output will be \$V_{sat-}\$

You would do the exact same procedure with an initial assumption that \$V_{out} = V_{sat-}\$ to see when you would see a transition going in the opposite direction.

Check out page 7 of Opamp circuits - Comparitors and Positive Feedback