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Validation evidence

The evidence behind each calculator's numbers, with the working shown.

Calculators
51
With independent evidence
40of 51
Independent cases agreeing
97 / 97
Checked against a stored result
11
Engine version
v1.26.4

Independent means a hand calculation or an engineering expectation. A stored result only shows the answer has not changed.

Calculators with independent evidence (40)

Ohm's Law & Power2 independent cases

Ohm's Law & Power

CaseExpectedComputedDifference
5 V across 250 ΩHand computation Ohm’s law, I = V/R: 5 / 250 = 0.02 A exactly.0.020.020 %
Dissipation at 5 V and 20 mAHand computation P = VI: 5 x 0.02 = 0.1 W exactly.0.10.10 %
dB, dBm & RMS Converter4 independent cases

dB, dBm & RMS Converter

CaseExpectedComputedDifference
0 dBm, as a powerHand computation One milliwatt by the definition of the dBm (IEC 60027-3): 10^(0/10)/1000 = 0.001 W exactly.0.0010.0010 %
0 dBm into 50 Ω, as an RMS voltageHand computation V = sqrt(P Z): sqrt(0.001 x 50) = sqrt(0.05) = 0.2236067977 V.0.2236070.223607< 0.001 %
+30 dBm is one wattEngineering expectation The RF anchor that catches a slip by a factor of 1000: 10^(30/10)/1000 = 1000/1000 = 1 W.110 %
1 V RMS of a sine, peak to peakHand computation For a sine, V_pp = 2 sqrt(2) V_rms: 2 x 1.414213562 x 1 = 2.828427125 V.2.828432.82843< 0.001 %
Voltage Divider2 independent cases

Voltage Divider

CaseExpectedComputedDifference
12 V across 10 k and 3.3 kHand computation V = Vin R2/(R1+R2): 12 x 3300 / 13300 = 2.977443609 V.2.977442.97744< 0.001 %
Source resistance of the same dividerHand computation R1 || R2: 10000 x 3300 / 13300 = 2481.203008 Ω.2481.22481.2< 0.001 %
LED Series Resistor1 independent case

LED Series Resistor

CaseExpectedComputedDifference
One 2.1 V LED at 10 mA from 5 VHand computation R = (V_S - V_F)/I: (5 - 2.1)/0.01 = 290 Ω exactly.2902900 %
I²C Pull-up Sizing3 independent cases

I²C Pull-up Sizing

CaseExpectedComputedDifference
Smallest allowed pull-up for a 3 mA sink at 0.4 V on 3.3 VHand computation R = (V_DD - V_OL)/I_OL: (3.3 - 0.4)/0.003 = 966.6666667 Ω.966.667966.667< 0.001 %
Largest allowed pull-up for 100 pF in fast modeHand computation NXP UM10204: R = t_r/(0.8473 C_b), where 0.8473 = ln(7/3), the 30 % to 70 % rise of an RC. With t_r = 300 ns: 300e-9/(0.8473 x 100e-12) = 3540.6585 Ω.3540.663540.66< 0.001 %
The E24 value at the centre of that windowHand computation Geometric centre sqrt(966.667 x 3540.659) = 1850.0 Ω. Of its E24 neighbours, 1.8 k and 2 k, it is nearer 1.8 k on a linear or a log scale.180018000 %
Op-Amp Gain & Bandwidth3 independent cases

Op-Amp Gain & Bandwidth

CaseExpectedComputedDifference
Non-inverting gain with 10 k over 1 kHand computation G = 1 + R_f/R_1: 1 + 10000/1000 = 11 exactly.11110 %
Closed-loop bandwidth of that stage on a 10 MHz partHand computation Gain-bandwidth product divided by the noise gain, which here equals the signal gain: 10e6/11 = 909090.909 Hz.909091909091< 0.001 %
An inverting stage of gain 10 still has a noise gain of 11Engineering expectation An inverting stage divides the gain-bandwidth product by 1 + R_f/R_1, not by its own gain. Same resistors, so the bandwidth is still 10e6/11.909091909091< 0.001 %
Op-Amp Noise & DC Error2 independent cases

