Enrique15
Member level 3
Hello again, my friends.
After realizing that normal batteries aren’t enough for the load that the speakers of my IC555-alarm need, I’m reviewing my notebooks about Power Supply design, of course, to design one myself, including the filters that are needed for a regular voltage and current.
I have copies from a text book used in my college that have some calculations about designing power supplies filters (sorry I can’t give you the book’s name, because I just have the copies, and didn’t wrote down the book’s name or author, my mistake).
So here I want to share some of these concepts for the people that maybe don’t know too much about it, hopping that I’m not telling the same things someone from this forum told before already. And besides the theory, I have some questions myself at the end of the topic.
OK. Here’s the thing:
First I start the analysis with a circuit like this one:
where there’s the AC voltage, a “10 to 1” (a = 0.1) transformer, a full wave rectifier and a load (5 Kohm, for this example).
So, with this circuit I have a rectified voltage wave in the load:
and the same happens with the current
(in my program, the load current is the only one that shows up inverted, so you’ll see it growing from zero to negative values, but we just care about the magnitude and shape, so don’t worry about this inverted position).
Of course, we know that there’s some voltage drop in the diodes, but the text book makes the analysis assuming ideal diodes, so the wave is a pure “rectified sinusoidal”, thus they can use Fourier analysis to get the DC and AC components.
The Fourier series for a rectified sine wave has a DC component and the even harmonics of the fundamental frequency as follows:
with Vm = peak amplitude of the sine wave.
So, the DC component is “2Vm / pi”.
In the load we have these frequency elements:
Knowing this, we want the filter to eliminate the harmonics and keep only the DC component.
The easiest filter is the one that uses a capacitor in parallel with the load:
But this simple filter is explained in terms of the capacitor's “charge and discharge”. So, with the capacitor discharged at the beginning, when the AC voltage starts to appear, we have a sine wave growing from zero to its peak amplitude, and the same sine wave is seen at the capacitor terminals. But when the sine wave starts to decrease, the voltage stored in the capacitor is bigger, so the diodes stop conducting (they’re cut), and the capacitor starts discharging through the resistor, decreasing its voltage, until the next period of the sine wave is bigger than the capacitor voltage, charging it again, and the cycle repeats itself.
So, the output voltage is determined by two expressions:
with
= angle when the diodes polarity is inverted.
As we can see, the maximum voltage Vm is at angle
, and the minimum voltage can be calculated knowing that the second pair of diodes (the ones active with the second half of the sine wave period) turn ON after the sine wave gets bigger than the voltage at the capacitor at the angle:
.
The voltage ripple is the difference between maximum and minimum:
The angle
can be calculated, but instead is preferred to approximate the minimum voltage using:
Assuming
, we can approximate the results like this:
For capacitors with a large discharge period, we can say:
. So, for the minimum voltage in angle
, we have:
And the ripple expression turns out:
The expression
can be approximated into:
And the ripple is:
This is were the famous expression
comes from.
Well, at least that’s the equation I was thought in college. I don’t know if you use or can tell me about other methods, but this is what I believe is the classic expression.
So, if I use this expression, and I want a ripple of 1%, with my load of 5 Kohms, the capacitor value will be “167 uF” (as seen in the schematic I showed you before). Thus I get a small ripple (almost a total DC voltage, “yellow line” of the next picture):
In my text book copies, for a filter made up with only one capacitor, this is the analysis. They didn’t mention anything about the harmonics or anything alike in the voltage. So the Fourier explanation I first told you about has nothing to do in this analysis.
BUT, when they talk about a “inductor-capacitor” filter, the book change the analysis, and forget about these last things I wrote, and come back with the Fourier analysis. Again they talk about the DC and harmonics components, but don’t set any ripple calculations. WHY IS THAT? Maybe the analysis is the same as the one explained before.
Well, back to the analysis, we have this circuit:
For the “inductor-capacitor” filter, there are 2 possibilities: or it has a “continuous” current, or it has a “discontinuous” current.
The condition for the current to be continuous (not touching the zero axis) is determined by saying that the amplitude of the harmonics current has to be less than the amplitude of the DC current flowing through the inductor. Because greater harmonics have less magnitude of current, an approximation will be using just the current at the second harmonic (fundamental frequency times 2: 60 x 2 = 120 hertz): I2 = [4Vm/3(pi)] /(2wL) = 2Vm/3(pi)wL. And the dc current is: I0 = 2Vm/(pi)R
Comparing them we have: I2 < I0 , thus: 2Vm/3(pi)wL < 2Vm/(pi)R , and the result is:
In a “continuous current” situation, the voltage at the load is:
, because the harmonics pass through the capacitor, and so at the resistor we have only the DC component of the rectified sine wave.
