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سجل التغييرات الكاملالمحرّك 1.26.4

أدلة التحقق

ما الذي يقف فعلياً خلف كل رقم في هذا الموقع، حالة بحالة، مع إظهار الحساب.

الحاسبات
51
لديها دليل مستقل
40من 51
حالات مستقلة متوافقة
97 / 97
اتساق ذاتي فقط
11
إصدار المحرك
v1.26.4

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

حاسبات لديها دليل مستقل (40)

Ohm's Law & Power2 independent cases

Ohm's Law & Power

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

dB, dBm & RMS Converter

الحالةالمتوقعالمحسوبالفرق
0 dBm, as a powerحساب يدوي 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 voltageحساب يدوي V = sqrt(P Z): sqrt(0.001 x 50) = sqrt(0.05) = 0.2236067977 V.0.2236070.223607< 0.001 %
+30 dBm is one wattتوقع هندسي 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 peakحساب يدوي 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

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

LED Series Resistor

الحالةالمتوقعالمحسوبالفرق
One 2.1 V LED at 10 mA from 5 Vحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Smallest allowed pull-up for a 3 mA sink at 0.4 V on 3.3 Vحساب يدوي 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 modeحساب يدوي 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 windowحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Non-inverting gain with 10 k over 1 kحساب يدوي G = 1 + R_f/R_1: 1 + 10000/1000 = 11 exactly.11110 %
Closed-loop bandwidth of that stage on a 10 MHz partحساب يدوي 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 11توقع هندسي 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

الحالةالمتوقعالمحسوبالفرق
Worst-case DC error of a gain-10 stage on a 10 k sourceحساب يدوي 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 kHzحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Error at 50 mA through a 10 mΩ shunt with a 50 µV offsetحساب يدوي 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, hotحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Resolution of a 10 kΩ, B 3435 K NTC under a 10 kΩ resistor at 100 °C, 12-bitحساب يدوي 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 noiseحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
5 V to 3.3 V at 250 mA with 50 µA quiescentحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
A 100 mA burst for 1 s every 60 s, with 5 µA betweenحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
3 % of 0.9 V under a 10 A stepحساب يدوي 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 4.3حساب يدوي 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 loadحساب يدوي 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 4.3حساب يدوي 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 togetherحساب يدوي 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 artefactحساب يدوي 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 helpsحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Peak inductor current of a 3.6 V to 12 V boost at 0.5 Aحساب يدوي 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 mΩحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Capacitance left in a 22 µF 6.3 V X5R part at 5 V, one year after reflowحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Characteristic impedance of a ring at 150 MHz that 1 nF slows to 75 MHzحساب يدوي 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 kHzحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
0.5 W in a 125 °C/W package at 70 °C ambientحساب يدوي T_J = T_A + P R: 70 + 0.5 x 125 = 132.5 °C exactly.132.5132.50 %
Allowable dissipation for the same packageحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Logic power of four channels at 5.25 V and 25 mAحساب يدوي 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 saturationحساب يدوي 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 Vحساب يدوي 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 °Cحساب يدوي P = (T_Jmax - T_A)/R: (150 - 85)/100 = 0.65 W exactly.0.650.650 %
Thermal Resistance Network2 independent cases

Thermal Resistance Network

الحالةالمتوقعالمحسوبالفرق
5 W through 1.5 + 0.5 + 4 °C/W from 50 °C ambientحساب يدوي 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 limitحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
One Foster stage of 1 °C/W and 1 s, after one time constantحساب يدوي 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 °Cحساب يدوي 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 stateحساب يدوي 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 caseحساب يدوي 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 %
عرض المسار2 independent cases

عرض المسار

الحالةالمتوقعالمحسوبالفرق
3 A at a 20 °C rise on an external 1 oz layerحساب يدوي 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 timesتوقع هندسي 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

الحالةالمتوقعالمحسوبالفرق
Resistance of 50 mm of 1 mm wide 1 oz copper at 25 °Cحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Copper in the barrel of a 0.3 mm hole plated 25 µmحساب يدوي 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 °Cحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
50 Ω microstrip on 0.11 mm prepreg over a planeتوقع هندسي 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 impedanceتوقع هندسي 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 thicknessحساب يدوي 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 %
مقاس السلك وهبوط الجهد2 independent cases

مقاس السلك وهبوط الجهد

الحالةالمتوقعالمحسوبالفرق
The cross-section of 18 AWGتوقع هندسي 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 °Cحساب يدوي 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 %
هندسة الثقب (via) والوسادة3 independent cases

هندسة الثقب (via) والوسادة

الحالةالمتوقعالمحسوبالفرق
Ring left by a 0.6 mm pad on a 0.3 mm holeحساب يدوي 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 centreحساب يدوي 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 fabحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Skin depth in copper at 5 GHzحساب يدوي 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 °Cحساب يدوي 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 GHzحساب يدوي 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 constantحساب يدوي 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 GHzحساب يدوي 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 lossتوقع هندسي 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

الحالةالمتوقعالمحسوبالفرق
Near-end crosstalk of two 0.15 mm microstrips at a 0.3 mm pitch, 20 mm coupledحساب يدوي 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_bحساب يدوي 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 % budgetحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Input impedance of a converter drawing 26.7 W at its lowest input, 24 Vحساب يدوي 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 filterحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Current of a 500 V surge through 42 Ω into a 15 V stand-off TVSحساب يدوي 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 pinحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
I²t of charging 470 µF from a 12 V step through 0.5 Ωحساب يدوي 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 ambientحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Return loss at 2.44 GHz of a series 6.8 pF and shunt 3.3 nH match into 25 − j15 Ωحساب يدوي 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 E24حساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Cut-off of 10 k and 100 nFحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Resonance of 10 µH with 22 µFحساب يدوي 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 pairحساب يدوي 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 kHzحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Target impedance for 3 % of 3.3 V under a 0.5 A stepحساب يدوي 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 inductanceحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
One LSB of a 12-bit converter on a 3.3 V referenceحساب يدوي LSB = V_FS/2^N: 3.3/4096 = 0.0008056640625 V exactly.0.0008056640.0008056640 %
The ideal signal-to-noise ratio of 12 bitsتوقع هندسي Ideal SNR = 6.02 N + 1.76 dB, from quantisation noise: 74 dB for 12 bits, the figure every data-conversion text gives.74740 %
سعة حمل البلورة2 independent cases

سعة حمل البلورة

الحالةالمتوقعالمحسوبالفرق
Load capacitors for a 12 pF crystal with 3 pF of strayحساب يدوي 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 seesحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
The corner of a 10 kHz unity-gain Sallen-Key Butterworth low-pass, with its parts roundedحساب يدوي 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 needsحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Peak at an open receiver, 100 mm of 50 Ω line driven by a 10 Ω, 3.3 V driverحساب يدوي 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 terminationحساب يدوي 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 MHzحساب يدوي 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

الحالةالمتوقعالمحسوبالفرق
Total link error of a 16 MHz AVR at 115 200 b/s, normal speedحساب يدوي 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/sحساب يدوي 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 MHzحساب يدوي 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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