[sdiy] Newbie question regarding the 4069 VCO

[Scott Gravenhorst]

"James Howe" <jwh at allencreek.com> wrote:
>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?

Because it is halfway between the voltages that represent the amplifier's
output extremes.  In fact, if you power a 4069 from say, -5v and +5v, this
equilibrium point is zero volts!.  (note that it may not be exactly zero
volts, just as when powered from a single supply it will not be exactly
Vcc/2, but should be very close.)  

Think about the gate as an amplifier again, but not as an integrator, rather
as an "opamp" with a feedback resistor (instead of the integrator's
capacitor) and no signal input connection at all.  What will the output
voltage be when power is applied?  Vcc/2.  [Note: This is actually more
accurately expressed as (Vdd-Vss)/2, but because Vss is often zero volts, we
say Vcc/2].  The output voltage is always the analog inversion of the input.
 If we assume that the starting condition (at power up) is that the output is
at zero volts - this means the input is also at zero volts because of the
feedback resistor, but it can't stay that way because it's an inverter, so
the output must rise (because the input is zero!) and does so until there is
an equilibrium (remember the feedback resistor supplies the output voltage
back to the input).  Just remember that in this example, the input voltage is
the same as the output voltage, a paradox for the inverter in the digital
domain, but that's not what's happening here, since we've employed the
amplifier in the analog domain.  Mathematically, what happens when you
"invert" Vcc/2?  You get exactly Vcc/2 and thus you also get stability or
equilibrium.  So no matter what the starting condition, the output will, by
presenting an inverted copy of the input, always seek Vcc/2.  This Vcc/2
point is also considered the system's "virtual ground", which in this case
isn't ground (or zero volts) at all.

Connection of an input signal through an input resistor now supplies current
into the "virtual ground" (the same way it does in an inverting opamp
configured similarly).  This current causes a voltage offset from Vcc/2 on
the amplifier's input and forces the output to respond with just enough
current through the feedback resistor to return the input to Vcc/2 (or
equilibrium).  Now we have an amplifier with it's gain set the same way we do
an opamp with feedback and input resistors!

You can look at this system (or an opamp summer) as a simple closed loop servo.

I hope this helps.


>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
>

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