[time-nuts] Squaring Tbolt 10Mhz output

James Fournier james at jfits.ca
Thu Mar 24 13:53:42 UTC 2011


The counter is an HP 5381A 80 Mhz. According to the manual it will take up
to 2Mhz as a reference input. Just for fun, i tried feeding it the 10Mhz and
it didn't like it.

On Thu, Mar 24, 2011 at 10:31 AM, Greg Broburg <semiflex at comcast.net> wrote:

> What is the model of the old HP counter?? Lets have a look
> at the receiving end of the arrangement. In my experience
> the use of an external frequency reference with HP test
> boxes has been painless, not needing any extra circuitry
> between the reference and the input.
>
> Regards;
>
> Greg
>
>
>
>
> On 3/24/2011 6:27 AM, James Fournier wrote:
>
>> 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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>>
>
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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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