Does this concept work: adding impedance in the form of a tuned parallel circuit in order to tame the current draw of a charging capacitor?
I think the harmonics are produced by capacitor charge spikes. I've seen this on this type of circuit before with a spectrum analyser.
The purpose is to find an efficient way to stop the many harmonics this sort of rectifying circuit produces.
The tank circuit might add ringing (?) but perhaps this would be easier to filter out than the big cap effect.
This is not for a specific circuit right now, just a concept. But down the road I want to power 30w 60v LEDs.



I'm not good with simulators.

Thanks, Mark

 

What are you actually seeing - say, on a scope - and what do you want to see instead?

 

By blocking the DC component of the rectifier output you have completely defeated the original purpose of making a DC power supply. The spikes you are seeing may or may not be real. They are sometimes artifacts of uncompensated scope probes.

 

By blocking the DC component of the rectifier output you have completely defeated the original purpose of making a DC power supply. the spikes you are seeing may or may not be real. They are sometimes artifacts of uncompensated scope probes.

So this would completely block the DC pulses? The circuit I drew is theoretical. I was thinking perhaps a tuned circuit that was not in perfect resonance would allow some current through, depending on how far out of resonance the tank circuit was. But I think what you're saying is the resulting current would be sine. Would a diode between the tank circuit and the smoothing cap work?

 

What are you actually seeing - say, on a scope - and what do you want to see instead?

Sorry for the confusion - I'll edit the post - what I drew is just to explore the theory, it's not the circuit that is actually producing the harmonic which I can see on a spectrum analyser - which is a much more complex circuit - but I think the current spikes come from the smoothing cap.

 

So this would completely block the DC pulses? The circuit I drew is theoretical. I was thinking perhaps a tuned circuit that was not in perfect resonance would allow some current through, depending on how far out of resonance the tank circuit was. But I think what you're saying is the resulting current would be sine. Would a diode between the tank circuit and the smoothing cap work?

What comes out of the rectifier still AC but at twice the original line frequency (100 or 120 Hz). This is still AC, and the filter capacitor in the load convert that AC to mostly DC with a bit of ripple. Your LC combo would create a voltage divider. I guess a simulation is worth paragraphs. Give me a few minutes.

 

Terminology clarification; what comes out of the rectifier will be DC, pulsed DC. To be AC it needs to cross the 0V line. Someone correct me if I'm mistaken here.

 

Was thinking the same thing about rectified AC. It wouldn't be AC anymore, it would be - as @MrSoftware said, DC ripple.

I'm no expert, and I'll post a diagram momentarily, but instead of that cap being across the choke it might be better if it were before the choke, not in series but parallel to the load, same as the second cap?

 

Like this?

 

I was mistaken, the inductor capacitor combination will not block the DC component of the AC waveform.
What you are doing will work - sort of but it has a rather nasty turn on behavior.
I'm happy to try other component values for you. Since this is an unregulated supply the output voltage will not be well controlled.

 

Was thinking the same thing about rectified AC. It wouldn't be AC anymore, it would be - as @MrSoftware said, DC ripple.

I'm no expert, and I'll post a diagram momentarily, but instead of that cap being across the choke it might be better if it were before the choke, not in series but parallel to the load, same as the second cap?

The problem with the pi-filter is that depending on component choices, along with the source impedance, and the load impedance it may make things worse. An unregulated supply is kinda like a pig in a pen. You can put lipstick on her, but she's still a pig.

 

Your LC will not pass the parallel LC tuned frequency but the inductor will pass DC and frequencies lower than the tuned LC circuit, and the capacitor will pass frequencies higher than the tuned LC circuit.

Why do you have harmonics? The low frequency AC electricity in your home is supposed to be a sinewave with no harmonics.
Are you using a squarewave or modified sinewave inverter to make your own electricity? They produce lots of harmonics.

 

To be AC it needs to cross the 0V line. Someone correct me if I'm mistaken here.

There have been threads arguing both sides of this question.
My opinion is that whether the signal crosses zero not does not determine whether it is AC (although the term alternating current does imply that).
Thus, any change in voltage with time is AC (in the broad sense of the term) as it has Fourier values (which can be seen on a spectrum analyzer).
If it has a DC offset where none of the signal goes through zero, that just means it is AC with a DC offset.
For example, if you start with an AC signal at the input to a linear amplifier, it doesn't cease to be AC if the signal is DC shifted above ground (as many amps do) as it goes through the amp.

Thus the rectified AC still has AC components, they are just different frequencies with added harmonics.

 

Some power supplies in sensitive equipment (such as audio amplifiers) place a small capacitor across each rectifier diode to suppress the sharp turn-off edge as the diode stops conducting, which can generate harmonics.
Perhaps that would help in your application.

 

Why do you have harmonics? The

After rectification it does.

Bon

 

Like this?

The makes a PI filter.
Actually it's better to remove the left capacitor and just have the right capacitor, as the minimizes the peak current through the transformer and diodes.
But that requires a large inductor that can carry the supply current.

 

Full wave rectification produces double the original frequency. The ripple is not a sinewave so of course it has some harmonics. Filter away the ripple then you will have no harmonics.

 

This might help:

 

What you are doing will work - sort of but it has a rather nasty turn on behavior.
I'm happy to try other component values for you. Since this is an unregulated supply the output voltage will not be well controlled.

Many thanks for the sim test! Apart from the start-up that looks good. It would be interesting to know what would happen if you tuned that tank circuit to resonance. At 50hz and 50mH the cap should be around 202uF for a tuned circuit. Theoretically current draw from the AC source would drop as the impedance of the tank circuit approaches infinity. The idea is that this tank circuit could throttle the current so that no sharp spikes are seen from the big cap charging. I can see me wanting to adapt this for a 240v 50hz AC input with a 60vdc output for a some 30w LEDs I'm using to grow plants.

 

Some power supplies in sensitive equipment (such as audio amplifiers) place a small capacitor across each rectifier diode to suppress the sharp turn-off edge as the diode stops conducting, which can generate harmonics.
Perhaps that would help in your application.

Thank you, I can see that might help. Would that be better than using soft recovery diodes?