Quick question for the builders...

[ldavid at lae.lad.gmeds.com]

> 
> On Tue, 9 Jan 1996 ldavid at lae.lad.gmeds.com wrote:
> 
> > Do you mean two zeners in series from the output of the opamp to gnd?
> > If so, wouldn't the total voltage drop be 18.2V when both diodes are
> > conducting; giving an output range from 0 to 18V and not -10 to 10?
> 
> Hmm...maybe I wasn't specific enough (and I'm not a fan of ascii graphics :),
> but I meant that the zeners should be connected "back-to-back", that is,
> cathodes together, one anode to the op-amp output, the other to the
> inverting input (assuming the non-inverting input is connected to gnd).
> This way the forward-biased zener will begin to conduct at 0.6V and the 
> reverse-biased zener at around 9V, giving a total treshold of 0.6 + 9 = 
> 9.6V. Because of symmetry, the negative treshold will be at -9.6V.
> 
> -joachim

O.k., I think I get it now.  You're talking about an opamp configured as a unity 
gain buffer except that the non-inverting input is tied to gnd (so the input 
voltage is 0) and there are two cathode-to-cathode zeners in the feedback loop; 
right?  I think they call two cathode-to-cathode zeners a "double-anode zener".  
After I wrote that top paragraph I found a simple limiter ckt with one of these 
double-anode zeners from the output voltage to gnd, and the new/limited output 
is taken across the zener; like this:

  Vout  o------------------o Vlim
                  |
                 \/
                 --
                  |
                 --
                 /\
                  |
                  |
                -----
                 gnd
                 
Vlim will be limited to +/-(forward thresh + reverse thresh).  The problem, I 
think, is that when Vout goes negative, whatever is "behind" it will have to 
sink current.  Hence the opamp ckt you describe; at least that's one reason for 
it.  If I understand it right, the summed cv voltage (the output voltage we want 
to limit) gets connected to the *output* of the opamp configured as described 
above.  As long as the voltage is within +/-9.6, the opamp does nothing.  If it 
goes >9.6, the inverting input sinks current; if it goes <-9.6, the output sinks 
current (assuming the output impedence of the opamp is << that of the cv 
summer).  Is this the idea?

ld
--
----------------------------------~~~~~~~*--------------~~~~~~~*    
Larry David                    "I no longer live, 
ldavid at lae.lad.gmeds.com        but Christ lives in me."