Does a choke always store some residual energy for immediate use even if the DC passing through it is already ripple free? An advantage of a choke is its stored energy can be delivered fast, I guess. So In a power supply if I replace a choke with a series pass transistor and CCS control regulator to form a "solid state choke"... I have no ripple anymore, but I lost my choke. If I now install a choke after the pass transistor do I regain any "stored energy"? Or is the choke now just acting like a big wire wound resistor because I got rid of all the ripple already? Excuse my noobyness, but intuitively I can't figure out if a choke needs both DC and pulsing DC components to store energy.
What a choke or inductor does is try to keep current flow the same.
If there is any change in current flow, it will counter the change with an opposing voltage.
So if your CCS works 100% then it will only add impedance.
If there is any change in current flow, it will counter the change with an opposing voltage.
So if your CCS works 100% then it will only add impedance.
Yes, for a DC [steady state] current of I, the energy stored in an inductor is L x I x I /2
For AC [sine] signals, the inductor is considered to have a steady state impedance of 2 X Pi x f x L
For transient [nonrepetitive] signals, differential equations are necessary to calculate the circuit behavior.
A series inductor after a regulator would raise the output impedance, which is not usually needed or wanted.
The larger the L value, the slower the current through the inductor will be able to change, with all else equal.
For AC [sine] signals, the inductor is considered to have a steady state impedance of 2 X Pi x f x L
For transient [nonrepetitive] signals, differential equations are necessary to calculate the circuit behavior.
A series inductor after a regulator would raise the output impedance, which is not usually needed or wanted.
The larger the L value, the slower the current through the inductor will be able to change, with all else equal.
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Thanks, it makes sense not to add it now to a power supply with a Transistor pass regulator. But You do see anode chokes in a place where the DC is already well filtered and maybe even delivered by a regulated PS. in this anode use of a choke the added impedance at the plate would be for what reason? Or is the reason at the anode to store /release energy demands fast? I have not learned the reasons for anode chokes, probably because I don't see them much. When I do see anode chokes in circuits they seem to have as high mH as they can afford for the mA needed, is this typical?
Take any "typical" resistor-loaded voltage amplifier, and some reasonable bass limit.
Say 100k (hi-gain tube) and 20Hz.
Simple math tells us this needs 833 Henries. 20Hz*6.28*833H = 100k.
Oh, we could load the tube with say 83H, but gain would fall below 200Hz, which is not "hi-fi".
While one 20Hz bass-cut in a system may be OK, a typical system has many bass-cuts. Three -3dB@20Hz networks makes about -3dB@40Hz, which is getting into the woofer's range.
Getting hundreds of Henries means either many-many-many turns of fine wire (and low current), or a huge core (but that adds capacitance which hurts the highs). Since 1933 this has been "the expensive way to do it".
In general the anode choke allows maximal output swing, almost mu. It does so by presenting a high impedance load to the tube. A resistive load can't deliver the same amplification and keep up with low distortion. When you observe the anode characteristics of a tube and draw your load line on the graphic, you end up either in the curvature at low anode currents or in high anode current at low grid voltage. The high impedance load line would resemble an almost horizontal line, avoiding the difficulties at either side of the tubes work point.
Because anode chokes have properties like real resistance and parasitic capacitance it takes great effort to arrive at usable specs, what makes them costly.
No. The energy storage of a choke is usually a nuisance, except in some SMPS. In conventional PSUs the choke is merely being used as a frequency-dependent impedance: low for DC and high for AC.Windcrest77 said:
DF96 puts it succinctly.
Think of a choke as a large (mechanical, hundred kilo!) flywheel on a stout but thin axle. Imagine that with your hands only on the shaft, you try to change the rotational speed of the flywheel. Trying hard, you can speed it up, or slow it down. The flywheel's rotational momentum is akin to a choke's current flow. The mass of the flywheel is likewise parallel to a choke's “henries” rating.
However, spinning (or not), you can see that a flywheel in 'series' with a power supply and a power consuming device acts kind of the opposite of what you might want: sudden increases in power demand are met with sudden decreases in force as the big ol' flywheel can't (actually won't) instantly speed up or slow down in response.
It is this fact that makes the choke very useful for power supplies to suppress the bump-bump-bump output of the rectifier stack 'pushing' the choke. The output side is much smoothed, in average, by the flywheel/choke. To counteract the opposite-of-what-you-really-want problem (pressure droop on instantaneous power demand), a reservoir capacitor is always placed downstream. In the analogy of flywheels, it is like an equally large spring under high pressure. Demand from the down-wind devices is met almost instantly by the spring. The flywheel catches up more slowly, while still suppressing the bump-bump-bump of the rectifier's output.
The most significant problem with chokes and capacitors is that they also form low-frequency resonant (“tank”) circuits. In the flywheel-and-spring case, it isn't hard to imaging that when the downwind power load suddenly (and for the duration) ceases, the flywheel will continue to fly! Where does the rotary momentum go? … into the spring, raising its tension, “pushing back” against the flywheel, which slows down. However, at the end, when the flywheel stops, the spring is under much higher tension, and is still pushing against the flywheel. So, the flywheel will reverse direction. And so on and so forth.