Op-Amp Noise & DC Error

CaseExpectedComputedDifference
Worst-case DC error of a gain-10 stage on a 10 k sourceHand computation NG = 1 + 9k/1k = 10. R+ = 10 kΩ and R- = 1k || 9k = 900 Ω. With I_os blank, taken as I_b = 10 nA: V_bias = 10e-9 x |10000 - 900| + (10e-9 / 2) x (10000 + 900) = 91 + 54.5 = 145.5 µV. V_dc = 10 x (100 µV + 1 µV/°C x 50 °C + 145.5 µV) = 2.955 mV.0.0029550.002955< 0.001 %
Peak-to-peak output noise of the same stage in 15.7 kHzHand computation MT-047 and MT-048, with 4kT = 4 x 1.380649e-23 x 298.15 = 1.6465620e-20: e_ni^2 = 5^2 + (0.5 x 10)^2 + (0.5 x 0.9)^2 + 164.65620 + 14.81906 = 229.67776 nV²/Hz, so e_ni = 15.155123 nV/√Hz. v_pp = 6.6 x 10 x 15.155123e-9 x sqrt(15700) = 125.32948 µV, 2.005 LSB of 62.5 µV.0.0001253290.000125329< 0.001 %
Current-Sense Resistor & Amplifier2 independent cases

Current-Sense Resistor & Amplifier

CaseExpectedComputedDifference
Error at 50 mA through a 10 mΩ shunt with a 50 µV offsetHand computation R = 50 mV / 5 A = 10 mΩ. At I = 50 mA the shunt rises 0.05^2 x 0.01 x 40 = 0.001 °C, so the TCR term is 100e-6 x |50 + 0.001 - 25| = 0.25001 %. The offset term is 50e-6 / (0.05 x 0.01) = 10 %. RSS with 1 % tolerance and 1 % gain error: sqrt(10^2 + 1^2 + 1^2 + 0.25001^2) = 10.10259892 %.10.102610.1026< 0.001 %
Error of the same shunt at 5 A, hotHand computation At 5 A the shunt dissipates 5^2 x 0.01 = 0.25 W and rises 0.25 x 40 = 10 °C, so it runs at 60 °C and the TCR term is 100e-6 x (60 - 25) = 0.35 %. The offset term is 50e-6 / (5 x 0.01) = 0.1 %. RSS: sqrt(0.1^2 + 1^2 + 1^2 + 0.35^2) = sqrt(2.1325) = 1.46030819 %.1.460311.46031< 0.001 %
NTC Thermistor to ADC2 independent cases

NTC Thermistor to ADC

CaseExpectedComputedDifference
Resolution of a 10 kΩ, B 3435 K NTC under a 10 kΩ resistor at 100 °C, 12-bitHand computation Beta model, R = R25 exp(B (1/T - 1/T25)): 10000 x exp(3435 x (1/373.15 - 1/298.15)) = 987.03677 Ω, and alpha = B / T^2 = 3435 / 373.15^2 = 0.024669526 per K. One code is (Rs + R)^2 / (2^N Rs alpha R) = 10987.03677^2 / (4096 x 10000 x 0.024669526 x 987.03677) = 0.12103418 °C, the coarsest step from 0 to 100 °C.0.1210340.121034< 0.001 %
Worst error of the same divider with 1 % parts and 1 LSB of noiseHand computation First order, added linearly, at 100 °C: (0.01 + 0.01) / 0.024669526 = 0.81071862 °C for R25 and Rs; 0.01 x |1/373.15 - 1/298.15| x 373.15^2 = 0.93866342 °C for B; 1 x 0.12103418 °C for the noise. The sum is 1.87041622 °C.1.870421.87042< 0.001 %
LDO Dissipation & Limits1 independent case

LDO Dissipation & Limits

CaseExpectedComputedDifference
5 V to 3.3 V at 250 mA with 50 µA quiescentHand computation P = (V_in - V_out) I_out + V_in I_q: (5 - 3.3) x 0.25 + 5 x 50e-6 = 0.42525 W exactly.0.425250.425250 %
Battery Runtime & Sizing1 independent case

Battery Runtime & Sizing

CaseExpectedComputedDifference
A 100 mA burst for 1 s every 60 s, with 5 µA betweenHand computation Duty is 1/60, so the average is 0.1 x (1/60) + 5e-6 x (59/60) = 1.6716e-3 A.0.00167160.00167158< 0.001 %
PDN Target Impedance7 independent cases

