[sdiy] Newbie question regarding the 4069 VCO

[Ryan Williams]

this is very long but I want to try and answer your question completely. 
I worked on the circuit about a year ago and this is what I ended up 
with. If anyone sees a mistake please correct me although this is how it 
makes sense to me.

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.  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?

to understand this part, you need to know a little about how the 
inverting opamp works, and also what a capacitor does.  If you don't 
understand some of the opamp stuff, read some online tutorials about 
inverting opamps.

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.

so why does it ramp you ask? well this is due to a capacitor being in 
the negative feedback loop of the amplifier. it's negative feedback 
because the output which is a negative of the input is fed back into the 
input. from the basic analysis rules for an inverting opamp (see art of 
electronics or many online tutorials) we know that the negative feedback 
will cause the opamps output to be such that both input terminals are 
equal or 0V in our case. the sum of the current coming into the input 
pin of the opamp and the current going out is 0A (Iin + If = 0) where If 
would be the current coming out through the feedback loop. so If=-Iin. 
Now look at the the equations for capacitor current: I=C*dV/dt, and if 
you don't know your calculus thats change in voltage per change in time. 
so then we have If=C*dV/dt then, Iin=-C*dV/dt. solving that for dV we 
get dV=-Iin*dt/C. Since we are using a current sink to adjust the ramp 
speed of the oscillator the current at the input of the opamp will be 
negative, and we'll say Icv=-Iin where Icv increases as the input 
control voltage increases. then we have dV=Icv*dt/C. As the time 
increases the change in voltage increases. The change in time dt will be 
the time now minus the time we started which is 0 seconds so we have dt 
= t - 0. to find instantaneous voltage V, we have V = V(0) + Icv*t/C 
when the oscillator starts up, I can't say what voltage will be present 
at time equals 0 seconds, but once it gets going the schmitt trigger 
will set initial voltage V(0) to be used for each time the ramp goes 
through it's cycle. the dV in the capacitor equation tells us the 
voltage difference across the capacitor. if one side of that capacitor 
is held at 0V like in our inverting opamp circuit, the only thing that 
will change is the other side of that cap and that is our opamp's output 
voltage.  so as time increases, the output voltage will do the same. 
eventually it will hit the trigger point of the schmitt trigger and that 
will reverse the process by changing the direction of the input current. 
This is like resetting time to 0 and then ramping in the other 
direction. To make the output look like a ramp waveform the resetting 
current has to be much larger than the current sink's current, if the 
reset current was twice the control current we'd have a triangle 
waveform instead.

If you already knew the circuit was an integrator, then you could think 
of it like that. In the standard inverting integrator opamp circuit you 
have a voltage input into the inverting opamp input through a resistor 
and a capacitor in the feedback loop. since the two input pins have 
equal voltage like before, the current through the resistor is 
(Vinput-0V)/R. So we are integrating the input with respect to time. 
with an actual integrator Voutput=integral(Vinput*dt) which is like this 
Voutput=Vinput*t+Vinitial. But our circuit doesn't have a perfect 
integrator and capacitor and resistor must come into play as well as the 
inverting characteristic... so it's really the negative integral 
multiplied by some constant and thats Voutput=-(Vinput/R)*t*C+V(0). if 
you take into account that the opamp's input resistor is a voltage to 
current converter then you'll see that Vinput/R = Iin from our first 
example and they are just the same. This really isn't helpful unless you 
already knew a bit about integrators and to come up with the equations 
you'll go through the same steps as before, but I think it's nice to see 
that it's really an integrator.


>> The specialty here is the way how the integrator gets reset. Normally  
>> you would have a switch (BJT FET or the like) to short the cap  
>> terminals, and discharging it. Here we push current into the invertor  
>> summing node, forcing the integrator to work into the other direction.
>> The diode current and the expo current have opposite direction.
> 
> 
> 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?  Is the summing node the capacitor/diode/resistor 
> portion  of the circuit?
> 

thats the top diode. when the schmitt trigger output changes from 0V to 
Vcc, the diode turns on.

hope that helps,

ryan