A technical guide to silicon photomultipliers (MPPC) - Section 3

We focus on studying photodetector S/N at the signal’s low end. We use the typical specs of a regular Hamamatsu MPPC like the S13360-3050 (3 × 3 mm, 50 µm pixels):

Using equation 2-2, we calculate charge output from a 10-photon input pulse: 10 × 0.4 × (1.7 × 10⁶) × (1.6 × 10^-19) = 1090 fC. The node sensitivity of a typical digitizing readout (like a QDC, which is the charge-digitizing equivalent of an oscilloscope with similar node sensitivity) is on the order of low 10s of fC/LSB (like 25 fC/LSB in the case of CAEN V965), and since 1090 fC > 25 fC, we conclude that S13360-3050 will be suitable for applications such as particle or nuclear physics in which use of flash digitizers (a QCD is simply a digitizer with on-board charge-to-voltage conversion) is prevalent; that is commonly the case if pulse shape information is required. However, such readout schemes are highly costly and power-intensive, making them unsuitable for common scientific, industrial, or consumer applications.

A cost-effective alternative readout scheme is use of photon-counting circuitry such as a multi-channel analyzer (MCA), consisting of discriminator/scaler, counter, and other signal processing functions. Towards using equation 2-8 to calculate S/N for this scheme, the desired counting integration time must be determined. For the sake of our discussion, let’s assume that is 1 ms for which S13360-3050’s DCR would yield 500 k × 1 m = 500 dark counts. Using equation 2-8, we set S/N = 1 and solve $\frac{N_{photon}\times 0.4}{\sqrt{\left(N_{photon}\times 0.4\right)+\left(500\right)}}=1$ to obtain $N_{photon}\approx 57$. At a min. pulse rate of 10 kHz, 10 photons per pulse would yield 100 photons in 1 ms. Alternatively, 100-photon S/N can be calculated using equation 2-8 to be $\frac{100\times 0.4}{\sqrt{\left(100\times 0.4\right)+\left(500\right)}}\approx 1.7$.

We repeat the same for S13360-3025 (3 × 3 mm, 25 μm pixels) with PDE @ 450 nm = 25% and typical DCR = 400 kcps to obtain an incident light level of $N_{photon}=82$ for S/N = 1 (by solving $\frac{N_{photon}\times 0.25}{\sqrt{\left(N_{photon}\times 0.25\right)+\left(400\right)}}=1$ ) or alternatively calculating S/N for 100 photons as $\frac{100\times 0.25}{\sqrt{\left(100\times 0.25\right)+\left(400\right)}}=1.2$.

It is noteworthy that “excellent” S/N is generally considered to be ≥ 10. Most instrument designers typically have a target performance of S/N > X in mind for which X is greater than 1 (even if less than 10). Thus, it is important to use the value of X that represents the instrument designer’s target S/N in the above calculations. For example, if X = 5, the above calculations would yield N_photon = 312 for S13360-3050 and N_photon = 453 for S13360-3025. These results would mean that a portion of the lower range of the expected signal levels could not be detected with the target S/N (= 5) performance in this case.

We now assess how linear a MPPC’s response would be under this application’s conditions:

- MPPC pulse-height linearity: The question we face here is: up to how many 450 nm photons would S13360-3050 or S13360-3025 be able to detect with 10% max. nonlinearity? To answer this, we first obtain the pixel capacitance⁴ by utilizing the nominal value of the MPPC’s typical gain, assuming a unit of electrons for it and converting it to charge in coulombs, and then dividing the resulting charge by the specified overvoltage corresponding to that gain value. We thus have: $\frac{\left(1.7E6 e-\right)\times \left(1.6E-19\frac{C}{e-}\right)}{3}\approx 91fF$. Using the quenching resistor value⁵ of a 50 μm MPPC pixel, we then calculate S13360-3050’s pixel recovery time to be 63 ns [approx. 4.6 × 91f × 150k where 4.6 = –ln(0.01) corresponds to a 99% MPPC recovery] and compare to the application’s light pulse width. Since the condition PW < T_recovery is met (8 ns < 63 ns), we use equation 2-13 to plot the MPPC’s expected response and compare that to its ideal response as obtained from equation 2-12 in Microsoft® Excel® and look for the point at which the 2 plots diverge by 10%. We perform the comparison by plotting the resulting nonlinearity using the combination of equation 2-10 and equation 2-11.

