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I am trying to learn about Buck converters from the below page.

https://www.allaboutcircuits.com/technical-articles/analysis-of-four-dc-dc-converters-in-equilibrium/

Below line is confusing me completely

from the steady-state perspective, magnitude of the inductor current increment during switch on is equal to the inductor current decrement during switch off; i.e. |(Imax−Imin)SWITCH−ON|=|(Imax−Imin)SWITCH−OFF|. ⇒|VS−VOLDT|=|−VOL(1−DT)| ⇒VO=DVS

I have few questions regarding this .

1)May I know what you mean by steady state here

2)How inductor current becomes (rising and falling) become equal during steady state.

Please reply

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  • \$\begingroup\$ Steady state means DC steady state. All the DC parameters remain constant. Vin, Iin, Vout and Iout are all constant (in the DC sense). There will always be ripple voltage and current in a buck converter. DC steady state does not mean there is no ripple, of course. \$\endgroup\$ – mkeith Apr 13 at 18:32
  • \$\begingroup\$ This is continuous mode where V+ =LdI/dt then the diode switches on to reverse the current slope as the average supplies the load current given a steady output voltage and load But startup is far different as well as load changes, so PWM must change as rapid as the load and input limit. \$\endgroup\$ – Tony Stewart EE75 Apr 13 at 20:01
  • \$\begingroup\$ If it falls a different amount than it rises, then it's not steady-state! \$\endgroup\$ – user253751 Apr 13 at 20:07
  • \$\begingroup\$ Unless Vin-Vout=Vout-Vf(diode) the +/- dI/dt ‘s are not equal (ignoring Losses) yet fig. Shows them equal which may be misleading. It’s the amplitude swings that are equal which change with duty cycle, \$\endgroup\$ – Tony Stewart EE75 Apr 13 at 20:19
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Take a look at figure 6 on that page (I've added the red lines): -

enter image description here

So, when the input supply voltage is constant and the output voltage and load current is constant (i.e. the circuit is operating in "steady state"), the two current changes (during inductor charge and discharge) are equal.

enter image description here

This mode of operation is called continuous conduction mode or CCM in case you didn't know.

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  • \$\begingroup\$ Notice that the +/- slopes will not be equal if I(min) actually repeated at the same level as the previous cycle. The + slope dI/dt =V(L)/L. Changes to the - dI/dt because the voltage drop on inductor to input to output is not the same as from Vf of diode thru inductor to output so here it was drawn wrong with the same slope so the i(min) rose a bit on the next cycle for the inductor, but “close enuf for gov’t work.” \$\endgroup\$ – Tony Stewart EE75 Apr 13 at 20:14
  • \$\begingroup\$ When the switch is ON the voltage across the inductor is (Vs-Vo) and when the switch is off the voltage is -Vs.Then may I know how the currents become equal. \$\endgroup\$ – HARI T O Apr 14 at 2:47
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    \$\begingroup\$ @Hari it's the change in current that is equal. \$\endgroup\$ – Andy aka Apr 14 at 6:04
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  1. Here, "steady state" means periodic steady state. In other words, all node voltages and branches currents have waveforms of the form f(t) = f(t+kT), where k is an integer and T is the period.
  1. During the periodic steady state, the (ideal) inductor current must increase linearly when a positive voltage is being applied and decrease linearly when a negative voltage is being applied. The current ramp rates are different because the positive and negative voltage magnitudes are different. But by definition, a periodic steady state requires that the start point and endpoint of one period are identical, which in turn forces the condition that the positive ramp and negative ramp must have the same change in magnitude.
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In this context, steady state means that the circuit has completed the start up phase, and if left undisturbed, it will continue to operate in the future as it does now. In other words, the behavior of the circuit will look the same now, 10 seconds from now, or 10 hours from now. Of course, a circuit will not in reality continue to operate the same way forever (it can overheat, components can fail, input power might fluctuate). However, steady state analysis is a useful way to think about the ideal operation of a system.

There will always be some ripple on the output caused by the increasing current when the fet is on and a decreasing current when the fet is off and the load is dissipating power. If the increasing current equals the decreasing current, the circuit is in steady state. If the current increment was larger than the current decrement, for example, the voltage at the output would be rising, and the circuit would not be in steady state.

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  • \$\begingroup\$ A significant change in load or input voltage results in a change of duty cycle and some period to adjust to a new steady state as well. Hmm on second thought that only applies to regulated buck converters. \$\endgroup\$ – K H Apr 14 at 8:38

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