ما الذي يقف فعلياً خلف كل رقم في هذا الموقع، حالة بحالة، مع إظهار الحساب.
بإمكان أي أحد كتابة حاسبة. السؤال الذي ينبغي أن يطرحه المراجع هو: كيف يعرف أحد أنها صحيحة؟ تجيب هذه الصفحة عن ذلك لكل حاسبة على حدة، وتفصل بين أمرين يُطبعان عادةً تحت عنوان أخضر واحد، لأن الخلط بينهما هو ما يجعل ملف التحقق بلا قيمة.
دليل مستقل
أن تطابق النتيجة شيئاً حُسب من دون هذا الكود: حساب يدوي من المعادلة المذكورة، أو مثال محلول منشور في مرجع قياسي أو في مذكرة تطبيقية، أو مقارنة مع حلّال حقل أو مع SPICE، أو نتيجة يتوقعها أي مهندس ذي خبرة. هذا النوع وحده دليل على أن الطريقة صحيحة. كل حالة أدناه تذكر مصدرها، وتُذكر قوة ذلك المصدر بدل أن تُذاب في متوسط.
الاتساق الذاتي
أن النتيجة لم تتغير منذ تسجيلها. لكل حاسبة هنا لقطة محفوظة، وتفشل مجموعة الاختبارات إذا تحرك رقم من دون قصد. هذا أمر يستحق الوجود، ولا يساوي شيئاً كدليل على أن الرقم كان صحيحاً من البداية: إنه المحرك يوافق نفسه في الماضي. لذلك لا يُحسب أبداً إلى جانب العمود المجاور.
هذه ليست شهادة اعتماد، ولم يدقق فيها أحد سوانا. إنها الحساب منشوراً كي تتمكن من مراجعته. وحيث لا يقف خلف حاسبة سوى لقطة محفوظة، تذكر هذه الصفحة اسمها بدل أن تحذفها من القائمة.
حال الأدلة
الحاسبات
36
لديها دليل مستقل
25من 36
حالات مستقلة متوافقة
65 / 65
اتساق ذاتي فقط
11
تذكر مصادرها
11
تذكر دقتها
13
إصدار المحرك
v1.18.3
كل حالة في هذه الصفحة شُغّلت أثناء بناء الصفحة، فعمود المحسوب هو ما ينتجه هذا الإصدار. وتُشغّل الحالات نفسها في مجموعة الاختبارات، فأي اختلاف يوقف البناء بدل أن يصل إليك.
Smallest allowed pull-up for a 3 mA sink at 0.4 V on 3.3 VR = (V_DD - V_OL)/I_OL: (3.3 - 0.4)/0.003 = 966.6666667 Ω.
حساب يدوي
966.667
966.667
< 0.001 %حتى الرقم الأخير
Largest allowed pull-up for 100 pF in fast modeNXP 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.
حساب يدوي
3540.66
3540.66
< 0.001 %حتى الرقم الأخير
The E24 value at the centre of that windowGeometric 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.
Non-inverting gain with 10 k over 1 kG = 1 + R_f/R_1: 1 + 10000/1000 = 11 exactly.
حساب يدوي
11
11
0 %حتى الرقم الأخير
Closed-loop bandwidth of that stage on a 10 MHz partGain-bandwidth product divided by the noise gain, which here equals the signal gain: 10e6/11 = 909090.909 Hz.
حساب يدوي
909091
909091
< 0.001 %حتى الرقم الأخير
An inverting stage of gain 10 still has a noise gain of 11An 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.
الطريقة منJEDEC JESD51-2, Integrated Circuits Thermal Test Method, still air
الدقةThe dissipation is exact arithmetic. Junction temperature and the current limit are only as good as R_thetaJA, which depends on the board as much as the package.
الحالة
نوع الدليل
المتوقع
المحسوب
الفرق
5 V to 3.3 V at 250 mA with 50 µA quiescentP = (V_in - V_out) I_out + V_in I_q: (5 - 3.3) x 0.25 + 5 x 50e-6 = 0.42525 W exactly.
الطريقة منSmith, Anderson, Forehand, Pelc and Roy, Power distribution system design methodology and capacitor selection for modern CMOS technology, IEEE Transactions on Advanced Packaging (1999); Smith and Bogatin, Principles of Power Integrity for PDN Design Simplified
الدقةA lumped network with exact arithmetic, so the error lies in the inductances you enter. Mounting inductance spans roughly 0.3 to 2 nH for ordinary pads and vias and sets the answer above about 30 MHz, so the verdict uses the worst-case value.
