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AN2986 датащи(PDF) 12 Page - STMicroelectronics |
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AN2986 датащи(HTML) 12 Page - STMicroelectronics |
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12 / 18 page ![]() Gate current variation AN2986 12/18 Doc ID 15727 Rev 1 2.3 Maximum current calculation The maximum current that the MCU can handle is based on the following opposite conditions: ● maximum supply voltage (VDDmax), ● minimum voltage drop of the gate junction (VGTmin), ● minimum gate resistor (0.99 x RG), ● minimum output low voltage (VOLmin). Equation 7 As previously stated, a load RMS current of 1 A for Z0103 and 0.8 A for ACS108-6T is used. Z01’s minimum VGT is then close to –0.1 V at 1.44 A peak load current in positive half-wave and for a 125 °C junction temperature (worst case). ACS108’s minimum VGT is close to 0.6 V for 1.15 A peak load current and for a 125 °C junction temperature (worst case). Two examples are shown. The same components are used as those given in Section 1.2.2. Example 5: MCU: STM8S with ACS108 -6T for power supply 5 V ±10% Specified values: RG = 180 Ω, VDD = 5 V, VGT min (0.8 A) = 0.6 V Calculated values: VDDmax = 1.1 x 5 V = 5.5 V, VOLmin IGT = 20 mA) = 0.8 V IGmax is then given by: Equation 8 This result exceeds the maximum MCU current capability for one high-sink I/O port. This IGmax current can be sunk with 2 I/O ports in parallel. Estimation of VOLmin has been provided for four high-sink I/O ports sinking at the same time 10 mA. VGTmin depends on load current. A lower load current means higher VGTmin and therefore lower IGmax. Example 6: STM32 + Z0103 for 3.3 V ±5% Specified values: RG = 300 Ω, VDD = 3.3 V, VGT min = –0.1 V Calculated values: VDD max = 1.05 x 3.3 V = 3.47 V VOLmin (IGT = 8 mA) = 0.4 V IG max is then given by: Equation 9 I Gma x V DDm ax V GTmin – V OLm in – 0.99 R G × ------------------------------------------------------------------------ < I Gm ax 5.5 0.6 – 0.8 – 0.99 180 × ------------------------------------- 23.2mA = < I Gmax 3.47 0.1 – () – 0.4 – 0.99 300 × ------------------------------------------------ 10.7mA = < |
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