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AN1353 データシート(PDF) 7 Page - STMicroelectronics |
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AN1353 データシート(HTML) 7 Page - STMicroelectronics |
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7 / 17 page ![]() AN1353 Gate current pulses Rev 2 7/17 When, Tx can be calculated as follows: Dividing by 256 is thus easily achieved simply by considering the Most Significant Byte of the multiplication result of T50Hz and Dx.The Dx variables are defined in the Constants list at the beginning of the software. To implement new pulse timings, only the values in this list have to be changed. For 60 Hz applications, the “10ms” used in the calculation should be replaced by 8.33. 4.3 Influence of the code time In fact, the timer is not launched exactly when it has to be. This is due to the time required by the MCU to perform some instructions between the last interrupt and the effective decrementation beginning. For example, before the launch of the first T1 decrementation, the program must run the NMI interrupt, save the DELTAT result, write PORT A and the start the timer. These 47 instructions all in all last 200 µs for a 4 MHz MCU clock frequency. Then, to ensure that the gate current will be applied on the Tr and Ts de-vices at the right time, it is better to cut off this delay from T1. Furthermore, this delay will vary depending on the MCU clock frequency (fcpu), which will vary according to the junction temperature and the supply voltage level. So, a method is required to remove the code delay whatever fcpu is. First, we know that the oscillator frequency is divided by 13 to drive the CPU core. Therefore, “N” CPU cy-cles last 13xN/fcpu seconds. Secondly, the CPU oscillator frequency is divided by 12 to drive the Timer, and then divided by the division factor programmed in the TSCR register. In our case, the division factor is 32 during the main program loop. Therefore, one unit timer counter equals 12x32/fcpu. Based on the previous formula, we can easily convert N cycles code execution time to the timer counter value (Tcode), as shown below. For example, the length of the code time for 47 cycles approximately equals one unit of the timer counter. Then, one must be subtracted from T1 in order to rectify the gate current pulse delay from the code execution delay. For T2 and T3, it does not matter if the code execution delay is removed or not. Indeed, it is not a problem that the gate current pulses last longer. For T4, it is important to begin the current measure at the right time. The 98 cycles must then be subtracted from T4. This leads to subtract 3 from the TSCR register value. When the half-cycle duration is calculated, the code execution time must also be added to T50Hz. As 110 cycles are performed, 4 must be added to the T50Hz value. Tx = T50Hz x Dx 256 x Tcode = = N x N x 13 13 12 x 13 fcpu 384 fcpu |
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