Increase filters on Rectifier tubes?

[SoulFetish]

??
At f(0) L is a short cirucuit, C is an open circuit.
At F(∞) L is an open circuit, C is a short circuit.

good catch! It's so "101" that I read it so fast, I didn't even realize that it was completely wrong
I'm guessing that he meant it the other way as well.

[Chuck H]

Probably the best of these situations is to have a small-to-moderate first filter cap, then some impedance, like maybe an inductor or a resistor, then a second and larger filter cap.

You mean, like a pi filter ?

Actually, I guess, a proper pi filter uses an inductor.?. But I've read about the circuit using a resistor as a pi filter for so long that I no longer discriminate.

[R.G.]

Yes, a pi filter with an inductor, but the thinking says that the first filter cap is small, and can't suppress all the ripple you want. It makes DC, but ripple-y DC. The inductor conducts DC all the time, and it's inductance is a high impedance for AC, so the pulses of incoming DC to the first filter cap don't get through the inductor well, and can't "reach" the second and possibly much bigger filter capacitor.

Another way to look at it is that a very large first filter cap is split into two parts, one a smaller first part that the rectifiers "see", and a bigger second part that the rectifiers are "protected from" by the inductor - or resistor.

The only difference between inductor and resistor is that (1) the inductor produces a much lower DC drop for the amount of smoothing it gives and (2) the inductor's impedance gives better and better filtering as frequency goes up. It's number 1 that makes it valuable in a power supply.

The inductance is different from the normal choke in a guitar amp. It's lower inductance (to keep the size and cost down) and has a bigger gap to tolerate the amount of DC that has to flow through it better.

[Old Tele man]

Sorry, it's function not frequency, ie: ƒ(t) at t=0 and t=∞, thus:

ƒ(0) is "...function at time zero..." and ƒ(∞) is "...function at infinite time..."

[Dave H]

Sorry, it's function not frequency, ie: ƒ(t) at t=0 and t=∞, thus:

ƒ(0) is "...function at time zero..." and ƒ(∞) is "...function at infinite time..."

Well, I'm feeling particularly thick today so you are going to have to explain it a lot better than that.

OK so it looks like f(0) is the value of some function f(t) when t is equal to 0 but what is the function f(t)?

[SoulFetish]

Sorry, it's function not frequency, ie: ƒ(t) at t=0 and t=∞, thus:

ƒ(0) is "...function at time zero..." and ƒ(∞) is "...function at infinite time..."

Boooooo.


(I'm just kiddin' Tele. I have no problem with this, I just wanted to give you a hard time is all )

[Old Tele man]

Circuit Theory 101:

ƒ_(t) = [ƒ₍₀₎ - ƒ_(∞)]·ε^(-α·t) + ƒ_(∞)

where: α = 1/RC or R/L and t = time (0 and ∞)

■ Make voltage-sources "shorts" and current-sources "opens."

At ƒ₍₀₎ L is open-circuit, C is short-circuit, because:
• current thru L is a current-source, but no current-flow yet, so make it "open."
• voltage across C is voltage-source, but no voltage-charge yet, so make it "short."

At ƒ_(∞) L is short-circuit, C is open-circuit, because:
• current thru L yields v = L·di/dt, a voltage-source, so make it "short."
• voltage across C yields i = C·dv/dt, a current-source, so make it "open."

The above summarizes what *happens* in RLC circuit at initial turn-on (time ZERO) and steady-state (time INFINITY)...in the OP context of power supply.

In other words, a simple "short-cut" way to VISUALIZE an RLC circuit at the two operating states: (a) turn-on and (b) steady-state.

[Dave H]

Thanks OTM. I get it now