PDN Target Impedance

CaseExpectedComputedDifference
3 % of 0.9 V under a 10 A stepHand computation Z = V r / dI: 0.9 x 0.03 / 10 = 0.0027 Ω exactly.0.00270.00270 %
50 cm² of plane pair on 0.1 mm of Er 4.3Hand computation C = e0 Er A / h: 8.8541878128e-12 x 4.3 x 50e-4 / 1e-4 = 1.90365038e-9 F.1.90365e-91.90365e-9< 0.001 %
Spreading inductance to a group 3 mm from the loadHand computation L = (mu0 h / 2 pi) ln(d / r_v): the prefactor is 2e-7 x 1e-4 = 2e-11 H and ln(3 mm / 0.15 mm) = ln(20) = 2.995732274, so L = 5.99146455e-11 H.5.99146e-115.99146e-11< 0.001 %
First cavity mode of a 100 mm plane on Er 4.3Hand computation f = c0 / (2 a sqrt(Er)): 299792458 / (2 x 0.1 x 2.073644135) = 7.22863805e8 Hz.722864000722864000< 0.001 %
The inductance floor of all three groups togetherHand computation Each branch is (ESL + L_mount)/N plus its spreading term, at the worst-case 2 nH mount: 1.89914645e-10, 4.62640533e-10 and 2.35596635e-9 H. In parallel, 1.27364474e-10 H. The capacitance derating does not enter: the floor is set by inductance alone.1.27364e-101.27364e-10< 0.001 %
The in-band peak used for the verdict, without the lumped-plane artefactHand computation Anti-resonance of the 2 µH regulator branch with the derated bulk capacitance: 2.442 Ω near 9.4 kHz, computed separately from the same network. The full sweep also peaks at 11.95 Ω near 320 MHz. That peak is the ideal lossless plane capacitor ringing against the bank inductance, an artefact of lumping a distributed plane into two elements. Remove the plane branch and the peak is still 2.442 Ω.2.4422.44220.00832 %
The frequency above which no capacitor on this board helpsHand computation f = Z_target / (2 pi L_node): 0.0027 / (2 pi x 1.27364474e-10) = 3.37392629e6 Hz. Both inputs are checked in the cases above.33739303373930< 0.001 %
Boost Converter Design2 independent cases

Boost Converter Design

CaseExpectedComputedDifference
Peak inductor current of a 3.6 V to 12 V boost at 0.5 AHand computation TI SLVA372, at 500 kHz, 8.2 µH and 90 % efficiency: D = 1 - 0.9 x 3.6 / 12 = 0.73; I_L = 12 x 0.5 / (0.9 x 3.6) = 1.85185185 A; dI_L = 3.6 x 0.73 / (500e3 x 8.2e-6) = 0.64097561 A; I_pk = I_L + dI_L / 2 = 2.17233966 A.2.172342.17234< 0.001 %
Output ripple of the same boost with 22 µF and 10 mΩHand computation TI SLVA372, dV = I_out D / (f_sw C_out) + I_pk ESR: 0.5 x 0.73 / (500e3 x 22e-6) = 33.181818 mV, plus 2.17233966 x 0.01 = 21.723397 mV, gives 54.905215 mV.0.05490520.0549052< 0.001 %
MLCC Effective Capacitance1 independent case

MLCC Effective Capacitance

CaseExpectedComputedDifference
Capacitance left in a 22 µF 6.3 V X5R part at 5 V, one year after reflowHand computation Four worst-case losses multiplied, C_min = C x (1 - tol) x (1 - b) x (1 - dT) x (1 - a x log10(t / t_ref)): 20 % tolerance, a 60 % DC-bias loss at 5 V (an example reading, as the reader takes it from the maker’s curve), the X5R class limit of 15 % (EIA-198), and 5 % per decade of ageing (KEMET) from 1000 h to 8760 h. 1 - 0.05 x log10(8.76) = 0.9528747947, so C_min = 22e-6 x 0.80 x 0.40 x 0.85 x 0.9528747947 = 5.70200277e-6 F.0.0000057020.000005702< 0.001 %
RC Snubber for Switch-Node Ringing2 independent cases

RC Snubber for Switch-Node Ringing

CaseExpectedComputedDifference
Characteristic impedance of a ring at 150 MHz that 1 nF slows to 75 MHzHand computation Kollman Power Tip #16, the add-a-capacitor method: C_par = 1e-9 / ((150 / 75)^2 - 1) = 333.333 pF; L_par = 1 / ((2 pi x 150e6)^2 x 333.333e-12) = 3.37737 nH. Ridley Design Step 3, R = Z_0 = sqrt(L_par / C_par) = 1 / (2 pi x 150e6 x 333.333e-12) = 10 / pi = 3.18309886 Ω.3.18313.1831< 0.001 %
Power in the snubber resistor at 24 V and 500 kHzHand computation Ridley Design Step 4, P_R = C_s V^2 f_sw with no factor of one half: C_s = 3 x 333.333 pF = 1.0 nF, rounded up in E12 to 1.0 nF, so P_R = 1e-9 x 24^2 x 500e3 = 0.288 W.0.2880.288< 0.001 %
Junction Temperature & Safe Limits2 independent cases

Junction Temperature & Safe Limits

CaseExpectedComputedDifference
0.5 W in a 125 °C/W package at 70 °C ambientHand computation T_J = T_A + P R: 70 + 0.5 x 125 = 132.5 °C exactly.132.5132.50 %
Allowable dissipation for the same packageHand computation P = (T_Jmax - T_A)/R: (150 - 70)/125 = 0.64 W exactly.0.640.640 %
Logic & Output Power, Duty Cycle4 independent cases