But, for a “discontinuous current” situation, the Vout is calculated using a procedure to approximate its value, using integration and some estimation steps.
In conclution:
1. For both “continuous” and “discontinuous” current situations, the harmonics go through the capacitor, and there’s just DC voltage at the output.
2. For “discontinuous” current, Vout is greater than the voltage for “continuous” current.
Using an example from my text copies: L=5mH, C=10 000uF, and R= 5 Ohm for “continuous current” and 50 Ohm for “discontinuous current”.
Continuous current:
And the results:
Current in the inductor (look that is a sine wave over zero axis)
Voltage in resistor:
Current in resistor (remember that in my program this current is upside down, from zero to negative values):
Discontinuous current:
Current in the inductor (the wave is cut by the zero axis)
Voltage in resistor (you can see is greater than the one for continuous current situation):
and current in resistor:
Well, after all this, I have some questions myself:
1. How is calculated the ripple in a “inductor-capacitor” filter?
2. What about a “capacitor, inductor, capacitor” filter as the one in the next picture:
How are its elements (capacitor1, capacitor2, inductor and load) calculated?
3. Do you know some information on internet with an easy-to-understand analysis for these filters, that can complement what I’ve commented here?
4. I see they always talk about a “load” resistor for the filter circuit, besides the REAL LOAD resistor which is going to be connected to the filter. For example, I see a schematic of a capacitor filter with a “resistor”, then comes a Voltage Regulator, and then another capacitor (for noise I suppose) and later my real load (my circuit or whatever I want to feed with the regulated voltage). So, why is used this “additional load” resistor ??? How is it calculated ? And how can it affect my calculations, because if it’s smaller than my circuit load, all the current will flow through it instead of going through my load, that's what I think will happen.
5. And finally: where can I find information about calculating filters for the AC line voltage? I mean, how to calculate the elements between the 120V AC line and the reduction transformer, to filter any noise in the AC line (like motors noise and any similar)?
Because differently from the explanations I gave here, I’ve seen in schematics that they use inductors in both AC lines (Hot and Neutral), one inductor each, and 2 or more capacitors, some in parallel, some in series, and another element they say is for “sparks” (the symbol is like two diodes, facing their cathodes one to another). I don’t know its name.
Well, hope to hear some comments about my topic, and any suggestions as well. I’ll appreciate that.
Regards.
After realizing that normal batteries aren’t enough for the load that the speakers of my IC555-alarm need, I’m reviewing my notebooks about Power Supply design, of course, to design one myself, including the filters that are needed for a regular voltage and current.
I have copies from a text book used in my college that have some calculations about designing power supplies filters (sorry I can’t give you the book’s name, because I just have the copies, and didn’t wrote down the book’s name or author, my mistake).
So here I want to share some of these concepts for the people that maybe don’t know too much about it, hopping that I’m not telling the same things someone from this forum told before already. And besides the theory, I have some questions myself at the end of the topic.
OK. Here’s the thing:
First I start the analysis with a circuit like this one:
where there’s the AC voltage, a “10 to 1” (a = 0.1) transformer, a full wave rectifier and a load (5 Kohm, for this example).
So, with this circuit I have a rectified voltage wave in the load:
and the same happens with the current
(in my program, the load current is the only one that shows up inverted, so you’ll see it growing from zero to negative values, but we just care about the magnitude and shape, so don’t worry about this inverted position).
Of course, we know that there’s some voltage drop in the diodes, but the text book makes the analysis assuming ideal diodes, so the wave is a pure “rectified sinusoidal”, thus they can use Fourier analysis to get the DC and AC components.
The Fourier series for a rectified sine wave has a DC component and the even harmonics of the fundamental frequency as follows:
with Vm = peak amplitude of the sine wave.
So, the DC component is “2Vm / pi”.
In the load we have these frequency elements:
Knowing this, we want the filter to eliminate the harmonics and keep only the DC component.
The easiest filter is the one that uses a capacitor in parallel with the load:
But this simple filter is explained in terms of the capacitor's “charge and discharge”. So, with the capacitor discharged at the beginning, when the AC voltage starts to appear, we have a sine wave growing from zero to its peak amplitude, and the same sine wave is seen at the capacitor terminals. But when the sine wave starts to decrease, the voltage stored in the capacitor is bigger, so the diodes stop conducting (they’re cut), and the capacitor starts discharging through the resistor, decreasing its voltage, until the next period of the sine wave is bigger than the capacitor voltage, charging it again, and the cycle repeats itself.
