Audio Thai Community

ir2153 cal

0 สมาชิก และ 1 บุคคลทั่วไป กำลังดูหัวข้อนี้

ออฟไลน์ surachit

  • *****
  • 1985
    • ดูรายละเอียด
ir2153 cal
« เมื่อ: 13,มีนาคม, 2019, 02:31:46 »
Calculation of the relative gate resistance

As known, the dynamic properties of a field-effect transistor are not characterized more accurately by the value of its parasitic capacitances, but from the total charge of the gate-Qg. The Qg parameter value is mathematically interconnected by a pulse current with the gate of the transistor switching time, thus allowing the developer to properly calculate the control node.
Take for example the MOSFETs IRF840 very common and present in the table.

With a drain current Id = 8 A, una tensione drain-source Vds= 400 V and a gate-source voltage Vgs = 10 V, is the gate charge Qg = 63 nC.

It should be specified that with the same Vgs, the gate charge decreases with an increase of the drain current Id and with a decrease in the voltage Vds, in the calculations seen that both voltages are constant for good take the value provided by the manufacturer, small changes do not affect the final result of the calculations.
We will calculate the parameters of the control circuit, provided that it is necessary to reach the switch-on time of the transistor ton = 120 ns. To do this, the driver control current must have the value:

Ig= Qg/ ton= 63 x 10-9/ 120 x 10-9= 0.525 (A) (1)

When the amplitude of the control voltage pulses on the gate Vg = 15 V, the sum of the driver output resistance and the resistance of the limiting resistor must not exceed:

Rmax= Vg/ Ig= 15 / 0.525 = 29 (Ohm) (2)

We calculate the output resistance in cascade output of the driver for the chip IR2155:

Ron= Vcc/ Imax= 15V / 210mA = 71,43 ohm
Roff= Vcc/ Imax= 15V / 420mA = 35,71 ohm

Taking into account the value calculated according to the formula (2) Rmax = 29 Ohm, we conclude that the specified speed of the transistor IRF840 It can not be obtained by the driver IR2155. If in the gate circuit, a resistor Rg is installed = 22 ohm, the ignition time of the transistor is defined as follows:

REon= Rg+ Rf, dove

RE = total resistance

Rf = the driver output impedance,

Rg = external resistance in the gate circuit of the power transistor

Reon = + 71,43 = 93,43 ohm;
Ion= Vg/ RE, dove

Ion = the drive current

Vg = value of control gate voltage

Ion= 15 / 93,43 = 160mA;
ton= Qg/ Ion= 63 x 10-9 / 0,16 = 392 nS
The sleep time can be calculated with the above formulas:

REoff= Rf+ Rg= 35,71 + 22 = 57,71 ohm;

Ioff= Vg/ Reoff = 15/58 = 259mA

toff= Qg/ Ioff= 63 x 10-9 / 0,26 = 242nS
To obtain the value of the real time it is necessary to add the time that physically employs the transistor to pass from one stage to another and that is 40ns for the on condition, and 80ns for that will be off the real-time

Your 392 + 40 = 432nS, e Toff 242 + 80 = 322nS.

Now it remains to be determined whether a power transistor will have time to completely close before the second beginning to open. A tal fine, we add Ton and Toff to get 432 + 322 = 754 nS, equivalent to 0,754 μS.

From the data shows that the DEAD TIME the IR2151 can not be used as it is 0,6 μS.

In the datasheet it says that Deadtime (tip.) It is fixed and depends on the model, but there is also a very embarrassing figure from which it emerges that DEAD TIME and the 10% of the control pulse duration:

face หนุ่มกิโลสี่ บุญยัง  0866209426  
[email protected]

ออฟไลน์ surachit

  • *****
  • 1985
    • ดูรายละเอียด
Re: ir2153 cal
« ตอบกลับ #1 เมื่อ: 13,มีนาคม, 2019, 02:35:10 »
in the scheme 1, there is no extra function, and the secondary is formed by two bipolar power rectifiers consisting of a pair of double Schottky diodes. The ability to 220 uF output to the bridge is calculated with the empirical formula of 1 uF per watt on the load. In this case it is used for a stereo amplifier 100W per channel. The two capacitors 2u2 on the primary of the transformer are placed in the range from 1 a 2u2 .

The power depends on the transformer core and the maximum current of the power transistors and in theory can reach 1500 watt. Practically, in this scheme it depends on the maximum current of the transistor temperature STP10NK60Z, the maximum current of 10 A If you have only 25 degrees. When the temperature of the silicon salt to 100 degrees is reduced to 5,7A, and talk about the temperature of the silicon, rather than the temperature of the heatsink.
Thus the maximum power must be chosen according to the divided transistor current 3, if you feed a power amplifier and divided by 4, if one feeds a constant load, such as incandescent lamps.
That said you can theoretically power an amplifier

10/3 = 3,3A 3,3A x 155V = 511W totali.

For a constant load 10/4 = 2,5 A 2,5 A x 155V = 387W.

From calculations reference is made to a fixed voltage 155V, where it comes from that value? It is derived from the effective voltage on the smoothing capacitor at maximum power, the value is empirical but, It is not very different from the real value and allows us to simplify our lives without too great deviations from real.

In both cases, it theorizes a yield of 100%, that can not be reached .

Furthermore, wanting to obtain the maximum power of 1500W given the need to 1 uF of the primary power supply capacity for each watt of power on the load, It needs of one or more capacitors to get to 1500 uF Total and to load them should be a soft-start in order not to jump the counter at each switch.

« แก้ไขครั้งสุดท้าย: 13,มีนาคม, 2019, 02:40:57 โดย surachit »
face หนุ่มกิโลสี่ บุญยัง  0866209426  
[email protected]