الحالة
نوع الدليل
المتوقع
المحسوب
الفرق
3 % of 0.9 V under a 10 A stepZ = V r / dI: 0.9 x 0.03 / 10 = 0.0027 Ω exactly.
حساب يدوي
0.0027
0.0027
0 %حتى الرقم الأخير
50 cm² of plane pair on 0.1 mm of Er 4.3C = e0 Er A / h: 8.8541878128e-12 x 4.3 x 50e-4 / 1e-4 = 1.90365038e-9 F.
حساب يدوي
1.90365e-9
1.90365e-9
< 0.001 %حتى الرقم الأخير
Spreading inductance to a group 3 mm from the loadL = (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.
حساب يدوي
5.99146e-11
5.99146e-11
< 0.001 %حتى الرقم الأخير
First cavity mode of a 100 mm plane on Er 4.3f = c0 / (2 a sqrt(Er)): 299792458 / (2 x 0.1 x 2.073644135) = 7.22863805e8 Hz.
حساب يدوي
722864000
722864000
< 0.001 %حتى الرقم الأخير
The inductance floor of all three groups togetherEach 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.
حساب يدوي
1.27364e-10
1.27364e-10
< 0.001 %حتى الرقم الأخير
The in-band peak used for the verdict, without the lumped-plane artefactAnti-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.The peak is taken on a discrete log grid, so it is quoted to four figures and checked to 0.1 %.
حساب يدوي
2.442
2.4422
0.00832 %±0.1 %
The frequency above which no capacitor on this board helpsf = 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.
الطريقة منAllegro MicroSystems AN295014 Rev. 1, Computing IC Temperature Rise, Eq. 1 (2022); JEDEC JESD51-2, Integrated Circuits Thermal Test Method, still air
الدقةThe equation is exact arithmetic. The uncertainty is all in R_θ, which depends on the board as much as the package: a datasheet figure from a JEDEC test board can be a factor of two off on yours.
الحالة
نوع الدليل
المتوقع
المحسوب
الفرق
0.5 W in a 125 °C/W package at 70 °C ambientT_J = T_A + P R: 70 + 0.5 x 125 = 132.5 °C exactly.
حساب يدوي
132.5
132.5
0 %حتى الرقم الأخير
Allowable dissipation for the same packageP = (T_Jmax - T_A)/R: (150 - 70)/125 = 0.64 W exactly.
الطريقة منAllegro MicroSystems AN295014 Rev. 1, Computing IC Temperature Rise, Eq. 2–4 (2022)
الحالة
نوع الدليل
المتوقع
المحسوب
الفرق
Logic power of four channels at 5.25 V and 25 mAP = n V_CC I_CC: 4 x 5.25 x 0.025 = 0.525 W exactly.
حساب يدوي
0.525
0.525
0 %حتى الرقم الأخير
Both stages together, driving 250 mA at 0.7 V saturationThe 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.
حساب يدوي
1.225
1.225
0 %حتى الرقم الأخير
OFF power: 7.5 mA of logic and 0.1 mA of leakage at 100 VP = 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.
حساب يدوي
0.205
0.205
0 %حتى الرقم الأخير
Allowable average power for the package at 85 °CP = (T_Jmax - T_A)/R: (150 - 85)/100 = 0.65 W exactly.
الطريقة منJEDEC JESD51-2, Integrated Circuits Thermal Test Method, still air
الدقةResistances in series, exact in steady state only. Each term is as good as its source: junction to case is the tightest, the interface the loosest, and sink to air depends on the real airflow.
الحالة
نوع الدليل
المتوقع
المحسوب
الفرق
5 W through 1.5 + 0.5 + 4 °C/W from 50 °C ambientSeries thermal resistances add: T_J = T_A + P(R_JC + R_CS + R_SA) = 50 + 5 x 6 = 80 °C exactly.The panel shows node temperatures to one decimal, so the check cannot be tighter.
حساب يدوي
80
80
0 %±0.1 %
The heat sink that holds a 25 W part 25 °C below its 150 °C limitR_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.
الطريقة منJESD51-14, Transient Dual Interface Test Method for the Measurement of the Thermal Resistance Junction to Case of Semiconductor Devices with a Single Heat Flow Path; JESD51-1, Integrated Circuit Thermal Measurement Method, Electrical Test Method
الدقةExact closed-form arithmetic, with nothing fitted. The error is in the R and tau set and how it was measured: a junction-to-case set assumes an isothermal case, a junction-to-ambient set the JEDEC test board in still air.