Thus the pair(s) form a resonant circuit, sloshing momentum back and forth between the springs and the flywheel(s) until the friction (resistance) of the circuit absorbs the momentum (current flow), and everything comes to a rest (quiescence).
Hope that helps. Its always helped me visualize what chokes and capacitors do.
Just saying,
GoatGuy ✓
Think of a choke as a large (mechanical, hundred kilo!) flywheel on a stout but thin axle. Imagine that with your hands only on the shaft, you try to change the rotational speed of the flywheel. Trying hard, you can speed it up, or slow it down. The flywheel's rotational momentum is akin to a choke's current flow. The mass of the flywheel is likewise parallel to a choke's “henries” rating.
However, spinning (or not), you can see that a flywheel in 'series' with a power supply and a power consuming device acts kind of the opposite of what you might want: sudden increases in power demand are met with sudden decreases in force as the big ol' flywheel can't (actually won't) instantly speed up or slow down in response.
It is this fact that makes the choke very useful for power supplies to suppress the bump-bump-bump output of the rectifier stack 'pushing' the choke. The output side is much smoothed, in average, by the flywheel/choke. To counteract the opposite-of-what-you-really-want problem (pressure droop on instantaneous power demand), a reservoir capacitor is always placed downstream. In the analogy of flywheels, it is like an equally large spring under high pressure. Demand from the down-wind devices is met almost instantly by the spring. The flywheel catches up more slowly, while still suppressing the bump-bump-bump of the rectifier's output.
The most significant problem with chokes and capacitors is that they also form low-frequency resonant (“tank”) circuits. In the flywheel-and-spring case, it isn't hard to imaging that when the downwind power load suddenly (and for the duration) ceases, the flywheel will continue to fly! Where does the rotary momentum go? … into the spring, raising its tension, “pushing back” against the flywheel, which slows down. However, at the end, when the flywheel stops, the spring is under much higher tension, and is still pushing against the flywheel. So, the flywheel will reverse direction. And so on and so forth.
Thus the pair(s) form a resonant circuit, sloshing momentum back and forth between the springs and the flywheel(s) until the friction (resistance) of the circuit absorbs the momentum (current flow), and everything comes to a rest (quiescence).
Hope that helps. Its always helped me visualize what chokes and capacitors do.
Just saying,
GoatGuy ✓
I design embedded controllers for diesel engines... So yes this analogy helps a lot!
The energy stored in an inductor depends only on the (instantaneous) current (squared).
This energy can be delivered extremely fast, in the form of a high voltage pulse if you try to interrupt the current flow suddenly. Lethal voltages can be produced from big chokes (even if used low voltage supply), they are not to be messed with idly.
A choke in an C-L-C filter never sees sudden interruptions in current (the caps can always absorb the current change), so the power transfers are smooth, but resonance is an issue to worry about (normally the load damps any oscillation if the filter is correctly designed).
Power supplies: Choke input gives 0.9 x the applied rms ACV.
Plate load: The tube, bias, and plate volts set the quiescent current in the choke.
1. The laminations of a choke that has current flowing in the windings Stores Magnetic Energy. The choke is a 'Load' of one polarity.
2. Reduce or shut off that current, and the magnetic energy collapses, which causes the choke to be a 'Source' of the opposite polarity, which causes the current to continue.
1, and 2, are what causes the choke to act as a constant current device.
As with all parts, there are parasitics. In the case of a choke, the parasitics are distributed capacitance, and DC resistance (and core losses when any external current is varied).
Here is an example of a power supply with a CLC filter that is resonant:
Input C = 0.5uf; L = 5H; output C = 200uF. The loop of CLC is: 0.5uF and 200uF are in series, and that is Very close to 0.5uF. Seems like a good design, right? Wrong . . .
Resonance = 1/(2 x pi x Root of (C x L)). Resonance is at 100.7Hz, which is very close to the 100Hz frequency of full wave rectified 50Hz Mains. Not so good. Resonance not only is dependent on the load on the output of CLC, it is dependent on the input C and L of the CLC. Check your power supplies.
Plate load: The tube, bias, and plate volts set the quiescent current in the choke.
1. The laminations of a choke that has current flowing in the windings Stores Magnetic Energy. The choke is a 'Load' of one polarity.
2. Reduce or shut off that current, and the magnetic energy collapses, which causes the choke to be a 'Source' of the opposite polarity, which causes the current to continue.
1, and 2, are what causes the choke to act as a constant current device.
As with all parts, there are parasitics. In the case of a choke, the parasitics are distributed capacitance, and DC resistance (and core losses when any external current is varied).
Here is an example of a power supply with a CLC filter that is resonant:
Input C = 0.5uf; L = 5H; output C = 200uF. The loop of CLC is: 0.5uF and 200uF are in series, and that is Very close to 0.5uF. Seems like a good design, right? Wrong . . .
Resonance = 1/(2 x pi x Root of (C x L)). Resonance is at 100.7Hz, which is very close to the 100Hz frequency of full wave rectified 50Hz Mains. Not so good. Resonance not only is dependent on the load on the output of CLC, it is dependent on the input C and L of the CLC. Check your power supplies.
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