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I am reading "Optoelectronic Sensors" by Didier Decoster and Joseph Harari.

At section 1.8.2 they start explaining the relationship between rise time and bandwidth, and they give a very weird formula that I don't understand:

Image showing section 1.8.2 of the referenced book

That's the whole thing, they don't explain anything more. Then they move on to some very interesting things about noise, but they mention BP in there and I don't understand what it is.

Now, my understanding about bandwidth and rise time came from this website, specifically equation 18 that basically says:

$$ t_r = \frac{0.35}{B} $$

where \$t_r\$ is rise time and \$B\$ is bandwidth.

With that in mind, here is what I don't understand about the equation 1.7 in "Optoelectronic Sensors":

  1. What is BP and how is it different from BW?

  2. Is \$\tau\$ (tau) the rise/fall time? Or is it something else?

  3. What are \$\tau_m\$ (tau_m) and \$\tau_d\$ (tau_d)? They didn't mention them anywhere before.

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The equations shown are all derived from the same formula which you already gave:

$$t_r = \frac{0.35}{B}$$

A single-pole system has the frequency response

$$H(s) = \frac{A}{1 + \tau\cdot s}$$

This sytem has a pole at

$$p_d = \frac{1}{\tau} \Rightarrow BW = \frac{1}{2\pi\cdot \tau}$$

The transient step response of this system can be calculated as

$$h_{out} = A\left(1 - e^{-\frac{t}{\tau}}\right)$$

From which you can calculate a timepoint for each output value:

$$t = -\tau\cdot\ln\left(1 - \frac{h_{out}}{A}\right)$$

If you'd rather use the rise/fall time instead of \$\tau\$, you can then easily calculate that

$$\tau_r = -\tau\cdot\ln(1-0.9) + \tau\cdot\ln(1-0.1) = \tau \ln(9) \approx 2.1972\cdot \tau$$

Hence

$$BW = \frac{1}{2\pi\cdot \tau} = \frac{\ln(9)}{2\pi\cdot \tau_r} = \frac{\ln(9)}{2\pi\cdot \tau_f}$$

The abbreviations are the ones used in French. "Bande Passante" (BP) means bandwidth, "Monter" (\$\tau_m\$) means to rise, "Descendre" (\$\tau_d\$) means to fall.

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For a first order system, the bandwidth is equal to the cut-off frequency: $$ B = f_c $$ The rise time \$ \tau_r\$ (respecting the symbols used on book): $$ \tau_r \approx \frac{0.35}{B}$$ The system time constant is \$ \tau\$, or: $$ \tau = \frac{1}{\omega_c} = \frac{1}{2 \pi f_c} = \frac{1}{2 \pi B} $$ Or: $$ B = \frac{1}{2 \pi \tau} $$ Replacing \$ B \$ in expression for rise time \$\tau_r \$ (similar for \$ \tau_f\$): $$ \tau_r \approx \frac{0.35}{\frac{1}{2 \pi \tau}} $$ Also: $$ \tau_r \approx 2.2\tau $$

In terms of \$ B \$:

$$ B = \frac{2.2}{2 \pi \tau_r} = \frac{2.2}{2 \pi \tau_f} $$

My guess is that the text seems to confuse some terms (see the last sentence, for example). If this is correct you can have \$ B = BP \$, \$ \tau_r = \tau_m \$ and \$ \tau_f = \tau_d \$.

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  • \$\begingroup\$ @Fernando Franco Félix: I didn't describe how the equation tr = 0.35/B was obtained, since would be unnecessary (it had already been done in the link you provided on question). \$\endgroup\$ – Dirceu Rodrigues Jr Feb 1 at 16:40
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Consider the basic photo-detector (photo-diode) below. Equivalent circuit of a photodiode

