7
\$\begingroup\$

I'm trying to understand this circuit and having trouble figuring out how the diodes will affect it:

enter image description here

I'm trying analyze this and understand how \$V_{\text{out}+}\$ and \$V_{\text{out}-}\$ will be affected by different values of \$V_i\$ but I am unsure. I believe \$V_{\text{out}+}\$ will see a higher voltage with lower \$V_i\$ values and once \$V_{\text{out}+}\$ is past 0.6V the current will go through both of the diodes and \$V_{\text{out}-}\$ will begin to see an increase. I'm unsure of what to think about the op amp in the center though.

\$\endgroup\$
  • \$\begingroup\$ Start using "ideal" diodes with Von=0. When Vin is +ve, VOA-out = -ve. Top diode conducts with 0 drop so you get an inverting amploifier with Vout = Vin x R/R1 as OA- must be forced to remain at )A+ potential = ground. | Now perform the same analysis for negative input. \$\endgroup\$ – Russell McMahon Nov 5 '15 at 7:56
8
\$\begingroup\$

The OpAmp is used in an inverting configuration, i.e. when you apply a positive Vi, the output Vo will be negative. So, the current flows through R3 and the upper path (R1, D1), while current through the lower path is blocked by D2.
So, for a positive input, Vo+ will be negative and Vo- will be zero.
The exact value on Vo+ is determined by the ratio of R3 and R1, as the output terminal Vo will be a value which cancels out the voltage at the negative input terminal.

To examine this a bit more, you can do a simulation, even here. I've drawn your circuit using the circuit designer of EE. Click on simulate this circuit and do a DC sweep for Vi.

schematic

simulate this circuit – Schematic created using CircuitLab

The following picture shows the result, blue is Vo+, orange Vo-. I've also added Vo (brown) just to see the effect of the diodes. (Click the image to enlarge) enter image description here

\$\endgroup\$
2
\$\begingroup\$

Andy's answer is spot-on if this circuit operates in isolation.

But if there happens to be a positive bias current source connected to Vo+, and a negative current source connected to Vo-, then it is something different. (For simplicity, these bias current sources can be resistors connected to +V and -V respectively.)

Then the outputs form three parallel lines, about 0.6V apart, with the central one (Vo = -Vi).

Vo+ and Vo- may be used to drive the bases of two complementary emitter followers, providing a crude power amplifier. Their emitters are connected together, and each base-emitter voltage is cancelled by the voltage across each diode. Therefore the final output (at the emitters) is approximately Vo = -Vi. But only approximately : errors here lead to crossover distortion.

schematic

simulate this circuit – Schematic created using CircuitLab

In its current form, it's not very efficient : if you think about what happens when VO+ = V+/2 you'll see that it runs out of bias current and starts clipping (with the current resistor values), but this illustrates the basic principle, and you'll often see something like this in an audio power amplifier.

Further circuitry is usually added to fix the defects in this basic configuration.

\$\endgroup\$
1
\$\begingroup\$

Compare this to the standard "ideal diode" op amp design.

If Vi is higher than A1+ (GND), the feedback loop will try to null the input voltage, driving the V+out line through the diode to "pull" A1- down.

If Vi is lower, the V-out line wil similarily be pulled up.

This circuit then behaves as a standard inverting amplifier, but with any negative output being delivered to the V+out line (for a positive input), and any positive to the V-out(for a negative input)

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.