AC noise affects the precision, stability, and functionality of various components. I have always assumed that high-quality coffee beans and an expensive coffee machine are all it takes to make the perfect espresso. After watching a world-class barista in action, I realized I was wrong. Apart from the sublime techniques, the water filter determines the end result of the espresso.
Witnessing this master was eye-opening in terms of how meticulous you need to be to make a perfect cup of coffee. The experience also highlights a similar and often overlooked aspect when designing a power supply.
There might be no water filter involved, but you will need to place an LC filter for the power supply. If you need a basic refresher, an LC filter is a low pass filter built with an inductor and capacitor. It is meant to prevent certain high-frequency AC components from passing through the circuit. At higher frequencies, the inductor acts as a choke, which blocks AC components from passing through. The idea of the LC filter is to prevent high-frequency AC components from flowing to the load.
In an ideal world, the voltage from the output of a DC power supply is a perfect horizontal line when observed on an oscilloscope. However, the world is far from ideal and the voltage is never a perfect line.
These multiple stages of filtering help to produce a very low ripple smooth DC output across the Pi filter. High-frequency Pi filters also provide surge immunities more than the silicon-based filters.
For instance, a silicon chip has a limit of voltage withstand capacity, whereas pi filters made using the passive components have much more immunity in terms of surges and harsh industrial environments. If this load current is relatively high, then the wattage of the Inductor also increases making it bulky and expensive. Also, the high current through the inductor increases the power dissipation across the inductor resulting in poor efficiency. High-value Input Capacitor Another major problem of the Pi filter is the large input capacitance value.
Pi filters require high capacitance across the input which became a challenge in space-constrained applications. Also, high-value capacitors increase the cost of the design. Bad Voltage Regulation Pi filters are not suitable where load currents are not stable and constantly changing. Pi filters provide bad voltage regulation when load current drifts a lot.
In such an application the filters with an L section are recommended. Generally, Pi filters are directly connected with the bridge rectifier and the output of the Pi filters is referred to as the High Voltage DC. This construction, from Bridge rectifier diode to the driver has a different operation with the working of Pi-Filter. First, this Pi filter provides smooth DC for the ripple-free operation of the overall driver circuit resulting in a low output ripple from the final output of the power supply, and the other one is for isolating main lines from the high switching frequency across the driver circuit.
A properly constructed line filter can provide Common-mode filtration A filter that rejects noise signal as if an independent single conductor and differential mode filtration differentiating two switching frequency noise, especially high-frequency noise that can be added into the mains line in a Power supply where Pi filter is an important component.
In the RF application, Pi filters are used in different operations and different configurations. This can result in oscillations and an unstable power supply.
Therefore, the first step to designing this filter is to choose how to damp the filter. Figure 4 shows three viable damping techniques. The damping resistor typically has little to no loss and can be small for even large power supplies. The drawback is that it significantly reduces the effectiveness of the filter by reducing the parallel impedance with the inductor. Technique 2 has the advantage of maximizing filter performance.
If an all ceramic design is desired, R D can be a discrete resistor in series with a ceramic capacitor. Otherwise a physically large capacitor with a high ESR is required. This additional capacitance C D can add significant cost and size to the design.
Damping Technique 3 looks very advantageous since the dampening capacitor C E is added to the output where it might help somewhat with transient response and output ripple.
However, this is the most expensive technique since the amount of capacitance required is much larger. In addition, the relatively large amount of capacitance on the output will lower the frequency of the filter resonance, which will reduce the achievable bandwidth of the converter—therefore Technique 3 is not recommended.
For the ADIsimPower design tools we use Technique 1 because of the low cost and relative ease of implementing it in an automated design process. Figure 4. ADP with an output filter with several different damping techniques highlighted. Another issue that needs to be dealt with is compensation. It may be counter intuitive, but it is almost always better to put the filter inside the feedback loop. This is because putting it in the feedback loop helps damp the filter somewhat, eliminates dc load shift and the series resistance of the filter, and gives a better transient response with less ringing.
Figure 5 shows the Bode plot for a boost converter with an LC filter output added to the output. The feedback is taken before or after the filter inductor. Since the control loop is affected with or without the filter in the feedback loop, one might as well compensate for it appropriately.
In general this will mean scaling back the target crossover frequency to a maximum of a fifth to a tenth of the filter resonant frequency F RES. The design process for this type of filter is iterative in nature since each component selection drives the selection of the others.
Step 1: Choose C 1 as if there was not going to be an output filter on the output. C 1 can then be calculated using Equation 8. Based on experience, a good value is between 0.
The inductor should be chosen for a high self-resonant frequency SRF. Larger inductors have larger SRFs, which means they are less effective for high frequency noise filtering. Smaller inductors will not have as much effect on ripple and will require more capacitance. The higher the switching frequency is the smaller the inductor can be. When comparing two inductors with the same inductance, the part with higher SRF will have lower interwinding capacitance.
The interwinding capacitances acts like a short circuit around the filter for high frequency noise. Once transformed into a series circuit, the analysis can be conducted in the usual manner. Previously in Lesson 4 , the method for determining the equivalent resistance of parallel are equal, then the total or equivalent resistance of those branches is equal to the resistance of one branch divided by the number of branches.
If the two or more resistors found in the parallel branches do not have equal resistance, then the above formula must be used. An example of this method was presented in a previous section of Lesson 4. By applying one's understanding of the equivalent resistance of parallel branches to a combination circuit, the combination circuit can be transformed into a series circuit. Then an understanding of the equivalent resistance of a series circuit can be used to determine the total resistance of the circuit.
Consider the following diagrams below. Diagram A represents a combination circuit with resistors R 2 and R 3 placed in parallel branches. This is shown in Diagram B. Now that all resistors are in series, the formula for the total resistance of series resistors can be used to determine the total resistance of this circuit: The formula for series resistance is. Once the total resistance of the circuit is determined, the analysis continues using Ohm's law and voltage and resistance values to determine current values at various locations.
The entire method is illustrated below with two examples. The first example is the easiest case - the resistors placed in parallel have the same resistance. The goal of the analysis is to determine the current in and the voltage drop across each resistor. As discussed above, the first step is to simplify the circuit by replacing the two parallel resistors with a single resistor that has an equivalent resistance. Thus, the total resistance is. In doing so, the total resistance and the total voltage or battery voltage will have to be used.
The 4 Amp current calculation represents the current at the battery location. Yet, resistors R 1 and R 4 are in series and the current in series-connected resistors is everywhere the same. For parallel branches, the sum of the current in each individual branch is equal to the current outside the branches. There are an infinite number of possible values of I 2 and I 3 that satisfy this equation.
Since the resistance values are equal, the current values in these two resistors are also equal. Therefore, the current in resistors 2 and 3 are both equal to 2 Amp.
These calculations are shown below. The second example is the more difficult case - the resistors placed in parallel have a different resistance value. The goal of the analysis is the same - to determine the current in and the voltage drop across each resistor. As discussed above, the first step is to simplify the circuit by replacing the two parallel resistors with a single resistor with an equivalent resistance. The 1. There are an infinite possibilities of I 2 and I 3 values that satisfy this equation.
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