How to make the output of a ring oscillator look more like a rectangular wave?

Question

How to make the output of a ring oscillator look more like a rectangular wave?

I have designed a differential ring oscillator whose output takes the following form:

The output of a ring oscillator should be in a rectangular shape. But the output of my ring oscillator resembles no rectangular shape. How can I correct this problem?

My design is a three-stage ring oscillator:

Each stage is a differential inverter:

I have tried the following tactics to make the output waveform looks more like a rectangular wave:

1. Reducing the output capacitor
2. increasing the source current in the differential inverters
3. Increase the width of the transistors in the differential inverters

But all the above methods cannot yield the rectangular waveform. What other techniques I can use?

Thanks a lot.

Answer 1

Easy peasy lemon squeezy. Just add a comparator that biases to the midpoint of the signal. They are basically op amps that run pole to pole. Here is the example circuit on the output.

^{simulate this circuit – Schematic created using CircuitLab}

Note that R1 and C1 should adjust according to how often, if at all, the DC bias of your signal changes. If however your signal is AC coupled and has no DC bias of any kind then you can eliminate R1 and C1 and put ground directly to the positive of the comparator.

Also keep in mind the comparator will output its square wave rail to rail, so you either need to use positive and negative poles to get a ac-coupled output or you need to put an appropriate capacitor on the output to reject the DC bias if that is what you want.

Here is the output from the simulation of the above circuit showing the conversion of the input sine wave into a square wave.

Answer 2

If you want to preserve the phaseshifts of each stage, then load EACH stage with the identical and additional tap-off circuit.

Avoid positive feedback in the first stage of the tap-off circuit, because any snappy/abrupt edges will inject charges onto the RingOscillator capacitors.

Have that tap-off be an over-driven differential stage, with no feedback. Then the downstream circuit in each tap-off can be a positive-feedback circuit to provide the 50 picosecond edges you want.

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regarding need for square edges: 10pF and 10picoSeconda at 3.3 volts will require

• I = C * dV/dT = 10pF * 3.3v/10pS ==3.3 amps