# Quasi-peak detector

This is the circuit of a quasi-peak detector:

simulate this circuit – Schematic created using CircuitLab

where $V_{\text{sig}}$ is the input signal.

When the diode is open-circuited, the capacitor discharges over $R_2$ with a time constant $\tau_D = R_2 C$.

When the diode is forward-biased, the capacitor charges with a different time constant. Textbook and EMC papers about quasi-peak detectors say that this time constant is $\tau_C = R_1 C$. This is an example (page 4: "EMI-Receiver charges the capacitor $C$ by the resistor $R_1$").

But as suggested in the answer to this question, we can easily compute the Thèvenin equivalent of $V_{\text{sig}}$, $R_1$ and $R_2$ for the charge process. The result is that the capacitor charges with a time constant

$$\tau_C = (R_1 || R_2) C = \frac{R_1 R_2}{R_1 + R_2} C$$

which is so different from $R_1 C$.

Is this wrong in some way?

• The only way it's different is that it accounts for R2 discharging C at the same time. – Ignacio Vazquez-Abrams Jun 24 '15 at 16:22
• @IgnacioVazquez-Abrams yes, exactly. So, why in the papers this is not considered? $R_2$ is not a so big resistance: it is comparable to $R_1$. – BowPark Jun 24 '15 at 16:44
• I don't see where the linked paper defines $\tau_C$ as $R_1 C$; perhaps $R_1$ is meant to be an abstraction of the actual computed resistance. – Ignacio Vazquez-Abrams Jun 24 '15 at 16:54
• Almost certainly, the paper assumes $R_2 >> R_1$ – WhatRoughBeast Jun 24 '15 at 17:32
• @IgnacioVazquez-Abrams it says that in the quote I wrote in the question: "EMI-Receiver charges the capacitor $C$ by the resistor $R_1$". – BowPark Jun 24 '15 at 18:05

## 2 Answers

This is interesting. Remember that in addition to the equivalent resistance change, the source becomes $$\\V_{thev} = V_{\text{sig}}\frac{R_2}{R_1 + R_2}$$ when you find the Thevenin equivalent. Then, when you perform some calculations using the time constant $$\tau_{thev} = \frac{R_1R_2}{R_1 + R_2}C = \tau_{paper}\frac{R_2}{R_1 + R_2}$$ it is similar to using simply $V_{\text{sig}}$ as the source with time constant $\tau_{paper} = R_1C$ , since the $R_2/(R_1 + R_2)$is common, and also because for $R_2>>R_1$, that factor is approximately 1. The scaling of the "equivalent" voltage equals the scaling of the "equivalent" resistance.

$R_2$ will limit the voltage across the cap to $V_{thev}$, since there will be a steady-state DC current through it. The paper's model is pretty similar, but not exactly the same. I'm not sure what I'm adding, but this can give you an idea of the difference between actual and the paper's model

$$V_C(t)=V_{thev}(1-e^{-t/\tau_{thev}})$$ (with some manipulation) $$V_C(t) =\frac{R_2}{R_1+R_2}V_{sig}(1-e^{-\frac{R_1t}{R_2\tau_{paper}}}e^{-\frac{t}{\tau_{paper}}})$$ The difference is the $R_2/(R_1+R_2)$ out front and the $e^{-\frac{R_1t}{R_2\tau_{paper}}}$. The approximation is very accurate as both factors approach 1, which happens when $R_2$ is large and\or $R_1$ is small. Imagine what happens when $R_1=R_2$. The factor out front becomes half, but the effective time constant is also cut in half. The result is the approximation seems like it could still be close without the condition $R_2>>R_1$, as long as you don't approach steady-state, since the half-as-small steady-state voltage might balance with the half-as-long time constant. But to be sure, we should check the derivative

$$\frac{dV_C}{dt}(t)=\frac{V_{thev}}{\tau_{thev}}e^{-t/\tau_{thev}}=\frac{V_{sig}}{\tau_{paper}}e^{-\frac{t}{\tau_{paper}}}e^{-\frac{R_1t}{R_2\tau_{paper}}}$$ Its value is very similar to the approximation, and notice how the resistor values divide out of the equation, except for the factor of $e^{-\frac{R_1t}{R_2\tau_{paper}}}$. Yet despite the first factor dividing out, the function depends on $R_2>>R_1$ in order to be a good approximation.

You are right in that the paper is incorrect. However I can see what their thinking was - if the value of $R_2$ is much larger than $R_1$ then your equation degenerates to $R_1C$. I think that is the assumption that have made although they haven't said so.

In the paper they give the time constants and the discharge time constant is at least 11 times the charging time constant and in the BAND C/D case is over 500 times longer. This supports the conjecture that $R_2 >> R_1$.