الحالة
نوع الدليل
المتوقع
المحسوب
الفرق
One Foster stage of 1 °C/W and 1 s, after one time constantZ = R(1 - e^(-t/tau)): 1 - e^-1 = 0.6321205588 °C/W. A wrong sign or an inverted tau breaks this first.
حساب يدوي
0.632121
0.632121
< 0.001 %حتى الرقم الأخير
20 W into that stage for 1 s, from 25 °CT = T_ref + P Z: 25 + 20 x 0.6321205588 = 37.64241118 °C.
حساب يدوي
37.6424
37.6424
< 0.001 %حتى الرقم الأخير
The same stage at 50 per cent duty, in steady stateThe 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.
حساب يدوي
0.731059
0.731059
< 0.001 %حتى الرقم الأخير
The application-note approximation for the same caseD 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.
الطريقة منIPC-2221B, Generic Standard on Printed Board Design, §6.2 (2012)
الدقةA curve fit to measurements on bare boards in still air. A plane underneath runs cooler; a sealed enclosure runs hotter. Use it as a starting width.
الحالة
نوع الدليل
المتوقع
المحسوب
الفرق
3 A at a 20 °C rise on an external 1 oz layerIPC-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.The hand figure is rounded to four places.
حساب يدوي
0.9029
0.9028
-0.0111 %±0.2 %
Doubling the current widens the trace about 2.6 timesThe 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.
Resistance of 50 mm of 1 mm wide 1 oz copper at 25 °CR = rho l/(w t), with rho = 1.724e-8 Ω m at 20 °C and alpha = 0.00393: 50 squares of 34.8 µm copper give about 24.8 mΩ at 25 °C.The hand figure rounds the copper thickness. Two per cent covers it.
الطريقة منIPC-2221B, Generic Standard on Printed Board Design, §6.2 (2012)
الدقةThe IPC-2221 trace curve applied to the barrel cross-section. It ignores the heat the barrel sheds into the planes it crosses, so the current it reports is conservative.
الحالة
نوع الدليل
المتوقع
المحسوب
الفرق
Copper in the barrel of a 0.3 mm hole plated 25 µmThe 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.
حساب يدوي
0.0255254
0.0255254
< 0.001 %حتى الرقم الأخير
Resistance of that barrel through a 1.6 mm board at 25 °CR = 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.The hand figure is rounded to six places.
الطريقة منHammerstad and Jensen, Accurate Models for Microstrip Computer-Aided Design (1980); Cohn, Characteristic Impedance of the Shielded-Strip Transmission Line (1954)
الدقةClosed-form fits to the exact static solution. Within their range, ε_r (quoted at one frequency, not yours) and etch tolerance move the answer more than model error does. Use the tolerance analysis to see the band.
الحالة
نوع الدليل
المتوقع
المحسوب
الفرق
50 Ω microstrip on 0.11 mm prepreg over a planeRule 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.
توقع هندسي
50
48.4
-3.2 %±12 %
A wider trace has a lower impedanceImpedance 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.
توقع هندسي
32
31.6
-1.25 %±25 %
Alumina, w/h = 1, zero copper thicknessHammerstad 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.
The cross-section of 18 AWGAWG 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.The panel and the wire tables both give three figures.
توقع هندسي
0.823
0.823
0 %±0.3 %
Resistance of 2 m of 18 AWG, there and back, at 20 °CR = 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.
Ring left by a 0.6 mm pad on a 0.3 mm holeHalf the difference of the diameters: (0.6 - 0.3)/2 = 0.15 mm exactly.
حساب يدوي
0.00015
0.00015
0 %حتى الرقم الأخير
The ring left when the drill lands 50 µm off centreThe 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.
حساب يدوي
0.0001
0.0001
< 0.001 %حتى الرقم الأخير
The smallest pad this hole can have and still pass the fabD + 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.
الطريقة منWheeler, Formulas for the Skin Effect, Proceedings of the IRE (1942); Hammerstad and Jensen, Accurate Models for Microstrip Computer-Aided Design (1980); Hammerstad and Bekkadal, A Microstrip Handbook, University of Trondheim
الدقةDielectric loss is exact for stripline. For microstrip the √ε_eff form leaves out the filling factor, so it overstates dielectric loss somewhat, on the safe side. Conductor loss uses Wheeler’s rule, which reproduces the exact coaxial result. Foil roughness and the loss tangent at your frequency move the answer most, so roughness is shown as a band and the verdict uses its rough end.
