Page 102 - Rashid, Power Electronics Handbook

88 I. Batarseh

From Eqs. (13) and (14), we obtain, v
GG
V GG ÿ v GS dv GS V (a)
; ¼ðC GS þ C GD Þ ð6:15Þ GG
R G dt

Solving Eq. (6.15) for v ðtÞ for t > t with v ðt Þ¼ 0, we t o
GS 0
0
GS
obtain,
v
GG
v ðtÞ¼ V ð1 ÿ e ðtÿt 0 Þ=t Þ ð6:16Þ v
GS GG GS
where
(b)
t ¼ R ðC þ C Þ V
G GS GD Th
The gate current i is given by
G
v GG ÿ v GS t o t 1 t 2 t 3 t
i ¼
G
R G
V GG ÿðtÿt 0 Þ=t ð6:17Þ i G
i ¼ e
G
R G
As long as v GS < V , i remains zero. At t ÿ t , v GS reaches (c)
D
1
Th
V , causing the MOSFET to start conducting. Waveforms for
Th
i and v GS are shown in Fig. 6.19. The time interval ðt ÿ t Þ is
1
G
0
given by - V GG - VTh
R G

V Th i D
Dt ¼ t ÿ t ¼ÿt ln 1 ÿ
10 1 0
V GG I
O
Dt 10 represents the ®rst delay interval in the turn-on process.
For t > t with v GS > V , the device starts conducting and (d)
Th
1
its drain current is given as a function of v GS and V . In fact,
Th
i D starts ¯owing exponentially from zero as shown in Fig.
6.19d. Assume the input transfer characteristics for the
v
MOSFET are limited as shown in Fig. 6.20 with the slope of DS
V
g given by DD
m
@i (e)
D
p 
I
@v GS 2 i DSS d I O r DS(ON)
g ¼ ¼ ð6:18Þ
m
I D V Th t t t t time
The drain current can be approximated as follows: FIGURE 6.19 Turn-on waveform switching.
i ðtÞ¼ g ðv GS ÿ V Þ ð6:19Þ
Th
m
D
For t > t the diode turns off and i I is as shown in Fig.
D
2
0
As long as i ðtÞ < I , D remains on and v DS ¼ V DD are as 6.18d. As the drain-current is nearly a constant, then the gate-
0
D
shown in Fig. 6.18c. source voltage is also constant according to the input transfer
The equation for v ðtÞ remains the same as in Eq. (6.16),
GS characteristic of the MOSFET, that is,
hence, Eq. (6.19) results in i ðtÞ given by
D
i ðtÞ¼ g ðV GG ÿ V Þÿ g V GG e ÿðtÿt 1 Þ=t ð6:20Þ
D
Th
m
m
i ¼ g ðv
D m GS ÿ V Þ I 0 ð6:22Þ
Th
The gate current continues to decrease exponentially as shown
in Fig. 6.19. At t ¼ t , i reaches its maximum value of I ,
2
D
0
turning D off. The time interval Dt 21 ¼ðt ÿ t Þ is obtained Hence,
1
2
from Eq. (20) by setting i ðt Þ¼ I .
D
0
2
g V GG I 0
m
Dt ¼ tLn ð6:21Þ v ðtÞ¼ þ V ð6:23Þ
21 GS Th
gmðV GG ÿ V Þÿ I 0 g m
Th