[time-nuts] Squaring Tbolt 10Mhz output

Bruce Griffiths bruce.griffiths at xtra.co.nz
Thu Mar 24 09:22:22 UTC 2011


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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