الحالة
نوع الدليل
المتوقع
المحسوب
الفرق
Skin depth in copper at 5 GHzd = 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.
حساب يدوي
9.34553e-7
9.34553e-7
< 0.001 %حتى الرقم الأخير
DC resistance of a 0.2 mm wide, 1 oz trace at 20 °CR = 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.
حساب يدوي
2.47701
2.47701
< 0.001 %حتى الرقم الأخير
Hammerstad roughness factor for 0.4 µm foil at 5 GHzK = 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.The panel shows three decimals, so the check cannot be tighter.
حساب يدوي
1.16
1.16
0 %±0.1 %
A stripline’s effective permittivity is its dielectric constantA 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.
حساب يدوي
4.3
4.3
0 %حتى الرقم الأخير
Dielectric loss of 8 inches of stripline at 2.5 GHza = 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.
حساب يدوي
1.91766
1.91766
< 0.001 %حتى الرقم الأخير
Against the 2.31 dB rule of thumb for dielectric lossRule 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.The rule is quoted to three figures, so a tighter band would test its rounding, not the physics.
الدقةExact for ideal parts. On a real board the capacitor is the main uncertainty: a class 2 ceramic loses much of its value under DC bias and over temperature, which moves the corner.
الحالة
نوع الدليل
المتوقع
المحسوب
الفرق
Cut-off of 10 k and 100 nFf = 1/(2 pi R C): 1/(2 pi x 10000 x 1e-7) = 159.1549431 Hz.
Resonance of 10 µH with 22 µFf = 1/(2 pi sqrt(LC)): sqrt(2.2e-10) = 1.4832397e-5, times 2 pi is 9.3194699e-5, reciprocal 10730.22 Hz.
حساب يدوي
10730.2
10730.2
< 0.001 %±0.001 %
Characteristic impedance of the same pairZ = sqrt(L/C): sqrt(10/22) = sqrt(0.4545454545) = 0.6741999 Ω.
حساب يدوي
0.6742
0.6742
< 0.001 %حتى الرقم الأخير
The capacitor that tunes 10 µH to 10 kHzC = 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.
الدقةA lumped model: N identical capacitors, each C, ESR and ESL in series. It ignores the plane pair, so it holds only below the first plane resonance.
الحالة
نوع الدليل
المتوقع
المحسوب
الفرق
Target impedance for 3 % of 3.3 V under a 0.5 A stepZ = dV/dI, with dV the allowed deviation: 3.3 x 0.03 = 0.099 V over 0.5 A = 0.198 Ω exactly.
حساب يدوي
0.198
0.198
< 0.001 %حتى الرقم الأخير
Self-resonance of 100 nF with 1.2 nH of loop inductancef = 1/(2 pi sqrt(L C)): sqrt(1.2e-16) = 1.0954451e-8, times 2 pi is 6.8828847e-8, reciprocal 14.5288 MHz.
One LSB of a 12-bit converter on a 3.3 V referenceLSB = V_FS/2^N: 3.3/4096 = 0.0008056640625 V exactly.
حساب يدوي
0.000805664
0.000805664
0 %حتى الرقم الأخير
The ideal signal-to-noise ratio of 12 bitsIdeal SNR = 6.02 N + 1.76 dB, from quantisation noise: 74 dB for 12 bits, the figure every data-conversion text gives.The panel shows one decimal, so the check cannot be tighter.
Load capacitors for a 12 pF crystal with 3 pF of strayThe 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.
حساب يدوي
1.8e-11
1.8e-11
0 %حتى الرقم الأخير
The load the crystal then seesTwo 18 pF in series is 9 pF. Add the 3 pF stray and the crystal sees 12 pF, its specified load.
حساب يدوي
1.2e-11
1.2e-11
0 %حتى الرقم الأخير
حاسبات لديها لقطة محفوظة فقط، حتى الآن (11)
هذه تطبّق طريقة موثّقة ومعظمها يذكرها، لكن لم يقارن أحد بعدُ مخرجاتها بمصدر خارج هذا المستودع. تعامل مع أرقامها كما تتعامل مع جدول زميل: صحيحة على الأرجح، وغير مؤكدة بصورة مستقلة. والمثال المحلول المنشور هو أثمن ما يمكن لأي أحد أن يساهم به هنا.