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LT1712: Need help understanding and modifying this oscillator

rsjawale24

I'm in need of an analog circuit that can generate a square wave output of 20MHz. My first thought was to use a schmitt trigger inverter with a crystal, but then I thought of using a fast opamp to make it more challenging and a fun to work on project. 

I came across this circuit made using LT1712 4.5ns comparator from Linear Technologies (now ADI). The circuit below generates a square wave output with a freq. of 1MHz which is filtered to obtain a sine wave of 1MHz. It uses a series resonant crystal to generate a stable 1MHz signal as shown in Fig.1 

I modified the circuit, by replacing the crystal with a  20MHz one and R3, C1 by using another circuit given in the datasheet of LT1016 for 10MHz to 25MHz signal as shown in Fig. 2. I made a PCB to test the chip. However, my output frequency is around 10MHz with a distorted square wave output. I have tried the circuit in LTspice where it gives proper 20MHz output. 

Hence, I need help in understanding how the series resonant crystal circuit is working, why the 210 ohm resistor? and how the R3 and C1affect the output freq? As the values are both different in the two circuit diagrams. I have also posted the LTspice circuit simulation and output as well as the measured output of my PCB. 

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Fig. 1: LT1017 1MHz square wave gen

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Fig.2: LT1016 10MHz to 25 MHz generator

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DSO output

  • 0

    It looks like you have a second, lower frequency crawling through your oscilloscope trace.  Do you have 0.1uF capacitors bypassing the power pins of your active devices?  Datasheets don't mention this and the circuit tends to work in simulation, but in the real world these caps are needed.

  • 0 in reply to dang74

    No, I forgot to put bypass caps on the PCB. But I will do it tomorrow, when I redo the measurements. But I need to understand the working to re modify the circuit to get 20MHz output. 

  • +1 in reply to rsjawale24

    I am not an expert on this type of oscillator but I've seen it before.  If you look at the negative feedback portion of the circuit, it could be said that when the output is positive capacitor C2 is charged through R3.  Conversely when the output is negative, capacitor C2 discharges through R3.  Now on the positive feedback side of things I believe the trip point at the noninverting input will be different for the two main phases of the output... and while I don't know the specifics R4 plays a role in this.  I suggest that you add two more probes on your simulation circuit... one at the inverting input and the other at the noninverting input.  From there play around with the value of R4 to see its effect.

  • 0

    I am not an expert on this at all. If you remove the crystal, this circuit will oscillate based on R3 and C2. If these component values are adjusted until the frequency is approximately correct, reinserting the crystal should lock the frequency to the crystal resonance?

  • 0

    Another non-expert here :)

    "why the 210 ohm resistor"

    Series resistor is to reduce the power that the crystal is having to deal with. It may be a small piece of hard rock (quartz), but it will crack if it's driven too hard. You need to choose it according to the spec of the crystal (which will give you a max figure) and a calculation of the power that you're delivering which will depend on the output characteristics of the comparator and its output swing on one side and the bias network on the other.

  • 0

    Hi, it's unusual to use such a circuit (especially that very expensive part). Personally I would have gone for maybe a CMOS logic IC. There are a few (I am mobile, not easy to check the specific part number) logic ICs that also contain built-in dividers, they provide a great quality square wave output. Otherwise, for more fun/challenge, check out Oscillator Design and Computer Simulation by Rhea (It's possible to read it for free by virtually borrowing it on Internet Archive), one particular chapter has various example circuits, the one here would provide a signal, then you'd need to square it using (depending on output level) say a logic IC or another transistor, you'd take the output from the arrow, using a capacitor). The left and right side of this circuit is supposed to be connected together, I.e. connect that arrow to the wire labeled 'loop'.

    image

  • 0 in reply to shabaz

    Thanks! But I needed a square wave o/p with fast pulse edges. I will check the mentioned book for more circuits!

  • 0 in reply to jc2048

    Got it. It is to drive the crystal with the specific drive power, which is less than equal to 100uW as per the datasheet

  • 0 in reply to rsjawale24

    I see.. the choice of oscillator doesn't affect that, it's how you square it (i.e. buffer it). There's a complete circuit that could be set to (not xtal oscillator level of stability) 20MHz by changing one resistor value:  Building a Fast Edge Square Wave Generator 

    Or you could just use a buffer stage perhaps like described there, with a crystal oscillator of your choice provided there's sufficient drive for the logic input. All depends how fast you want the edges, that one achieved 500ps rise time which is pretty fast, but you may have a need for even higher, in which case a different circuit would be needed. What's the use-case and desired rise and fall times?

  • 0 in reply to shabaz

    Ah got it. I thought the device being used for the oscillator is important for the rise time. 500ps is more than enough for my application. I'm working on a TDR project, 500ps would suffice as I'm only detecting open circuit (distant to fault measurement) of a not so long coax cable.