High Pass Filter Working Principle

[ARoyC]

TL;DR Summary

How does a High Pass Filter Capacitor work in an AC to DC converter?

A few days ago, I learned to make an AC to DC converter. One question is troubling my mind. How does a High Pass Filter Capacitor work? Going through some websites I got a sketchy idea.

As the current after passing through the bridge rectifier comes to capacitor, the capacitor starts charging. After reaching the peak voltage, it starts discharging through the load. But the time for discharge is longer than charging. So as it discharges a bit, the input voltage reaches the value for the capacitor to start charging again. Like this, the upper part of the AC voltage gets filtered out towards the load.

Is this idea correct? What modifications are needed if it is somewhat correct? Could you please help me out with this?

 

[berkeman]

I don't see where you associate a HPF function with an AC Bridge rectifier circuit...

https://www.electronics-tutorials.ws/wp-content/uploads/2018/05/diode-diode23.gif
https://www.electronics-tutorials.ws/wp-content/uploads/2018/05/diode-diode23.gif

 

[ARoyC]

My question is why the Resultant Waveform is like that with a smoothening capacitor. What is the working principle of that capacitor?

 

[berkeman]

I

What is the working principle of that capacitor?

It is just acting as an energy storage element, kind of like how a car engine flywheel stores energy from the short impulses from the pistons, and smooths out the rotation of the crankshaft between those energetic pulses...

 

[ARoyC]

I

It is just acting as an energy storage element, kind of like how a car engine flywheel stores energy from the short impulses from the pistons, and smooths out the rotation of the crankshaft between those energetic pulses...

But why isn't the output current from the capacitor an AC? Why is it almost constant DC? Is it because of the time taken by the capacitor to get discharged? Discharging is much slower?

 

[berkeman]

The smoothing capacitor is charged to a DC voltage value during those charging pulses, so it will present a decaying positive voltage during the discharge periods. If the capacitor is large compared to the load current, yes you get almost a constant DC voltage value out of it. In practical circuits, you will always have some measurable ripple.

 

[Averagesupernova]

@ARoyC you are mistakenly calling it a high pass filter. It helps form a low pass filter.

 

[DaveE]

This is a surprising complex, yet common circuit. Very non-linear behavior. This is the sort of circuit where circuit simulators can be really helpful. While it may be a bit complex compared to your question, this old National Audio/Radio Handbook has a nice treatment of it in the appendix.

 

[Tom.G]

Another good, low priced ($15), resource is the ARRL Handbook published by the Amateur Radio Relay League. The current versions are rather expensive ($65) but e-bay has several of the older ones. Those from the mid 1960's to the mid 1970's would be a good bet, some of the later ones tend to leave out some of the basic theory.
https://www.ebay.com/b/arrl-handbook/bn_7024902254

Also, @berkeman (one of the Hams [amateur radio operator] here) may have some more ideas for you.

Cheers,
Tom

 

[ARoyC]

Okay. Thanks. I am not sure if my doubt is fully cleared. Maybe I have to study myself more.

 

[ARoyC]

@ARoyC you are mistakenly calling it a high pass filter. It is helps form a low pass filter.

Okay, sorry. Thanks for pointing out the mistake.

 

[ARoyC]

This is a surprising complex, yet common circuit. Very non-linear behavior. This is the sort of circuit where circuit simulators can be really helpful. While it may be a bit complex compared to your question this old National Audio/Radio Handbook has a nice treatment of it in the appendix.

I will check it out. Thank you.

 

[ARoyC]

Another good, low priced ($15), resource is the ARRL Handbook published by the Amateur Radio Relay League. The current versions are rather expensive ($65) but e-bay has several of the older ones. Those from the mid 1960's to the mid 1970's would be a good bet, some of the later ones tend to leave out some of the basic theory.
https://www.ebay.com/b/arrl-handbook/bn_7024902254

Also, @berkeman (one of the Hams [amateur radio operator] here) may have some more ideas for you.

Cheers,
Tom

I am from India. Probably, I won't get that book here. Even though thanks for your input.

 

[Windadct]

In the case of the voltage here - we have a DC component ( typically an average) and a voltage "ripple" that is really the part you are looking at.

It is the rectifier that "makes" the DC from the AC source, basically each half cycle of the AC is now a positive hump.

You can look at the capacitor two ways ( and understanding both is ideal)

1) A low pass filter,
2) An energy storage element

As pointed out - this is a very good circuit to simulate in something like LSpice ( its free)

And build the circuit up element by element and look at the resulting voltage AND current waveforms.

AC Voltage source + Load
then add rectifier and move the load to the "DC" Side
then add the cap

then add more capacitance.

(then add inductors to the AC and DC sides (before the caps, after the caps)

While this may seem like rudimentary circuits - it is surprising how few EEs have actually gone through the exercise and seen the variety of waveforms and effects. The ripple voltage, and resulting current, are the critical factor in selecting the actual capacitors needed.

 

[jsgruszynski]

Summary:: How does a High Pass Filter Capacitor work in an AC to DC converter?

A few days ago, I learned to make an AC to DC converter. One question is troubling my mind. How does a High Pass Filter Capacitor work? Going through some websites I got a sketchy idea.

As the current after passing through the bridge rectifier comes to capacitor, the capacitor starts charging. After reaching the peak voltage, it starts discharging through the load. But the time for discharge is longer than charging. So as it discharges a bit, the input voltage reaches the value for the capacitor to start charging again. Like this, the upper part of the AC voltage gets filtered out towards the load.

Is this idea correct? What modifications are needed if it is somewhat correct? Could you please help me out with this?

This is where understanding a "frequency domain" view of signals can be helpful. A sine wave that is chopped off per a full or half-wave rectifier has had some of the sine wave energy converting into frequencies at DC and at frequencies above the original AC frequency.

You want the DC but none of the rest (even the original frequency) so you want a low-pass filter (capacitor to ground) to send those currents to ground instead of the output. This is what diodes can do because they are "nonlinear" - a purely linear component may change the input level on the output but NEVER creates new frequencies while nonlinear devices can.

Capacitors look like shorts to "sufficiently high frequency" and opens to "sufficiently low frequency". The opposite is true for inductors so you sometimes see linear power supplies with LC or Pi ladders.

You can also view the capacitors as charging but that doesn't take you as far as a frequency domain explanation.

 

[sophiecentaur]

Okay. Thanks. I am not sure if my doubt is fully cleared. Maybe I have to study myself more.

Let me step in with a really 'Noddy' explanation. As few hard words as possible:
The AC source will feed charge into the Capacitor whenever the voltage of the source waveform exceeds the 'downstream volts'. That's what the (one way) diode does. The charge flows very easily from the source into the C and quickly boosts the volts across it until peak supply volts are reached. The Load keeps taking current all the time and the volts drop a bit after each peak and the diode stops current flowing back into the supply. The bigger the C value, the slower the volts drop and the lower the load resistance, the faster they will drop (discharge). The ripple is due to both those factors and the design of any AC/DC 'adaptor' allows for a certain load (or available current). The 'Low Pass' action of the R and C, wipes off a lot of the AC voltage variation and sets the minimum value it will reach.

Too much current demand and the ripple will be too great and affect the quality of the DC supply.

PS If there is no load connected, the volts on the Capacitor will stay at a DC level equal to the peak AC volts.