DC-DC step down converter working principles

Thread Starter

nqchanh194

Joined Jul 5, 2013
27
I've read some materials about the working principle of the DC-DC step down converter, for example this link , especially section "What does steady-state mean?". I haven't understood based upon which theories the report concludes about the Steady State definition, or it's just what the designer tries to control so as to reach it.
Hope to getting comments from you.
 

MrAl

Joined Jun 17, 2014
11,342
I've read some materials about the working principle of the DC-DC step down converter, for example this link , especially section "What does steady-state mean?". I haven't understood based upon which theories the report concludes about the Steady State definition, or it's just what the designer tries to control so as to reach it.
Hope to getting comments from you.
Hi,

The simplest definition of steady state response is the response after all the one-time-only transients have died down.

In most circuits there are turn on transients that only occur when the circuit is coming up from zero energy input. These transients die down after a while because they are usually in a form with a multiplier like e^(-a*t) so after a certain time t anything that is multiplied by that exponential goes almost to zero, leaving the other terms to dominate.

For example:
v(t)=2*sin(w*t)*e^(-t/2)+5

After about 10 seconds the exponential becomes small, and so the first term becomes much smaller than the second term. In fact, the first term becomes about 0.013v peak and the second is always 5, so the peak would be 5.013 volts. But after a little longer the first term becomes even smaller, so v(t) becomes very close to 5 volts. After an even longer time, it stays at 5v meaning all the one-time-only transients have died down and we are left with a constant 5 volts.

The result doesnt always end up being a constant however. Consider this example:
v(t)=2*sin(w*t)*e^(-t/2)+10*cos(w*t)

Here we have the same first term, so it dies off after about 10 seconds or maybe a little more. Then we are left with only the second term:
10*cos(w*t)

and that varies forever. So this time we are left with a changing signal but the part that was a factor of the exponential died out so that part does not influence the response any longer. So the steady state for this example is 10*cos(w*t) which is a sinusoidal wave.

It's also possible that the exponential term reaches a steady value instead of going to zero as in:
v(t)=5*(1-e^(-t/3))+10*sin(w*t)

where after about 15 seconds we have approximately:
5+sin(w*t)

because the first term went to a constant 5 volts.

Still yet it is possible for the exponential term to reach a constantly changing value as in:
v(t)=5*cos(w*t)*(1-e^(-t/2))+10*sin(w*t)

where after about 10 seconds we end up with approximately:
5*cos(w*t)+10*sin(w*t)

In all of these cases after we reach the steady state anything left that changes changes in a constant way so as to form a cyclic response. The cyclic response is often easier to analyze so understanding the steady state response helps in many cases.

The true definition means time must be forced to go to infinity, but often there is a much shorter time where all those temporary exponentials die out.
 
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