Now, let’s assess APD’s S/N at those signal levels above which MPPC linearity falls short of the application’s linearity requirement. We choose a blue-enhanced APD of suitable size, like Hamamatsu’s S8664-30K, and utilize its characteristics ($QE @ 450nm=75\%, I_d=1 nA, M_{opt}=50, F={50}^{0.2}\approx 2.2$) in equation 2-7 to calculate S/N. Considering that the application is photometric (i.e. measuring the amount of incident light in absolute terms for which a charge amplifier is required), we also use the readout noise spec of a sufficiently fast charge amplifier (one that can resolve a single pulse at the max. pulse rate); we note 993 e- in case of Analog Devices AD8488. Note that the charge amplifier’s bandwidth (as inverse of its integration time) must be at least twice that of the application’s max. expected pulse rate, and hence, we will use 2 MHz as the min. amplifier bandwidth limit. We proceed to calculate $S_{dark}=\frac{I_d}{q\times △f}=1n/\left(2M\times 1.6\times {10}^{-19}\right)=3125e-$. Using equation 2-7, we thus have $S/N=\frac{0.75\times 50\times 2000}{\sqrt{2.2\times \left[\left(0.75\times {50}^2\times 2000\right)+\left(50\times 3125\right)\right]+{993}^2}}\approx 24$ for 2000 photons and $S/N=\frac{0.75\times 50\times 12000}{\sqrt{2.2\times \left[\left(0.75\times {50}^2\times 12000\right)+\left(50\times 3125\right)\right]+{993}^2}}\approx 63$ for 12000 photons.    

As mentioned before, excellent S/N is typically considered to be ≥ 10. Therefore, in such a photometric application using a charge amplifier, we conclude that S8664-30K can perform well at those signal levels at which S13360-3050 and S13360-3025 exhibit excessive nonlinearity. Thus, it is imperative to use S13360-3050 (instead of S13360-3025) in photon-counting mode to detect the smaller pulses in this application but then consider using S8664-30K for pulses > 2000 ph.

As a side exercise, we calculate the APD’s S/N for making a relative measurement (requiring a resistive trans-impedance amplifier for readout) under the same conditions. First, let’s explore the case that the APD and its output amplifier circuit are intended for use as a pulse counter, which would make the signal’s highest-frequency component to be the max. expected pulse rate of 1 MHz. Since the output amplifier’s bandwidth must be at least twice that of the measurement frequency (which would be the frequency of the signal’s highest-frequency component that is to be detected), we adopt 2 MHz as our min. amplifier bandwidth limit (or cutoff frequency in other words). Using equation 2-6 along with S8664-30K’s characteristics [$\Phi @ 450 nm\approx 0.3 A/W$ [derived from QE by equation 2-5], $I_d=1 nA, M_{opt}=50, F ={50}^{0.2}\approx 2.2$] and Texas Instruments amplifier OPA380 characteristics (readout noise spec of $10\frac{fA}{\sqrt{Hz}}$ resulting in an estimated readout noise of approx. 14 pA at 2 MHz), we have:

$S/N=\frac{0.3\times 8.8\times {10}^{-16}\times 1E6\times 50}{\sqrt{2\times 1.6E-19\times 2E6\times {50}^2\times 2.2\times \left[\left(0.3\times 8.8E-16\times 1E6\right)+{10}^{-9}\right]+\left(1.96\times {10}^{-22}\right)}}~6.2$ for detecting pulses of 2000 photons or 0.88 fJ of incident 450 nm light per pulse at a rate of 1 MHz, and

$S/N=\frac{0.3\times 5.28\times {10}^{-15}\times 1E6\times 50}{\sqrt{2\times 1.6E-19\times 2E6\times {50}^2\times 2.2\times \left[\left(0.3\times 5.28E-15\times 1E6\right)+{10}^{-9}\right]+\left(1.96\times {10}^{-22}\right)}}~26$ for detecting pulses of 12000 photons or 5.28 fJ of incident 450 nm light per pulse at a rate of 1 MHz.

Now, let’s study the case that the APD and its output amplifier circuit will be used to perform pulse-shape discrimination (PSD) on signal pulses, which would require the signal’s highest-frequency component to be obtained from its rise time (since shorter than the fall time) by the following approximation $\frac{0.35}{3 ns}\approx 117MHz$. Like before, since the output amplifier’s cutoff frequency must be at least twice that of the measurement frequency, we adopt 234 MHz as our min. amplifier bandwidth limit. Using equation 2-6 along with S8664-30K’s characteristics $\left(\Phi @ 450nm\approx 0.3A/W, I_d=1 nA, M_{opt}=50,F={50}^{0.2}\approx 2.2\right)$ and Analog Devices amplifier AD8015 specifications (indicating a readout noise spec of $3\frac{pA}{\sqrt{Hz}}$ resulting in an estimated readout noise of approx. 46 nA at 234 MHz), we have:

$S/N=\frac{0.3\times \frac{8.8\times {10}^{-<!--\; -\; -->16}}{8n}\times 50}{\sqrt{2\times 1.6E-<!--\; -\; -->19\times 234E6\times {50}^2\times 2.2\times [(0.3\times \frac{8.8\times {10}^{-<!--\; -\; -->16}}{8n})+(8n\times {10}^{-<!--\; -\; -->9}\left)\right]+(2.1\times {10}^{-<!--\; -\; -->15})}}\sim <!--\; \sim \; -->13$ for 2000 photons or 0.88 fJ of incident 450 nm light per pulse with a pulse width of 8 ns, and

$S/N=\frac{0.3\times \frac{5.28\times {10}^{-<!--\; -\; -->15}}{8n}\times 50}{\sqrt{2\times 1.6E-<!--\; -\; -->19\times 234E6\times {50}^2\times 2.2\times [(0.3\times \frac{5.28\times {10}^{-<!--\; -\; -->15}}{8n})+(8n\times {10}^{-<!--\; -\; -->9}\left)\right]+(2.1\times {10}^{-<!--\; -\; -->15})}}\sim <!--\; \sim \; -->33$ for 12000 photons or 5.28 fJ of incident 450 nm light per pulse with a pulse width of 8 ns.