Logic & Output Power, Duty Cycle

CaseExpectedComputedDifference
Logic power of four channels at 5.25 V and 25 mAHand computation P = n V_CC I_CC: 4 x 5.25 x 0.025 = 0.525 W exactly.0.5250.5250 %
Both stages together, driving 250 mA at 0.7 V saturationHand computation The output stage adds n V_CE(sat) I_C = 4 x 0.7 x 0.25 = 0.7 W, so the instantaneous ON power is 0.525 + 0.7 = 1.225 W exactly.1.2251.2250 %
OFF power: 7.5 mA of logic and 0.1 mA of leakage at 100 VHand computation P = n V_CC I_CC(off) + n V_off I_leak: 4 x 5.5 x 0.0075 + 4 x 100 x 0.0001 = 0.165 + 0.04 = 0.205 W exactly.0.2050.2050 %
Allowable average power for the package at 85 °CHand computation P = (T_Jmax - T_A)/R: (150 - 85)/100 = 0.65 W exactly.0.650.650 %
Thermal Resistance Network2 independent cases

Thermal Resistance Network

CaseExpectedComputedDifference
5 W through 1.5 + 0.5 + 4 °C/W from 50 °C ambientHand computation Series thermal resistances add: T_J = T_A + P(R_JC + R_CS + R_SA) = 50 + 5 x 6 = 80 °C exactly.80800 %
The heat sink that holds a 25 W part 25 °C below its 150 °C limitHand computation R_SA = (T_des - T_A)/P - R_JC - R_CS, with T_des = 150 - 25 = 125 °C: (125 - 45)/25 - 1.0 = 3.2 - 1.0 = 2.2 °C/W exactly.2.22.20 %
Transient Thermal Impedance4 independent cases

Transient Thermal Impedance

CaseExpectedComputedDifference
One Foster stage of 1 °C/W and 1 s, after one time constantHand computation Z = R(1 - e^(-t/tau)): 1 - e^-1 = 0.6321205588 °C/W. A wrong sign or an inverted tau breaks this first.0.6321210.632121< 0.001 %
20 W into that stage for 1 s, from 25 °CHand computation T = T_ref + P Z: 25 + 20 x 0.6321205588 = 37.64241118 °C.37.642437.6424< 0.001 %
The same stage at 50 per cent duty, in steady stateHand computation The periodic form gives (1 - e^-1)/(1 - e^-2). The denominator factorises as (1 - e^-1)(1 + e^-1), which leaves 1/(1 + e^-1) = 0.7310585786 °C/W.0.7310590.731059< 0.001 %
The application-note approximation for the same caseHand computation D R + (1 - D) Z(t_p): 0.5 x 1 + 0.5 x 0.6321205588 = 0.8160602794 °C/W, against the exact 0.7310585786. The page shows both.0.816060.81606< 0.001 %
Trace Width2 independent cases

Trace Width

CaseExpectedComputedDifference
3 A at a 20 °C rise on an external 1 oz layerHand computation IPC-2221 solved for area: A = (3/(0.048 x 20^0.44))^(1/0.725) = 48.70 mil²; at 1.37 mil thick that is 35.55 mil, or 0.9029 mm wide.0.90290.9028-0.0111 %
Doubling the current widens the trace about 2.6 timesEngineering expectation The IPC exponent is 1/0.725, so width scales as I^1.379: 2^1.379 = 2.60. A result near 2 or near 4 would be wrong.0.9029 mm x 2.60. A loose band: it checks the shape of the law, not the digits.2.3482.3480 %±2 %
Trace Resistance & IR Drop1 independent case

Trace Resistance & IR Drop

CaseExpectedComputedDifference
Resistance of 50 mm of 1 mm wide 1 oz copper at 25 °CHand computation R = rho l/(w t), with rho = 1.724e-8 Ω m at 20 °C and alpha = 0.00393: 1.724e-8 x 1.01965 x 50 / 34.8e-6 = 25.256 mΩ at 25 °C.0.0252560.02525680.00336 %
Via Current & Resistance2 independent cases

Via Current & Resistance

CaseExpectedComputedDifference
Copper in the barrel of a 0.3 mm hole plated 25 µmHand computation The barrel is an annulus, taken at its mean diameter: A = pi (d + t_p) t_p = pi x 0.325 mm x 0.025 mm = 0.02552544 mm².0.02552540.0255254< 0.001 %
Resistance of that barrel through a 1.6 mm board at 25 °CHand computation R = rho l/A, with rho = 1.724e-8 Ω m at 20 °C and alpha = 0.00393: 1.75788e-8 x 1.6e-3 / 2.552544e-8 = 1.10188e-3 Ω.0.001101880.00110188< 0.001 %
Controlled Impedance & Delay3 independent cases

