Electronotes/AN-23 - The CA3080 as a voltage-controlled resistor: Difference between revisions
Electronotes/AN-23 - The CA3080 as a voltage-controlled resistor (view source)
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[[File:AN23 fig 1a.jpg|thumb|right|250px|Fig. 1a.jpg]][[File:AN23 fig 1b.jpg|thumb|right|250px|Fig. 1b.jpg]][[File:AN23 fig 1c.jpg|thumb|right|250px|Fig. 1c.jpg]][[File:AN23 fig 1d.jpg|thumb|right|250px|Fig. 1d.jpg]][[File:AN23 fig 1e.jpg|thumb|right|250px|Fig. 1e.jpg]][[File:AN23 fig 2a.jpg|thumb|right|250px|Fig. 2a.jpg]][[File:AN23 fig 2b.jpg|thumb|right|250px|Fig. 2b.jpg]]
<math>\frac{V_{diff}}{I_{out}} = \frac{1}{(19.2 \cdot I_{c
We can thus look at the circuit of Fig.
<math>I_{out} = 19.2 \cdot I_{c} \cdot V_{diff} = \frac{I_c \cdot V_{in}}{23.7} = \frac{V_{in}}{R_{eq}}</math>
where <math>R_{eq} = \frac{23.7}{I_{c}}</math>, and where we have made use of the values shown for the [[voltage divider]] [[attenuator]]. What kind of a VCR is this? Well, we can see that the ground point receives current as though it were being supplied from a voltage <math>V_{in}</math> through a resistor <math>R_{eq}</math> is shown in Fig.
Next we would like to look to see if we can make a VCR that looks like a resistive load. That is, we want to have a VCR that actually draws different currents depending on the voltage across it - which after all is what a real resistor does. This can be implemented as shown in Fig. 2a. Ignore for the moment the upper op-amp in Fig. 2a, which is just a [[operational amplifier buffer|voltage follower]] to drive the attenuator on the input of the CA3080. The output of the CA3080 is thus
<math>I_{out} = I_{c} \cdot \frac{V_{in}}{23.7} = \frac{V_{in}}{R_{eq}}</math>
which is the same thing we had above. But here the current is drawn from the input (<math>I_{in} = I_{out}</math>) so the VCR looks like a resistor to ground, as seen in Fig. 2b. We can thus use this sort of VCR to implement the [[high pass filter|high-pass filter]] structure shown in Fig. 2c. The implementation is shown in Fig. 2d, and the 3dB frequency is <math>\frac{1}{2 \pi R_{eq}C}</math>.
<div><ul>
<li style="display: inline-block;">[[File:AN23 fig 2c.jpg|thumb|right|250px|Fig. 2c.jpg]]</li>
<li style="display: inline-block;">[[File:AN23 fig 2d.jpg|thumb|right|250px|Fig. 2d.jpg]]</li>
<li style="display: inline-block;">[[File:AN23 fig 3a.jpg|thumb|right|250px|Fig. 3a.jpg]]</li>
Since we have implemented the simple R-C high-pass filter (with an output buffer), it is of interest to ask if the corresponding R-C low-pass filter can be realized. It might at first seem that the VCR of Fig. la would be the answer, but if we look at this closely we see that it supplies a current that is indeed proportional to an input voltage, but the other end is always taken to be ground, while in the simple R-C low-pass, the voltage "x" (see Fig. 3a) is not in general zero. Thus, we need a VCR which sees two voltages, <math>V_{in}</math>, and the voltage "x" which is the same as the output voltage. The circuit of Fig. 3b is the proper realization. This is easy to show.▼
<li style="display: inline-block;">[[File:AN23 fig 3b.jpg|thumb|right|250px|Fig. 3b.jpg]]</li>
</ul></div>
▲Since we have implemented the simple R-C high-pass filter (with an output buffer), it is of interest to ask if the corresponding R-C [[low pass filter|low-pass filter]] can be realized. It might at first seem that the VCR of Fig.
<math>I_{out} = 19.2 \cdot I_{c} \cdot V_{diff} = 19.2 \cdot I_{c} (V_{+} - V_{-}) = \frac{I_{c}(V_{in} - V_{out})}{23.7} = \frac{(V_{in} - V_{out})}{R_{eq}}</math>
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<math>I_{R} = \frac{(V_{in}-x)}{R} = \frac{(V_{in}-V_{out})}{R}</math>
Thus we have implemented three types of VCR's. The first (Fig.
A more general form of a floating resistor is shown in Fig. 4, and is a circuit first suggested by G. Wilcox.
<div><ul>
[[File:AN23 fig 4.jpg|thumb|250px|Fig. 4.jpg]][[File:AN23 fig 5.jpg|thumb|250px|Fig. 5.jpg]] ▼
<li style="display: inline-block;">[[File:AN23 fig 4.jpg|thumb|250px|Fig. 4.jpg]]</li>
▲
</ul></div>
First assume that <math>I_{c}=0</math>, and thus the two CA3080's are effectively out of the circuit. The resulting resistance between <math>V_{1}</math> and <math>V_{2}</math> is just 100k + 100k + 220 which is approximately 200k. Now, if a current <math>I_{c}</math> is flowing in both CA3080's, it can be shown that the current I is given by:
<math>I = (V_{2}-V_{1})(\frac{1}{200k}+\frac{I_{c}}{47.3})</math>
Thus, as the current <math>I_{c}</math> is increased, the effective resistance goes down, starting from a value of 200k (or whatever attenuator is used). For a completely linear system, the circuit of Fig. 5 can be used to get rid of the
== References ==
*
[[Category:Electronotes Application Notes]]
[[Category:CA3080]]
[[Category:OTA]]
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