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How to design active filters for standard definition video channels

today's high-speed operational amplifiers make it easier to design active filters than passive inductor capacitor (LC) filters in video applications. With the introduction of high-speed amplifiers and video filters by Fairchild Semiconductor, active filters for video channels have become a very cost-effective solution

why use video filters

any interference signal in the video frequency band will produce some visible display distortion problems, which can be eliminated by using video filters. Signal aliasing is an obvious distortion in sampled video systems. When high-frequency signals (such as external wireless transmission signals or local clock signals) beyond the range of video frequency band are aliased back to the video frequency band through the sampling process of analog-to-digital converter, aliasing will occur. This distortion can be prevented by placing an anti aliasing filter in front of the ADC. When the digital to analog converter is used to reconstruct the digitized video signal, the video signal will be copied at a higher frequency, which will also cause image distortion. This kind of distorted signal can be eliminated by adding a video filter after the digital to analog converter

why use active video filter

the active filter realized by operational amplifier will have better frequency response, flatter bandwidth and better matching between channels than the passive filter realized by inductance. The bandwidth and frequency response characteristics of a passive filter depend on the accuracy of its inductance and capacitance. The active filter uses operational amplifiers and resistors to replace inductors. Because the accuracy of active filter depends on the accuracy of resistance and capacitance values, rather than the accuracy of inductance and capacitance values, the accuracy is fully improved. It is also worth mentioning that the accuracy of cheap resistors is much higher than that of cheap inductors. In addition, the price of operational amplifiers is also very low, so an active filter is often cheaper than a passive filter realized by inductance

design active video filter

video filter requires good phase linearity, that is, it should have a very constant phase delay in the whole video frequency band, and the flatness of amplitude response is good. These two parameter requirements make Butterworth filter a good choice. A fourth-order filter can achieve good stopband suppression at high frequencies

4-order Butterworth filter is composed of two operational amplifiers in salen key mode. The design in this paper uses Fairchild's high-speed dual operational amplifier (fhp3230) to construct a filter that can be placed in a narrow printed circuit board. The gain bandwidth of fhp3230 is 60MHz. It is an amplifier with single power supply and rail to rail output. These features make the display of F at this time unstable to 0.0 (± 1.0kn is the allowable value); Hp3230 becomes the best choice for this kind of application

Figure 1 shows a fourth-order filter

Figure 1: active video filter with fhp3230 dual operational amplifier (+5v single power supply, AC coupled input)

in the following paragraphs, we will discuss how to set the values of each element in the filter. It should be noted that the filter operates under a +5v single power supply and assumes that the input signal is AC coupled. If the input DC level bias level is compatible with the supply voltage and the input common mode voltage range of the amplifier, the input signal may be DC coupled

design Butterworth active video filter

the filter has two stages, and each stage is realized by a fhp3230 operational amplifier (therefore, the whole filter requires a dual operational amplifier device). Each stage provides two poles to form a complete 4th order filter. There are many ways to select the component value of this filter. In this example, the first stage is the unit gain, and the Q value is 0.54; The gain of the second stage is +2, and the Q value is 1.3. These Q values form a Butterworth filter. In order to ensure the flatness of the filter in the whole video frequency band, the cut-off frequency fo is 6mhz. This will result in a gain of -1db for the entire Butterworth waveform at 5MHz

the capacitance should be large enough so that the parasitic capacitance of PCB will not affect the capacitance value. But the capacitance should not be too large, otherwise the resistance value will be very small, so that the amplifier will be difficult to drive such a small resistance. The parasitic capacitance of PCB is about 1pf in the case of short wiring (the material of PCB directly affects the parasitic capacitance). If we take 18pF as the basic order of capacitance elements, we can minimize the influence of parasitic capacitance

the following is the conversion function of the first stage Sallen key filter:

the conversion function of each filter stage in Figure 1 can be expressed by the above formula, and only additional subscripts (a or b) are needed to distinguish different filter stages (for example, replacing R1 in the equation with R1a or r1b, etc.)

the following are some key parameters in the conversion function:

k=1+r4/r3 or K = 1 (for the first level)

these equations are not the most effective forms. For example, given the fo and Q values, we hope to find the best resistance and capacitance values. At present, there are many different combination schemes of resistance and capacitance values that can define filter shaping. In order to simplify the selection, the resistance and capacitance can be expressed as a scale factor of a common value, that is:

R1 = R × a. R2 = R, C1 = C and C2 = C × b。 Where C = 18pF

after replacement, you can get:

design the first stage of the filter

if the capacitance value used in the first stage is the same, and this stage is a unit gain stage, the maximum Q value can only reach 0.5. And because the required Q value is 0.54, the capacitance of the first stage must not be equivalent. This requires the selection of a standard capacitor greater than 18pF, and the 22pf capacitor just meets the requirements. Of course, there are many other capacitance values that are also very suitable. In this way, the scale factor B of the first stage is 22/18 (i.e. 1.22), and the scale factor K of the first stage is 1, so it can be obtained:

since q = 0.54 is known, a can be solved from the above formula to obtain a = 1.53

based on the scale factor A, the R value can be solved from the equation of the cut-off frequency fo:

substitute fo = 6mhz into the above formula, and get: r = 1079 Ω, R1 = 1650 Ω, R2 = 1079 Ω

the value of the nearest (1%) standard resistance is: R1 = 1.65k Ω, R2 = 1.07k Ω

design the second stage of the filter

the gain of the second stage amplifier is +2, so an equivalent capacitance can be used (although its Q value is higher). Assuming C1 = C2 = C = 18pF, the resistance value is also expressed as a scale factor A, that is:

R1 = R × A and R2 = R

we can get:

since Q of the second stage is known to be 1.3, we can solve a from the above formula and get a = 1.69. Substitute fo = 6mhz into the equation of the cut-off frequency fo, and solve the R value: r = 1134 Ω, R1 = 1916 Ω, R2 = 1134 Ω. The value closest to (1%) the standard resistance is: R1 = 1.91k Ω, R2 = 1.1k Ω

adjust the active filter to adapt to the amplifier gain bandwidth

there is a problem with the component values derived from the above equation, that is, they all assume that the bandwidth of the operational amplifier is infinite (if there is such an operational amplifier, it will be very expensive). Using the component values derived above to build the filter circuit, it will be found that the circuit has a peak at 0.6dB gain, and the bandwidth will be reduced by about 18% when the gain is -3db. Therefore, these component values must be adjusted to compensate for the phase shift delay in the amplifier. Fortunately, this can realize the effect of self-regulation. The adjustment process is very simple

first, measure the bandwidth of the constructed circuit at -3db. Then, adjust the resistance of all four filters (R1a, R2A, r1b, r2b) according to the ratio of the actual measured bandwidth to the expected bandwidth, that is:

here

adjusting the resistance in this way will change the filter frequency, but will not have much impact on the Q value. Table 1 shows the calculated resistance value after measuring the circuit. Table 1: the R & D and production of the resistance tester calculated after measuring the circuit are the values we complete

change the resistance in the circuit into the adjustment value calculated according to the circuit (these values may be different for specific circuits). After changing the resistance value, measure the bandwidth. The current bandwidth will be closer to the requirements, but it may not be completely correct, but there is no need to worry about it. The peak value of the filter may be slightly larger than before, but this will not cause problems

secondly, adjust the peak value (i.e. Q value). Almost all peaks in the filter are generated by the second stage, because the Q value and gain of this stage are higher (smaller loop transmission and narrower bandwidth). Therefore, we focus the adjustment of Q value on this stage. The peak value can be adjusted by reducing r1b resistance. First reduce by about 20%. If there is still a peak, further reduce the resistance value. If the frequency response caused by the reduction of resistance value is too large, the resistance value should be increased appropriately. A good way to make the frequency response waveform of the filter closest to Butterworth filter is to observe the frequency at -1db and adjust r1b until the frequency at -1db is 5MHz (when this value is reached, the frequency response waveform should be clean and flat). By reducing the r1b value from 1.58k Ω to 1.24k Ω, the frequency response waveform in the filter circuit matches an ideal Butterworth filter very well. Table 2 shows the final resistance values used in the circuit. Table 2: after adjusting the peak value of the final resistance value

used in the circuit (the resistance value should be based on the measurement results, and it does not have to be the same as the resistance value in Table 2), the frequency at -3db will be close to the expected value of 6mhz. Figure 2 shows the frequency response curve measured in the circuit. The initial frequency response, the result after frequency adjustment, the result after peak adjustment (final circuit) and the frequency response of the ideal Butterworth filter are also given in the figure

Figure 2: the frequency response curve measured in the circuit, including the initial frequency response, the result after adjusting the frequency, the result after adjusting the peak value, and the frequency response of the ideal Butterworth filter

filter performance

the frequency response measurement result of the filter is shown in Figure 2 (that is, the result after adjusting the peak value, represented by the red curve). Figure 2 also shows the frequency response curve (green curve) of an ideal 6mhz Butterworth filter. The frequency response result is measured on the circuit constructed by this method. As shown in the figure, the filter has almost no peak in the video frequency band, good flatness, and good stopband suppression effect. The differential gain and phase are also good. Table 3 summarizes the performance details of the filter circuit. Table 3: performance details of an ideal 6mhz Butterworth filter circuit

source impedance and load impedance

this filter works well when the source impedance is 100 Ω. If the source impedance is higher, it must be taken into account when calculating the resistance value of the first resistance (R1a). For example, if the source impedance is 150 Ω, reduce the resistance of R1a according to this impedance

the gain of the filter in Figure 1 is +2v/v, which is used to drive the line terminated with a series output resistance (75 Ω in the figure), so as to provide back-end termination. The back-end termination divides the signal into two, so the total gain from the input to the video load at the far end of the cable becomes +1v/v

circuit layout

components should be as close as possible and short wiring should be used. It is more important to keep the input short circuit of operational amplifier than the output short circuit. Note: remove the grounding copper layer near the input end of the operational amplifier to reduce the parasitic capacitance of the circuit board. The parasitic capacitance near the inverted input will cause the peak value of the operational amplifier, and if the parasitic capacitance depends on

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