Controlled Impedance & Delay

CaseExpectedComputedDifference
50 Ω microstrip on 0.11 mm prepreg over a planeEngineering expectation Rule of thumb: a 50 Ω microstrip on FR-4 is about twice as wide as its dielectric is thick. This checks the band, not the digits.The rule of thumb is good to about twelve per cent, no better.5048.4-3.2 %±12 %
A wider trace has a lower impedanceEngineering expectation Impedance falls as the trace widens, in every transmission-line model. This catches a sign error.A wide band: it checks the direction and the order of magnitude.3231.6-1.25 %±25 %
Alumina, w/h = 1, zero copper thicknessHand computation Hammerstad and Jensen’s equations, worked in a separate script without the engine. This checks the code against the method, not the method itself.One per cent: the tightest check on this calculator.49.2949.30.0203 %±1 %
Wire Gauge & Voltage Drop2 independent cases

Wire Gauge & Voltage Drop

CaseExpectedComputedDifference
The cross-section of 18 AWGEngineering expectation AWG is a geometric series, d = 0.127 mm x 92^((36-n)/39): 18 AWG is 1.0238 mm across, so pi d^2/4 = 0.8232 mm². Wire tables give 0.823 mm², and a gauge off by one would show at once.0.8230.8230 %
Resistance of 2 m of 18 AWG, there and back, at 20 °CHand computation R = rho l/A over 4 m of conductor: 1.724e-8 x 4 / 8.2316e-7 = 0.083775 Ω. Handbooks give about 20.95 Ω/km for 18 AWG, which is 0.0838 Ω for 4 m.0.0837750.08378620.0134 %
Via & Pad Geometry3 independent cases

Via & Pad Geometry

CaseExpectedComputedDifference
Ring left by a 0.6 mm pad on a 0.3 mm holeHand computation Half the difference of the diameters: (0.6 - 0.3)/2 = 0.15 mm exactly.0.000150.000150 %
The ring left when the drill lands 50 µm off centreHand computation The whole hole shifts, so the ring loses the full registration on one side: 0.15 - 0.05 = 0.10 mm exactly. Halving the shift is the optimistic mistake.0.00010.0001< 0.001 %
The smallest pad this hole can have and still pass the fabHand computation D + 2(ring_min + t_reg): 0.3 + 2 x (0.125 + 0.05) = 0.3 + 0.35 = 0.65 mm exactly. That is larger than the 0.6 mm pad drawn, so the drawn pad is too small.0.000650.000650 %
AC Resistance, Skin Effect & Trace Loss6 independent cases

AC Resistance, Skin Effect & Trace Loss

CaseExpectedComputedDifference
Skin depth in copper at 5 GHzHand computation d = sqrt(rho / (pi f mu0)), with rho = 1.724e-8 Ω m: sqrt(1.724e-8 / (pi x 5e9 x 4pi e-7)) = 9.34552622e-7 m. At 1 GHz the same formula gives 2.09 µm, the textbook figure.9.34553e-79.34553e-7< 0.001 %
DC resistance of a 0.2 mm wide, 1 oz trace at 20 °CHand computation R = rho / (w t), with 1 oz of finished copper taken as 34.8 µm: 1.724e-8 / (2e-4 x 3.48e-5) = 2.47701149 Ω/m.2.477012.47701< 0.001 %
Hammerstad roughness factor for 0.4 µm foil at 5 GHzHand computation K = 1 + (2/pi) arctan(1.4 (D/d)^2): D/d = 0.4/0.934552622 = 0.42801, squared is 0.183193, times 1.4 is 0.256470, arctan is 0.250996 rad, times 2/pi is 0.159795, so K = 1.15980.1.161.160 %
A stripline’s effective permittivity is its dielectric constantHand computation A stripline sits in one uniform dielectric, so it is a TEM line and its effective permittivity equals the dielectric constant exactly: 4.3 in, 4.3 out. Any other value would be wrong.4.34.30 %
Dielectric loss of 8 inches of stripline at 2.5 GHzHand computation a = pi f sqrt(e_eff) tand / c0 in Np/m, times 8.685889638 for dB/m. A stripline has e_eff exactly 4.3, so nothing here is fitted: pi x 2.5e9 x 2.07364414 x 0.02 / 299792458 x 8.685889638 = 9.43729887 dB/m, and 8 inches gives 1.91765913 dB.1.917661.91766< 0.001 %
Against the 2.31 dB rule of thumb for dielectric lossEngineering expectation Rule of thumb: loss = 2.31 dB per inch x f in GHz x tand x sqrt(e_eff). Here 2.31 x 2.5 x 0.02 x 2.073644 x 8 inches = 1.9160 dB, within a tenth of a per cent of the computed 1.9177 dB. The gap is the rounding in 2.31: the exact coefficient is 2.3119.1.9161.917660.0866 %
Crosstalk & Spacing3 independent cases

