[sdiy] Newbie question regarding the 4069 VCO

[René Schmitz]

Hi James,

> I think I understand this.  I remember reading that a simple current  
> source is just a voltage and a resistor as long as the input voltage is  
> substantially larger than the load voltage.  I guess this holds true  
> because the voltage at the transitor is quite small relative to the  
> control voltages?

Exactly.

>> 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.  I think I'm hung up on how an 
> invertor  works and I'm not taking into account other things, but here 
> is what I  don't get.  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'.  I think of 
> 'low' as 0v and 'high'  as +V where V is the value of the positive rail 
> (I think I'm using the  correct terminology).  If this is the case, how 
> does this integrator  circuit produce a voltage ramp?

Well what you say is right of course, thats the way an invertor is 
usually used. But, actually it is just an inverting amplifier, which is 
just usually used overdriven, so you don't notice that its infact an 
amplifier. This also explains why there is a "forbidden zone" for 
digital operation. But inside this region, the invertor behaves as a 
more or less linear normal inverting amplifier. And thus it is acting 
like an opamp in this circuit. This works because the 4069 has only a 
single gain stage, whereas this wouldn't work with different invertors 
like the ones from the TTL series.

Imagine an ideal opamp, without any feedback network, and its + terminal 
tied to GND. When you apply any positive voltage, its output would snap 
to the negative supply, and if you apply a positive one, it snaps to the 
negative supply. That would be an "opamp invertor". Actually there is a 
region as well where the opamp would behave as a linear amplifier, when 
you apply tiny voltages (in the uV range) you would see them amplified 
at the output. (As well as any noise....)
That is the situation with a "invertor" where there is no feedback 
element. Now if we add a cap as feedback element, we get the normal 
integrator.

Ok, now that we've seen that the "invertor" can act as a linear 
amplifier, and the opamp can act as an invertor, we can do the same as 
with the opamp, by adding a feedback network. The only difference is 
that the invertor doesn't have a + input, but if having a feedback 
element, it tries to keep its input terminal more or less halfway 
between the supplies.

If you know the opamp "Golden Rules", they apply here as well. (With the 
constraint that the invertor is further away from an ideal opamp than 
real opamps are. Lower open loop gain for example.)

> I'm a little fuzzy on this operation.  When you talk about the 'diode  
> current', is this the top diode, or the one which is parallel to the  
> 680ohm resistor?  

Yes, thats the top diode.

> Is the summing node the capacitor/diode/resistor 
> portion  of the circuit?

Yes.

> I need to ponder this one some more and actually try calculating some  
> voltages.

See a textbook on schmitt triggers, once you understand how they work 
with comparators or opamps, you'll find this fairly easy.

Cheers,
  René

-- 
uzs159 at uni-bonn.de
http://www.uni-bonn.de/~uzs159