[time-nuts] Squaring Tbolt 10Mhz output

James Fournier james at jfits.ca
Thu Mar 24 12:27:58 UTC 2011


My intention is to divide the signal by 10 and feed it as  an external
frequency reference into my old HP counter. Hopefully this will increase
it's stability.

As for the circuits i have tried, there  have been so many. Most of them are
variations of each other as i experimented on a breadboard. However, a few
examples are the inverter and diff. amp. circuits from the wenzel site. The
inverter (4049) produced a small .1vpp sine wave. The amp produced another
sine wave of of basically the same magnitude as the input. I also replaced
the inverter with a buffer (4050) and Schmidt trigger buffer. The buffer
produced the same result as the inverter and the Schmidt produced no output.
I tried some small signal diodes, can't remember the #, to try and rectify
the signal and just got a high output. I tried a comparator LM339 (i think)
and i got no response from the output. I tried everything with and without
an input  capacitor (.1uf) and retried most of the experiments with a 10k
pot between 5v and ground to replace the biasing resistors to allow a finer
adjustment of the input.

I have a feeling my problem is two fold: small signal with the forward
voltage drop of many of the devices i have tried and the speed of the
signal. I'm not sure everything can handle the 10Mhz signal.

On Thu, Mar 24, 2011 at 7:42 AM, Bob Camp <lists at rtty.us> wrote:

> Hi
>
> What is the resulting square wave going to be used for?
>
> A simple biased ACMOS gate is adequate for a lot of applications.  A 0.1 uf
> cap to couple the signal to the input. A 120K to B+ and a 100K to ground for
> bias on the same input. Square wave comes out the other side.  One usually
> terminates the line with 50 ohms ahead of the blocking cap. If the rest of
> your hex inverter is used for other things in the circuit, it's definitely
> the bang for the buck champion.
>
> That said, it's not the phase noise champion, or the highest dynamic range
> circuit in the group. Which brings us back to - what are you using it for?
>
> Bob
>
>
> On Mar 24, 2011, at 5:22 AM, Bruce Griffiths wrote:
>
> > The attached circuit uses lower capacitance Schottky diodes than the
> BAT45 to reduce the capacitive feedthrough so that a much smaller value
> compensation capacitor can be used.
> > It also draws a relatively constant current from the supply and the
> capacitive coupling between the diodes ensures that the effect of transistor
> and diode mismatch has little effect on the switching thresholds.
> > Faster switching will occur if the pnp transistor (Q2, Q3) emitter
> current has a minimum value of a few mA whilst the diode current actually
> goes to zero however this requires a negative supply to ensure that the
> output signal actually switches to ground. Additional unswitched current
> sources for the pnp transistor emitters (Q2, Q3) are also required.
> >
> > The Wenzel circuits lacking the constant current sources have a
> significant pulsed current flowing in the supply bypass system.
> > This can be reduced by adding an inductor in series with the emitter
> resistor, however this has the drawback that the value of the emitter
> resistor required depends on the input signal amplitude.
> >
> >
> > Bruce
> >
> > Charles P. Steinmetz wrote:
> >>
> >>> One problem that is evident when a simple longtailed pair (differential
> amplifier) is used to convert a sine wave to a square wave is the tilt that
> is evident in the waveform when the output transistor is conducting. This is
> due to feedthrough from the input signal via the emitter base capacitance of
> the input transistor to the emitter of the output transistor.
> >>> The attached circuit schematic illustrates one classical method of
> minimising this tilt.
> >>> Compensation isn't perfect due to the voltage dependence of the emitter
> base capacitance but the tilt can be significantly reduced,
> >>
> >> I have used the attached circuit, which is a bit simpler, to the same
> end.  For the reason you stated, the compensation is not perfect, but it is
> surprisingly good.  The compensation slows the rise and fall times by about
> 1 nS, from about 7.5 nS to about 8.5 nS.
> >>
> >> This circuit produces 5 Vpp output -- for 3.3 Vpp output, using a 121
> ohm tail resistor should work.
> >>
> >> Best regards,
> >>
> >> Charles
> >>
> >>
> >>
> >>
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-- 
Best Regards,

James Fournier
I.T. Services
6a Ave. de Lourdes
Pointe-Claire, QC
H9S-4R2

514-562-0645
james at jfits.ca
www.jfits.ca


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