Crosstalk & Spacing

CaseExpectedComputedDifference
Near-end crosstalk of two 0.15 mm microstrips at a 0.3 mm pitch, 20 mm coupledHand computation Hammerstad and Jensen at w/h = 1.5, e_r = 3, t = 35 µm, worked in a separate script without the engine, give e_eff = 2.2217732. L_sat = t_r c / (2 sqrt(e_eff)) = 1e-9 x 299792458 / (2 x 1.4905613) = 0.10056361 m = 100.56361 mm. K_b = 0.5 x 0.48 x exp(-0.96 x 0.15 / 0.1) = 5.6862662 %. NEXT = 5.6862662 x 20 / 100.56361 = 1.1308795 %.1.130881.13088< 0.001 %
A saturated stripline run: the near-end crosstalk is the whole K_bHand computation In one dielectric e_eff = e_r = 4, so L_sat = 0.3e-9 x 299792458 / (2 x 2) = 0.02248 m = 22.48 mm, shorter than the 50 mm run, and NEXT = K_b = 0.5 x 0.347 x exp(-2.9 x 0.1 / 0.3) = 0.1735 x 0.38034869 = 6.5990509 %.6.599056.59905< 0.001 %
The pitch that puts twice the near-end crosstalk at a 5 % budgetHand computation From the first case, the coupling needed is (5 % / 2) / (20 / 100.56361) = 12.570451 %. The edge gap is (0.1 / 0.96) x ln(0.24 / 0.12570451) = 0.0673651 mm, so the pitch is 0.15 + 0.0673651 = 0.2173651 mm.0.2173650.21740.0161 %
Input EMI Filter & Damping2 independent cases

Input EMI Filter & Damping

CaseExpectedComputedDifference
Input impedance of a converter drawing 26.7 W at its lowest input, 24 VHand computation Erickson and Maksimović, well below its loop crossover frequency a converter draws constant power, so its input is the negative resistance -V_in^2 / P_in, smallest at the lowest input and full power (Kollman, Power Tip #4): |Z_N| = 24^2 / 26.7 = 576 / 26.7 = 21.5730337 Ω.21.57321.573< 0.001 %
Corner of a 2.2 µH and 10 µF input filterHand computation The LC corner, f_c = 1 / (2 pi sqrt(L_f C_f)): sqrt(2.2e-6 x 10e-6) = 4.69041576e-6 s, so f_c = 1 / (2 pi x 4.69041576e-6) = 33931.94788 Hz.33931.933931.9< 0.001 %
TVS & Surge Clamp2 independent cases

TVS & Surge Clamp

CaseExpectedComputedDifference
Current of a 500 V surge through 42 Ω into a 15 V stand-off TVSHand computation Straight clamp line through the datasheet maxima, V_BR,max 18.5 V and V_C 24.4 V at I_PP 24.6 A: R_dyn = (24.4 - 18.5) / 24.6 = 0.2398373984 Ω. The IEC 61000-4-5 generator is V_s behind R_src, so I = (500 - 18.5) / (42 + 0 + 0.2398373984) = 11.39919161 A.11.399211.3992< 0.001 %
Voltage the same TVS leaves at the protected pinHand computation The clamp line at the surge current above, V = V_BR,max + I R_dyn: 18.5 + 11.39919161 x 0.2398373984 = 21.23395246 V, which is 0.708 of the 30 V pin maximum.21.23421.234< 0.001 %
Fuse Sizing, Inrush & I²t2 independent cases

Fuse Sizing, Inrush & I²t

CaseExpectedComputedDifference
I²t of charging 470 µF from a 12 V step through 0.5 ΩHand computation Fuseology Pulse Energy Model, the exponential wave of an ideal RC charge, I²t = 1/2 x I_pk^2 x tau: I_pk = 12 / 0.5 = 24 A and tau = 0.5 x 470e-6 = 235 µs, so I²t = 0.5 x 24^2 x 235e-6 = 0.06768 A²s.0.067680.067680 %
Smallest fuse rating for 1.5 A with the fuse re-rated to 90 % at its ambientHand computation Fuseology checklist items 1 and 3, the fuse runs at no more than 75 % of its re-rated current: I_min = I_op / (0.75 x k_T) = 1.5 / (0.75 x 0.90) = 1.5 / 0.675 = 2.222222222 A.2.222222.22222< 0.001 %
L-Network Impedance Matching2 independent cases

