[sdiy] Newbie question regarding the 4069 VCO

[James Howe]

Thanks for the explanation.  This information, combined with the  
information I received from others (Rene, Harry) has been very helpful.   
Your explanation on why the inverter is basically an amplifier was most  
helpful.  I work with software so I tend to think in a more 'digital'  
sense and so I wasn't thinking about the process by which a 'low' voltage  
would become a 'high' voltage.  Your explanation gave me the whack in the  
head that made me understand that in order for a voltage to go from low to  
high it would have to be amplified! (duh)  The one thing that I'm still a  
little unclear on is why in the 4069 case I should be thinking in terms of  
Vcc/2.  I read a Fairchild application note on using an inverter as a  
linear amplifier last night and it said the same thing.  What I don't  
understand is why it's Vcc/2?

Your explanation on the way the reset operation works was equally  
helpful.  It hadn't fully sunk into my head that the current derived from  
the input voltages would be negative (even though the convertor was  
described as a sink) and that the Schmitt trigger would generate a current  
to offset this when it fired (even though Rene said basically the same  
thing in another e-mail)  It also explains to me why the ramp would be  
positive instead of negative.  In all the examples I had seen in various  
books, they showed a positive current going into the integrator and  
generating a negative ramp.  Of course, it makes sense that a negative  
input would generate a positive ramp.

Anyway, I now think I have a pretty good understanding of how this circuit  
works, and why.  My next step is to actually build it.

Thanks for the help.


On Tue, 09 Nov 2004 18:58:31 -0600, Ryan Williams <destrukto at cox.net>  
wrote:

> [...]
>
> James Howe wrote:
>>> The integrator behaves the same as the textbook version with an  
>>> opamp.  Just think of the invertor as an opamp running on single  
>>> supplies, with  its + input tied to Vb/2. The invertor input being the  
>>> - terminal.
>>>
>> Ok, this I don't quite understand.  [...]
>> From my reading on how an invertor operates, it takes an input   
>> voltage, and if the voltage is 'low', the output is 'high' and if the   
>> input is 'high', the output is 'low'. [...]
>> If this is the case, how does this integrator  circuit produce a  
>> voltage ramp?
>
> [...]
>
> You seem to understand what an inverter does in normal operation. If an  
> input is low then the output is high. This describe the behavior of a  
> very high gain amplifier and that is exactly what an opamp is. consider  
> an opamp without feedback. the output is the voltage difference of input  
> ((Vin+) - (Vin-)) amplified by the gain of the particular opamp. when  
> the gain is very high, this will output the maximum voltage the  
> amplifier can output when the differential input is positive, and the  
> minimum when the  input is negative. by setting the noninverting input  
> of an opamp to a constant 0V your differential input would be the  
> negative of whatever the input voltage was on the inverting input (0V -  
> (Vin-)).  this circuit would work just like an inverter. I'll just  
> continue with the noninverting input set to 0V even though when using  
> the 4069 inverter it would be Vcc/2. If Vcc/2 was tied to the  
> noninverting input of the opamp it would be a level shifted version of  
> what we had before. I'm going to continue using an opamp instead because  
> it's probably a little easier to understand at first but hopefully you  
> see that it just like the inverter.
> [...]
>
> ryan
>



-- 
James Howe

Contact: http://public.xdi.org/=James.Howe