Signal Rectifier Circuit -- clarification questions from The Art of Electroics

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Discussion Overview

This discussion revolves around the functioning and components of a signal rectifier circuit as presented in The Art of Electronics. Participants seek clarification on the roles of specific components, particularly resistors and diodes, in the context of signal processing and biasing.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the necessity of resistor R2, suggesting that the output voltage could be measured directly from the LED without it.
  • Another participant explains that in the first figure, the diode requires a minimum voltage of 0.5V to forward bias, while the second figure allows any signal above 0V to pass due to a resistor network that biases the diode.
  • A different participant states that R2 serves as the output impedance of the circuit, indicating that without it, the output voltage could not decrease.
  • One participant notes that grounding the right side of the circuit would result in a constant 0V output, hence the need for a resistor to ground, but expresses confusion about how the biasing works in the second circuit.
  • Another participant describes the current flow through R3 and D1, establishing a voltage reference based on the forward-biased diode voltage drop.
  • A participant advises keeping the two diodes thermally coupled to mitigate self-heating effects and suggests that higher frequencies or voltages may necessitate buffering the reference.
  • One participant shares their understanding of the bias effect through mathematical reasoning, explaining how the circuit allows the signal to pass if certain conditions are met.

Areas of Agreement / Disagreement

The discussion contains multiple competing views and remains unresolved regarding the necessity and function of certain components, particularly R2 and the biasing mechanism in the second circuit.

Contextual Notes

Participants express uncertainty about the exact roles of components and the implications of circuit configurations, indicating a need for further clarification on specific assumptions and definitions related to the circuit operation.

pierce15
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I am studying The Art of Electronics and am confused about this particular part about a signal rectifier. In the first picture above, the first cap and resistor function differentiate the signal, and the signal passes the diode if the voltage is > .6V. The second circuit improves this by allowing any signal of above 0 V to pass.

I am confused about two things. First, why is R2 necessary? It seems like we could just measure the voltage coming out of the LED without the resistor there... Also, I do not understand how circuit 2 makes it so any signal > 0 V will pass. Can someone explain these two things?
 
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In fig (1) the diode's anode sits at 0V DC so to forward bias it requires that the signal supply at least 0.5V.

In fig (2) the resistor network biases the diode at approx. 0.5V so then any rise/fall brought in by the coupling capacitor is superimposed on this DC bias.
 
R2 is the output impedance of the circuit. Without R2 the output voltage could never fall.
 
OK i now see that the right side of the circuit needs to be completed but it cannot just be grounded because then the output voltage would always be 0V, hence the resistor to ground. I still don't understand how the part below the R1 in circuit 2 is biasing D1, however. I understand that it would be biased if there were a .6 V battery in front of D2 in circuit 1. However I don't understand how the new setup is achieving that.
 
Current flows down through R3, then to ground through D1 which sets the anode voltage of D1 at about 0.6V.
That makes a voltage reference of one forward biassed diode voltage drop.
R1 then biasses the point in front of D2 from that 0.6V reference voltage.
 
For accuracy keep the two diodes thermally coupled. You will get some self heating due to the bias current on D1. Higher frequencies or voltages may require buffering of the reference.

BoB
 
Alright I figured it out. The only way I am able to see how the bias effect works is by writing out the math. Let V be the voltage right after the cap and I the current through the cap. Then ## I = -C \frac{d}{dt} (V - V_{in}) ##. Ignoring the load of D2 and R2 for now, the cap and R1 are in series, so the current through the resistor is the same as the current through the cap and it is ## (V - .6)/R_1##. Hence ## V = RC \frac{dV_{in}}{dt} + .6##, since ##\frac{dV}{dt}## is small by hypothesis. Adding the load, the signal ## \frac{dV_{in}}{dt} ## will pass if it is ## \ge 0##.
 
Wow thanks for this, was just wondering the exact same thing
 
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You are welcome.
That is what PF is here for.
 
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