So, the output voltage is determined by two expressions:
with
As we can see, the maximum voltage Vm is at angle
The voltage ripple is the difference between maximum and minimum:
The angle
Assuming
For capacitors with a large discharge period, we can say:
And the ripple expression turns out:
The expression
And the ripple is:
This is were the famous expression
Well, at least that’s the equation I was thought in college. I don’t know if you use or can tell me about other methods, but this is what I believe is the classic expression.
So, if I use this expression, and I want a ripple of 1%, with my load of 5 Kohms, the capacitor value will be “167 uF” (as seen in the schematic I showed you before). Thus I get a small ripple (almost a total DC voltage, “yellow line” of the next picture):
In my text book copies, for a filter made up with only one capacitor, this is the analysis. They didn’t mention anything about the harmonics or anything alike in the voltage. So the Fourier explanation I first told you about has nothing to do in this analysis.
BUT, when they talk about a “inductor-capacitor” filter, the book change the analysis, and forget about these last things I wrote, and come back with the Fourier analysis. Again they talk about the DC and harmonics components, but don’t set any ripple calculations. WHY IS THAT? Maybe the analysis is the same as the one explained before.
Well, back to the analysis, we have this circuit:
For the “inductor-capacitor” filter, there are 2 possibilities: or it has a “continuous” current, or it has a “discontinuous” current.
The condition for the current to be continuous (not touching the zero axis) is determined by saying that the amplitude of the harmonics current has to be less than the amplitude of the DC current flowing through the inductor. Because greater harmonics have less magnitude of current, an approximation will be using just the current at the second harmonic (fundamental frequency times 2: 60 x 2 = 120 hertz): I2 = [4Vm/3(pi)] /(2wL) = 2Vm/3(pi)wL. And the dc current is: I0 = 2Vm/(pi)R
Comparing them we have: I2 < I0 , thus: 2Vm/3(pi)wL < 2Vm/(pi)R , and the result is:
In a “continuous current” situation, the voltage at the load is:
But, for a “discontinuous current” situation, the Vout is calculated using a procedure to approximate its value, using integration and some estimation steps.
In conclution:
1. For both “continuous” and “discontinuous” current situations, the harmonics go through the capacitor, and there’s just DC voltage at the output.
2. For “discontinuous” current, Vout is greater than the voltage for “continuous” current.
Using an example from my text copies: L=5mH, C=10 000uF, and R= 5 Ohm for “continuous current” and 50 Ohm for “discontinuous current”.
Continuous current:
And the results:
Current in the inductor (look that is a sine wave over zero axis)
Voltage in resistor:
Current in resistor (remember that in my program this current is upside down, from zero to negative values):
Discontinuous current:
Current in the inductor (the wave is cut by the zero axis)
Voltage in resistor (you can see is greater than the one for continuous current situation):
and current in resistor:
Well, after all this, I have some questions myself:
1. How is calculated the ripple in a “inductor-capacitor” filter?
2. What about a “capacitor, inductor, capacitor” filter as the one in the next picture:
How are its elements (capacitor1, capacitor2, inductor and load) calculated?
3. Do you know some information on internet with an easy-to-understand analysis for these filters, that can complement what I’ve commented here?
4. I see they always talk about a “load” resistor for the filter circuit, besides the REAL LOAD resistor which is going to be connected to the filter. For example, I see a schematic of a capacitor filter with a “resistor”, then comes a Voltage Regulator, and then another capacitor (for noise I suppose) and later my real load (my circuit or whatever I want to feed with the regulated voltage). So, why is used this “additional load” resistor ??? How is it calculated ? And how can it affect my calculations, because if it’s smaller than my circuit load, all the current will flow through it instead of going through my load, that's what I think will happen.
5. And finally: where can I find information about calculating filters for the AC line voltage? I mean, how to calculate the elements between the 120V AC line and the reduction transformer, to filter any noise in the AC line (like motors noise and any similar)?
Because differently from the explanations I gave here, I’ve seen in schematics that they use inductors in both AC lines (Hot and Neutral), one inductor each, and 2 or more capacitors, some in parallel, some in series, and another element they say is for “sparks” (the symbol is like two diodes, facing their cathodes one to another). I don’t know its name.
Well, hope to hear some comments about my topic, and any suggestions as well. I’ll appreciate that.
Regards.