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سجل التغييرات الكاملالمحرّك 1.26.4

Logic & Output Power, Duty Cycle

Thermal

Driver IC dissipation from its logic and output stages, and the duty cycle the package allows.

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المدخلات

ملف اللوحة

Fills Ambient temperature وJunction temperature limit from your board profile.

V
A
25 mA
V
A
250 mA
%
°C/W
°C
°C
OFF-state losses
A
7.5 mA
V
A
100 µA

النتائج

1 ناجحة · 1 للمراجعة
Average power at D = 41 %
618.8mW
ناجح

المعيار: PD ≤ 1.25 W, the allowable average at TA = 25 °C
Allowable average power PD
1.25W
Junction temperature
86.9°C
كل النتائج (4)
Logic power, ON PI
Output power, ON PO
Instantaneous ON power PON
Instantaneous OFF power POFF
PD = D·PON + (1−D)·POFF

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من أين تأتي الطريقة
  • Allegro MicroSystems AN295014

Allegro MicroSystems AN295014, Computing IC Temperature Rise المعادلة

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Waveform

المبدأ

A driver IC dissipates power in its logic and in its output stage. Add the two for the ON state and for the OFF state, then average over the duty cycle. For short pulses, that average sets the junction temperature.

Logic-gate (supply) power
PI=n (VCCICC)P_I = n\,(V_{CC} I_{CC})
Use maximum datasheet values for the worst case.
Output-stage power
PO=n (VCE(sat)IC)P_O = n\,(V_{CE(sat)} I_{C})
Total instantaneous power
PON=PI+POPOFF=PI(off)+PO(off)P_{ON} = P_I + P_O \qquad P_{OFF} = P_{I(off)} + P_{O(off)}
The OFF-state output term is leakage times OFF-state voltage: small, but not negligible at high voltage.
Duty-cycle averaged power
PD=D PON+(1−D) POFFP_D = D\,P_{ON} + (1-D)\,P_{OFF}
Rearranged for the allowable duty cycle
D=PD−POFFPON−POFFD = \frac{P_D - P_{OFF}}{P_{ON} - P_{OFF}}
Here PD is the allowable average, (TJ(max) − TA)/Rθ.
  • nnnumber of gates or channels
  • VCC, ICCV_{CC},\,I_{CC}logic supply voltage and ON-state supply current
  • VCE(sat)V_{CE(sat)}output saturation voltage at the load current
  • ICI_{C}output load current
  • DDduty cycle, 0…1
  • PON,POFFP_{ON},P_{OFF}instantaneous dissipation in each state
تفاصيل إضافية

Average versus peak

The package's thermal mass averages short pulses, so the junction follows the average power. That is why a low duty cycle allows a peak far above what the package could sustain continuously.

Only for short pulses. Averaging holds while each ON time is short against the package's thermal time constant. The classic rule of thumb for a small IC is under about 0.5 s. Longer, and the junction reaches steady state within one pulse, so design for the peak power.

Modern parts

For a MOSFET output, use ID²·RDS(on) in place of VCE(sat)·IC. The same averaging applies to PWM LED, solenoid, relay and motor drivers. The thermal time constant grows with package size, copper and heat sinking.

Worked example, how hard may I drive this part?
A four-channel power driver, R_θ = 100 °C/W, ambient 85 °C, 250 mA per output. Step 1, what the package can shed, on average: P_D(allowed) = (150 − 85)/100 = 0.65 W Step 2, what it burns while ON: P_I = 4 × (5.5 V × 26.5 mA) = 583 mW P_O = 4 × (0.7 V × 250 mA) = 700 mW P_ON = 1.283 W ← twice the allowance Step 3, what it burns while OFF (leakage is not zero): P_I = 4 × (5.5 V × 7.5 mA) = 165 mW P_O = 4 × (100 V × 0.1 mA) = 40 mW P_OFF = 205 mW Step 4, solve the averaging equation for D: D = (0.65 − 0.205)/(1.283 − 0.205) = 41 % The part cannot run continuously at this load, but it can run 41 % of the time, if each ON pulse stays under about 0.5 s.
Method and example parameters adapted from Allegro MicroSystems AN295014, Computing IC Temperature Rise, and its safe-operating-limit examples. The explanations are our own.

إصدار المحرك ⁨1.26.4⁩