These results show that the above signal amplitudes are too low for relative detection at such high bandwidths using S8664-30K and the aforementioned amplifiers. So, let’s calculate at what signal levels we could attain S/N = 1 and S/N = 10 using S8664-30K and the above amplifiers at the same bandwidths.

For the earlier scenario of detecting and counting 450 nm light pulses at a rate of 1 MHz using S8664-30K, we have:

$\frac{0.3\times S_{input}\times 50}{\sqrt{2\times 1.6E-19\times 2E6\times {50}^2\times 2.2\times \left[\left(0.3\times S_{input}\right)+{10}^{-9}\right]+\left(1.96\times {10}^{-22}\right)}}=1$ which yields S_input ≈ 127 pW or ≈ 288 450 nm photons per pulse at 1MHz of pulse rate.

$\frac{0.3\times S_{input}\times 50}{\sqrt{2\times 1.6E-19\times 2E6\times {50}^2\times 2.2\times \left[\left(0.3\times S_{input}\right)+{10}^{-9}\right]+\left(1.96\times {10}^{-22}\right)}}=10$ which yields S_input ≈ 1.51 nW or ≈ 3412 450 nm photons per pulse at 1MHz of pulse rate.

For the latter scenario of performing PSD on 450 nm light pulses with a rise time of 3 ns (and thus, amplifier bandwidth of 234 MHz) using S8664-30K, we have:

$\frac{0.3\times S_{input}\times 50}{\sqrt{2\times 1.6E-19\times 234E6\times {50}^2\times 2.2\times \left[\left(0.3\times S_{input}\right)+{10}^{-9}\right]+\left(2.1\times {10}^{-15}\right)}}=1$ which yields S_input ≈ 3.6 nW or ≈ 66 450 nm photons per pulse with a width of 8ns.

$\frac{0.3\times S_{input}\times 50}{\sqrt{2\times 1.6E-19\times 234E6\times {50}^2\times 2.2\times \left[\left(0.3\times S_{input}\right)+{10}^{-9}\right]+\left(2.1\times {10}^{-15}\right)}}=10$ which yields S_input ≈ 71 nW or ≈ 1283 450 nm photons per pulse with a width of 8ns.

We now discuss how linear S8664-30K’s response would be under this application’s conditions:

- Pulse-rate linearity: For this, a calculation of response cutoff frequency using equation 2-15 would be made based on the terminal capacitance values specified in Hamamatsu’s APD datasheets for a load resistance of 50 Ω. However, APD cutoff frequency values (calculated in the same way) are provided in Hamamatsu’s APD datasheets (so no need to calculate!). In the case of S8664-30K, the specified cutoff frequency is 140 MHz, which far exceeds this application’s max. pulse rate of 1 MHz.

- Pulse-height linearity: For calculating the upper limit of pulse-height linearity, one would calculate the APD’s charge storage capacity by using Q = C_t x V_bias in which V_bias is the APD’s reverse bias voltage for the desired gain; please note that plots of APD gain and terminal capacitance vs. reverse voltage are provided in Hamamatsu APD datasheets. For our case study, we will use S8664-30K’s terminal capacitance of 22 pF and the reverse bias voltage of 360 V for the optimal gain of M_opt = 50 to obtain a charge storage capacity of 7.9 nC. Back-calculating from that charge storage capacity by taking S8664-30K’s QE (0.75 @ 450 nm) and optimal gain (M_opt = 50) into account, we arrive at a photon count of 1.3 × 10⁹, which is larger than the max. photon count of 10⁶ per pulse in this application. Thus, this application is within S8664-30K’s pulse-height linearity.

Furthermore, one also needs to take the readout circuitry into account. For example, Hamamatsu’s H4083 charge amplifier has a 2 pF storage capacitor biased up to the rail voltage of 12 V, which by Q = C × V yields a max. charge storage capacity of 24 pC or 1.5 × 10⁸ electrons (or alternatively obtained by max. output voltage of 12 V divided by gain of 0.5 V/pC). Like the previous step, back-calculating from that amount of charge by taking S8664-30K’s QE (0.75 @ 450 nm) and optimal gain (M_opt = 50) into account, we arrive at a photon count of 4 × 10⁶, which is larger than the max. photon count of 10⁶ per pulse in this application. Thus, pulse-height linearity would not be limited by the charge amplifier circuitry if H4083 is utilized in this case.