Dr Bruce Griffiths wrote:
> Dr Bruce Griffiths wrote:
>   
>> Jason Rabel wrote:
>>   
>>     
>>> What voltage goes to the crystal? I may have had my scope scale too high and
>>> just didn't notice, but at first glance I don't think it was getting power.
>>>
>>> Jason
>>>  
>>>
>>> -----Original Message-----
>>>
>>> Yes, the black wires are the thermistor.
>>>
>>> The crystal has three wires.  Two are actually connected
>>> to the crystal electrodes.  The third is the case.  The
>>> case is fastened to the casting with a screw.  The only
>>> way the casting gets an electrical ground is by way of the
>>> crystal ground screw and ground wire to the board.
>>> The wires are welded to the crystal, so if you want to
>>> disconnect one of them, you have to unsolder it from the
>>> PC board.
>>>
>>> Rick Karlquist N6RK
>>>
>>>
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>>>
>>>   
>>>     
>>>       
>> Jason
>>
>> There should be no dc (with respect to ground) on any of  the crystal leads.
>> The ac voltage developed across the crystal is determined by the 
>> parallel combination of the effective crystal inductance at the 
>> operating frequency and the crystal ESR plus the impedance of the 
>> trimmer network. Since the crystal current is 1mA rms (see HP journal 
>> March 1981 p24) and the crystal ESR is around 50 ohms the AC voltage 
>> across R4 in the schematic will only be about 50 mv rms.
>>
>> Bruce
>>
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>>   
>>     
> Jason
>
> Since the oscillator circuit is checked by substituting a 10uH inductor 
> for the crystal then the crystal impedance must be around
> 50 + j628 ohms at the circuit resonance.
> The trimmer network has an impedance of -j568 ohms and the common base 
> transistor has an input impedance of about 8 ohms plus an inductance of 
> a few nanohenries.
> with a 1mA crystal current the voltage across R4 is consequently about 
> 80mV rms.
>
> Bruce
>
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>   
Jason

Table 8-4 p61 in the service manual is probably a more accurate guide to 
the AC and dc voltages expected at the emitter base and collector of 
various transistors in the oscillator circuit.

Bruce

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