L-Network Impedance Matching

CaseExpectedComputedDifference
Return loss at 2.44 GHz of a series 6.8 pF and shunt 3.3 nH match into 25 − j15 ΩHand computation Pozar, the capacitor next to the load and the inductor at the source: w = 2 pi x 2.44e9 = 1.5330972e10 rad/s; X = -1 / (w x 6.8e-12) = -9.5922700 Ω and B = -1 / (w x 3.3e-9) = -0.019765890 S. 1 / (25 - j24.592270) = 0.020328844 + j0.019997296 S; adding jB gives 0.020328844 + j0.000231407 S, so Z_in = 49.184816 - j0.559879 Ω. |Gamma| = |Z_in - 50| / |Z_in + 50| = 0.0099704529, and RL = -20 log10 |Gamma| = 40.025702 dB.40.025740.0257< 0.001 %
Mismatch loss of the low-pass match the page designs for the same load, in E24Hand computation Pozar, with R_L below R_s, the + sign: X = sqrt(25 x 25) + 15 = 40 Ω gives L = 40 / w = 2.6091 nH, nearest E24 2.7 nH, and B = sqrt(25 / 25) / 50 = 0.02 S gives C = 0.02 / w = 1.3045 pF, nearest E24 1.3 pF. With those parts at 2.44 GHz, X = 41.393625 Ω and B = 0.019930264 S; 1 / (25 + j26.393625) = 0.018916130 - j0.019970609 S; adding jB gives Z_in = 52.864697 + j0.112752 Ω, |Gamma| = 0.027870719, and ML = -10 log10 (1 - |Gamma|^2) = 0.0033748105 dB.0.003374810.00337481< 0.001 %
RC / RL Filter & Time Constant1 independent case

RC / RL Filter & Time Constant

CaseExpectedComputedDifference
Cut-off of 10 k and 100 nFHand computation f = 1/(2 pi R C): 1/(2 pi x 10000 x 1e-7) = 159.1549431 Hz.159.155159.155< 0.001 %
LC Resonance & Damping3 independent cases

LC Resonance & Damping

CaseExpectedComputedDifference
Resonance of 10 µH with 22 µFHand computation f = 1/(2 pi sqrt(LC)): sqrt(2.2e-10) = 1.4832397e-5, times 2 pi is 9.3194699e-5, reciprocal 10730.22 Hz.10730.210730.2< 0.001 %
Characteristic impedance of the same pairHand computation Z = sqrt(L/C): sqrt(10/22) = sqrt(0.4545454545) = 0.6741999 Ω.0.67420.6742< 0.001 %
The capacitor that tunes 10 µH to 10 kHzHand computation C = 1/(L (2 pi f)^2): 2 pi x 10000 = 62831.853, squared is 3.9478418e9, times 1e-5 is 39478.418, reciprocal 2.5330296e-5 F.0.00002533030.0000253303< 0.001 %
Decoupling & Bulk Capacitance2 independent cases

Decoupling & Bulk Capacitance

CaseExpectedComputedDifference
Target impedance for 3 % of 3.3 V under a 0.5 A stepHand computation Z = dV/dI, with dV the allowed deviation: 3.3 x 0.03 = 0.099 V over 0.5 A = 0.198 Ω exactly.0.1980.198< 0.001 %
Self-resonance of 100 nF with 1.2 nH of loop inductanceHand computation f = 1/(2 pi sqrt(L C)): sqrt(1.2e-16) = 1.0954451e-8, times 2 pi is 6.8828847e-8, reciprocal 14.5288 MHz.1452880014528800< 0.001 %
ADC Resolution & Noise Budget2 independent cases

ADC Resolution & Noise Budget

CaseExpectedComputedDifference
One LSB of a 12-bit converter on a 3.3 V referenceHand computation LSB = V_FS/2^N: 3.3/4096 = 0.0008056640625 V exactly.0.0008056640.0008056640 %
The ideal signal-to-noise ratio of 12 bitsEngineering expectation Ideal SNR = 6.02 N + 1.76 dB, from quantisation noise: 74 dB for 12 bits, the figure every data-conversion text gives.74740 %
Crystal Load Capacitance2 independent cases

Crystal Load Capacitance

CaseExpectedComputedDifference
Load capacitors for a 12 pF crystal with 3 pF of strayHand computation The two capacitors are in series across the crystal, so C_L = C/2 + C_stray and C = 2(C_L - C_stray): 2 x (12 - 3) = 18 pF, an E24 value.1.8e-111.8e-110 %
The load the crystal then seesHand computation Two 18 pF in series is 9 pF. Add the 3 pF stray and the crystal sees 12 pF, its specified load.1.2e-111.2e-110 %
Active Filter: Sallen-Key / MFB2 independent cases

