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In general, it can be assumed that there will be a short delay
time before the device begins to turn ON. This delay time is
approximately
( 12-9)
has an inverted polarity to that of the input.
EXAMPLE 1:
A typical JFET inverter is shown in Fig. I 2-10. Compute
the rise, fall, and delay times given the device parameters belo w
which is necessary to allow the gate-source voltage to reach zero
volts and the device to be fully ON. After this delay, the output
voltage will fall to V 0 s(on) with the time
tr = 2.2 C 1 d IR,
+rd.
(on) I
rds(on)
@
lo = OmA = I 00 n
Cgs
@
1.0 MHz
Cgd
@
1.0 MHz = !OpF
(12-10)
= ISpF
which is the same as the rise-time.
Since the MOSFET has nearly the same equivalent circuit as
the JFET, except for the input voltage polarity, then the
expressions for the rise and fall-times are approximately identical
to those in Eqs. I 2-8 and 12-1 J. It must be remembered, though,
that the MOSFET turns ON with a positive voltage and turns
OFF with a zero or negative voltage.
I
OFF
/Ov +
Fig. 12-10. Example of a JFET Inverter.
I~
The rise and fall-times are given in Eqs. 12-8 and I 2-1 0 as
t.,r = 2.2 C 11 d
ON
lrdl ,, I
Fig. 12-9. Input and Output Voltage Waveforms for the MOSFET
Inverter.
Fig. I 2-8 shows typical input and output waveforms for the
JFET inverter and Fig. 12-9 shows the same for the MOSFET
circuit. The switch movements are assumed to have zero rise and
fall-times. Notice that for each set of waveforms the output pulse
ADAPTING
-l
IR, + rd 5 (on))
tr.f
=2.2· 10 · 10- 12
tr,f
= 2.2 JJSec
•
[100· 10 3 + 100) sec
The delay time is given by Eq. 12-9 as
~
= 2.2 C 11 , R 11
td
= 2.2 • 15 • 10- 12
td
= 3.3 JJseC
•
100 · 10 3 sec
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PLAY METER, June 1, 1985