The Turn-on Transition

Since Vgs Is less than Vt, Drain Current Is zero, and Drain Voltage is fixed at Vin. 

Since Drain Voltage is fixed, there is almost no current injected into Gate (through Cgd) attributable to any variation of Vd. 

There is a small Cgd-current due to the increasing Vgs. But this is factored in by the use of Cg=Cgd+Cgs instead ofjust Cgs in the time constant Tg. 

Vgs continues exactly as in t1, it we assume Lik/Rdrive is very small. 

There is a small voltage spike on Vd node, as determined by V=Ld(ld)/dt, because voltage at the switching node is clamped. There is thus only a very small current injected through Cgd, and only a small perturbation in Vgs, which we ignore. 

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This results in watts for the loss seen only during the turn-on transitions of the power switch. One would add the turn-off and conduction loss to this amount to arrive at the total loss within the power switch. 

What are the reasons for clock instability High-frequency noise is always generated at turn-on and turnoff in any switcher. This noise can infiltrate into the IC via various pins. It can be very hard to filter out and control. You may need to ultimately simply avoid turning the Fet OFF too dramatically. In most switchers, the turn-on transition is traditionally delayed (or slowed) just a little, so as to allow the output/catch diodes to recover... 

In Figure 5-5, we first consider the turn-on transition (on the left). Just prior to this, the diode is obviously carrying the full inductor current (circled 1 ). Then the switch starts to turn ON, trying to share some of this inductor current (circled 2 ). The diode current therefore must fall correspondingly (circled 3 ). However, the important point is that while the switch... 

This is the correct result for the energy lost in the switch, during a resistive turn-on transition. 

Now, suppose we had ramped up the gate voltage at a rate of 1 V per second as before, but ramped down faster, say, at the rate of 2 V per second. Then the turn-on time and the turn-off transition times would be different. So in that case we need to split up the crossover... 

Note Here we seem to be indirectly suggesting that the drive resistance is the same for tum-on and turn-off. That need not be so. All the equations we will present can easily take any existing difference in the turn-on and turn-off drive resistances into account. So in general, we will have different crossover times for the turn-on and turn-off transitions. Also note, that in general, within a certain crossover interval (tum-on or turn-off), the actual time it takes for the voltage to transit need not be the same as the time the current takes (unlike the case of a resistive load). 

Finally, with all this background information, we can start looking closely at what actually happens during the turn-on and turn-off transitions. 

With help of the four-level diagram of the =I= system (see figure BL15.8 two conniion ways for recording ELDOR spectra will be illnstrated. In freqnency-swept ELDOR the magnetic field is set at a value that satisfies the resonance condition for one of the two EPR transitions, e.g. 4<- 2, at the fixed observe klystron frequency, The pump klystron is then turned on and its frequency, is swept. When the pump... 

ZINDO/I is based on a modified version of the intermediate neglect of differential overlap (INOOh which was developed by Michael /ern cr of the Quan turn Th cory Project at th e Lin iversity oIFIorida. Zerner s original IXDO/1 used the Slater orbital exponents with a distance dependence for the first rorv transition metals only. (See Thvorel. Chirn. Ada (Bed.) 53, 21-54 (1979).) However, in HyperChem con stan I orbital expon en ts are used for all the available elements, as recommended by. Anderson, Kdwards, and /.erner, /norg. Chern. 25, 2728-2732,1986,... 

The fact that the probability of excitation from i to f grows linearly with the time T over which the light source is turned on implies that the rate of transitions between these two states is constant and given by ... 

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