Active Filter: Sallen-Key / MFB

CaseExpectedComputedDifference
The corner of a 10 kHz unity-gain Sallen-Key Butterworth low-pass, with its parts roundedHand computation TI SLOA088 with a = 1.4142, b = 1 and C1 = 1 nF: C2 at least 4 x 1 nF / 1.4142^2 = 2.0000 nF, so 2.2 nF (E12); R1,2 = (1.4142 x 2.2e-9 -/+ sqrt((1.4142 x 2.2e-9)^2 - 4 x 1e-9 x 2.2e-9)) / (4 pi x 10e3 x 1e-9 x 2.2e-9) = 7861.0 Ω and 14646.7 Ω, so 7.87 kΩ and 14.7 kΩ (E96). Built: f0 = 1 / (2 pi sqrt(7870 x 14700 x 1e-9 x 2.2e-9)) = 9976.1451 Hz and Q = sqrt(7870 x 14700 x 1e-9 x 2.2e-9) / (1e-9 x 22570) = 0.70684764. A second-order low-pass is 3 dB down at f0 x sqrt((p + sqrt(p^2 + 4)) / 2), p = 2 - 1/Q^2 = -0.00146671: 9976.1451 x 0.99963339 = 9972.4878 Hz.9972.499972.49< 0.001 %
The op-amp gain-bandwidth the same stage needsHand computation TI SLOA088 for a second-order stage with Q below 1: f_T = 100 x Gain x f_Ci, f_Ci the stage's own corner. At unity gain and the 9972.4878 Hz corner above, 100 x 1 x 9972.4878 = 997248.78 Hz.997249997249< 0.001 %
Line Termination & Reflections3 independent cases

Line Termination & Reflections

CaseExpectedComputedDifference
Peak at an open receiver, 100 mm of 50 Ω line driven by a 10 Ω, 3.3 V driverHand computation Johnson and Graham, the reflection lattice. t_d = 5.96 ps/mm x 100 mm = 596 ps, longer than half the 1 ns edge, so the first arrival ends before the second begins. V_launch = 3.3 x 50 / (50 + 10) = 2.75 V, and an open end doubles it: 2.75 x (1 + 1) = 5.5 V.5.55.50 %
Driver DC current into a 50 Ω parallel terminationHand computation With R_p = Z0 = 50 Ω to ground, the high level is 3.3 x 50 / (10 + 50) = 2.75 V, and the driver carries (3.3 - 2.75) / 10 = 55 mA.0.0550.055< 0.001 %
Power in the resistor of a 50 Ω + 82 pF AC termination at 50 MHzHand computation C = 3 x 2 x 596 ps / 50 Ω = 71.52 pF, rounded up in E12 to 82 pF. The periodic steady state of a 0-to-3.3 V square wave into R and C: P = C V^2 f tanh(1 / (4 f R C)) = 82e-12 x 10.89 x 50e6 x tanh(1 / 0.82) = 44.6490 mW x 0.8395102 = 37.48329 mW.0.03748330.0374833< 0.001 %
UART Baud Error & Timer Prescaler3 independent cases

UART Baud Error & Timer Prescaler

CaseExpectedComputedDifference
Total link error of a 16 MHz AVR at 115 200 b/s, normal speedHand computation ATmega328P datasheet, UBRR = round(16e6 / (16 x 115200)) - 1 = round(8.680556) - 1 = 8. The rate is 16e6 / (16 x 9) = 111111.111 b/s, so e = 111111.111 / 115200 - 1 = -3.5493827 %. With 50 ppm (0.005 %) of clock tolerance and 1 % at the far end, the total is 3.5493827 + 0.005 + 1 = 4.5543827 %.4.554384.55438< 0.001 %
Baud rate of an STM32 fractional divider at 16 MHz and 115 200 b/sHand computation RM0008 fractional baud rate generation, USARTDIV = 16e6 / (16 x 115200) = 8.680556: mantissa 8, fraction round(16 x 0.680556) = round(10.889) = 11, so BRR = 16 x 8 + 11 = 139 = 0x8B and the rate is 16e6 / 139 = 115107.9137 b/s.115108115108< 0.001 %
Steps per period of a 16-bit timer making 50 Hz from 72 MHzHand computation RM0008 TIMx_PSC and TIMx_ARR, f = f_clk / ((PSC + 1)(ARR + 1)): N = 72e6 / 50 = 1440000 = 2^8 x 3^2 x 5^4. The largest divisor of N that a 16-bit counter holds (65536 at most) is 60000 = 2^5 x 3 x 5^4, so ARR = 59999 and PSC + 1 = 1440000 / 60000 = 24.60000600000 %

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