# Please explain this formula for the relationship between bandwidth and the temporal response of a photodetector

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:

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.

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.

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 \$$.

• @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). – Dirceu Rodrigues Jr Feb 1 '19 at 16:40

Consider the basic photo-detector (photo-diode) below.

Image credit: Wikipedia

Now this equivalent circuit can be simplified into:

simulate this circuit – Schematic created using CircuitLab

This can be equivalently modeled again as:

simulate this circuit

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

• what is $R_P$ in the equivalent circuit? – Fernando Franco Félix Feb 5 '19 at 14:24
• 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. – John Brookfields Feb 5 '19 at 14:43

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.

• Using the equation firmly establishes that the response is a single-pole filter, not "a number of internal poles" (unless the number is one). – WhatRoughBeast Feb 1 '19 at 19:36
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\$$

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