Image credit: Wikipedia

Now this equivalent circuit can be simplified into:

schematic

simulate this circuit – Schematic created using CircuitLab

This can be equivalently modeled again as:

schematic

simulate this circuit

Where, \begin{equation} I_1 = I_D \pm I_{PH} \pm I_R \end{equation} \begin{equation} V_1 = I_1 \times R \end{equation} \begin{equation} R = R_P || (R_S+R_L) \end{equation} \begin{equation} C_1 = C_S \end{equation} Now, applying the mesh equation (Kirchoff's Voltage Law) in the second image, \begin{equation} V_1 = I_1 R + \frac{1}{C} \int I_1(t) \text{dt} \end{equation} where, \$ V_0 \$ is the output voltage taken across the capacitor. Taking Laplace transform on both sides, \begin{equation} V_1 (s) = I_1 (s)R + \frac{I_1(s)}{Cs} \end{equation} Assuming the capacitor in uncharged initially, \begin{equation} \frac{V_1(S)}{V_o(s)} = RCs + 1 \end{equation} Thus, the transfer function is given by \begin{equation} G(s) = \frac{1}{1+RCs} = \frac{V_0(s)}{V_1(s)} \end{equation} Assuming that the input light intensity is constant and thus giving a constant current and hence giving a equivalent constant modeled voltage \$ V_1(s) = \frac{V}{s} \$, we get, \begin{equation} V_0(s) = \frac{V}{s(1+RCs)} \end{equation} Taking inverse Laplace transform, we get \$ V_0(t) = V\big(1-e^{-\frac{t}{RC}}\big) \$ Or equivalently, \begin{equation} V_0(t) = V\Bigg(1-e^{-\frac{(R_S+R_P+R_L)t}{R_P(R_S+R_L)C_S}}\Bigg) \end{equation} Now rise time is defined as the time taken for the output to reach \$ 90%\$ of the output from \$ 10%\$ of its value. That is, difference between time when \$ V_0(t) = 0.9 V_1(t) \$ and time when \$ V_0(t) = 0.1 V_1(t) \$ Thus, \begin{equation} 0.9 = 1-e^{-\frac{t_1}{RC}} \end{equation} \begin{equation} 0.1 = 1-e^{-\frac{t_2}{RC}} \end{equation} \begin{equation} t_1 = -\ln(0.1)\tau \end{equation} \begin{equation} t_2 = -\ln(0.9)\tau \end{equation} The rise time is \begin{equation} t_r = t_1-t_2 = \ln(9) \tau \end{equation} where, \$\tau = RC\$. Since, this time constant \$ \tau = \frac{1}{2\pi\times\text{Bandwidth}} \$, \begin{equation} t_r = \frac{\ln(9)}{2\pi\times BW} = \frac{0.3496991526}{BW} \approx \frac{0.35}{BW} \end{equation}

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  • \$\begingroup\$ what is \$R_P\$ in the equivalent circuit? \$\endgroup\$ – Fernando Franco Félix Feb 5 at 14:24
  • \$\begingroup\$ As you see the construction of the photodiode as considered here, it has a p-n junction. Intrinsically, there is a depletion layer that has several components which are parasitic capacitance, series and parallel resistances namely \$R_S\$ and \$R_P\$ , with dark current \$I_D\$, noise current \$I_R\$ and photocurrent \$I_{PH}\$ and diode capacitance \$C_S\$. \$R_L\$ is the load resistance. \$\endgroup\$ – John Brookfields Feb 5 at 14:43
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Tektronix Corporation suggested their scopes, perhaps rated at 10MHz, with a number of internal poles establishing that 10MHz bandwidth, would have a risetime of 0.35/10MHz.

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  • \$\begingroup\$ Using the equation firmly establishes that the response is a single-pole filter, not "a number of internal poles" (unless the number is one). \$\endgroup\$ – WhatRoughBeast Feb 1 at 19:36
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  1. What is BP and how is it different from BW?

Probably means "Bandwidth of Photodetector", regardless the author wrote poorly and did not define the term. They also wrote a bad equation and didn't explain where they got the rise time factor. If you compare the detective equation 1.11 to the equation for detectivity a better definition would be \$\frac{1}{2\pi\tau}=BP=\Delta f\$

enter image description here
source: https://www.picotech.com/library/oscilloscopes/rise-time

Is tau the rise/fall time?, or is it something else?

Tau is the time it takes to get from 10% to 90%, or the rise/fall time.

\$\log(0.90)-\log(0.10)=2.197\$ or 2.2

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