2ND ORDER NARROW BAND FILTER WITH GAIN

A high Q narrow band filter with gain is shown in Fig. 1. This filter can be used to filter steady state signals but it is not suitable for audio use or with signals that have frequent transient signals. Its high Q and gain will cause it to ring or be unstable with such transients. The high Q, high gain circuit is with R5 not populated. The circuit can be used with audio signals if Q and gain is reduced with the " Q spoiler" resistor R5 set to approximately 12.5K maximum. With R5 in place the Q is approximately 1.3 and the gain is approximately 22.5db at the pass frequency, in the case shown, of 2KHZ. Without R5 populated, the Q is set to 10 and the gain is approximately 40db or 100 at the pass frequency. A 0.05 volt peak 2KHZ input signal will produce a 5 volt peak output signal.

The design shown has been simulated for AC response and transient response using the LTC6244 rail-to-rail 5MHZ op-amp. Fig.1 is the actual simulation schematic.

To design the circuit, we must first decide on the pass band center frequency Fo.
Next set the Q value. Then set a practical value for C1 and C2, e.g., 0.01UFD. Next we will calculate a value called Beta which is the ratio of the center frequency in radians to the Q value. Set Fo to 2KHZ and set Q to 10.

Beta = 2 * Pi * Fo / Q

Beta = 1256.6 with an Fo of 2KHZ and a Q of 10.

In the case of the circuit shown we can now calculate the value of R4 as

R4 = 2 / ( Beta * C ) = 159.16K with C=C1=C2= 0.01UFD

The closest standard value would then be selected from the standard resistance decade table.

Also we can compute the bandwidth of the filter measured at the -3db points as

BW = Beta / ( 2 * Pi ) = Fo / Q = 200. The bandwidth for this circuit is 10% of the center frequency with Q set at 10.

It is also known for the circuit of Fig. 1 that the circuit component values are related to the center frequency as

Fo = 1 / ( 2 * Pi * ( R2||R3 * R4 )^0.5 * C )

where R2||R3 is the parallel value of R2 and R3. We can re-write the equation as


R2||R3 = 1 / ( Wo^2 * C^2 * R4 ) = 397.9

where

Wo = 2 * Pi * Fo

Assuming that R2 = R3, we can then say that R2 and R3 are 2 * (R2||R3) = 795.8 each as shown in Fig. 1. The closest standard value would be selected from the standard resistance decade table.

We have now completely designed the circuit of Fig. 1.

The performance of the circuit was verified using a SPICE circuit simulator. The center frequency was at 2KHZ. Bandwidth was 200 HZ. Gain was 40db or 100. With a 50 millivolt peak 2KHZ input signal, the output signal was 5 volts peak. AC analysis showed that symmetry was excellent.

2ND ORDER BAND PASS FILTER

Fig. 1 shows a second order band pass filter that is designed to have a center frequncy of 2KHZ and is stable with transient signals or noise inputs. The schematic is the actual simulation schemtaice used including the biasing circuit as the basic circuit. Note that a practical value of 0.01UFD is set for the capacitors to start with but the R values are calculated as shown below and the calculated values are shown to 5 or 6 digits for use in the simulation but standard resistor decade values that are close will provide acceptable results.

We want this filter to have a maximally flat response in the pass band so the damping factor d is set to the square root of 2, as

d = ( 2 )^2 = 1.1412...

If we wanted a maximally flat time delay instead, such as might be required for an audio cross-over network, we would set d to the square root of 3 instead.

Once we have decided on d we can then calculate a gain that will be stable for the circuit. For an ordinary wide band pass filter, like the circuit in Fig. 1, we don't need a lot of gain and we need to be careful not to use too much gain as the filter may become unstable and ring on any kind of input transient, making it useless. We will use a standard formula to calculate the gain K that will work. We will treat the gain K as a positive value although the real gain is actually negative due to the inverting amplifier circuit we are using.

K = 3 - d = 1.5857

We can round off the gain to 1.59.

Now there is a criteria for setting the minimum value of the quality factor of the filter Q. Again, we don't want too high a Q for this simple wide band filter as the circuit may not then be time stable. The criteria is:

Q > ( K / 2 )^0.5 = 0.89

So what ever Q we choose it needs to be greater than 0.89. So lets try a Q = 1.

Also we set the center of the pass band to 2KHZ for demonstration.

Fo = 2000

Now we can calculate the resistor values for the filter:

R3 = Q / ( 2 * Pi *K * C * Fo ) = 1 / ( 2 * 3.14159 * 1.5857 * 0.01*10^-6 * Fo )

R3 = 5004.9

R4 = Q / (( 2 * Q^2 - K ) * 2 * Pi * C * Fo ) = 19409

From the equation for R4 we can see that 2Q^2 must be greater than the absolute value of K, and 2Q^2 cannot equal K, or R4 must be removed from the filter and then we have a different circuit and performance. I have found that including R4 makes a better filter.

R2 = 2 * Q / ( 2 * Pi * C * Fo ) = 15915.5

R1 is simply a dummy load for the op-amp and has no theoretical effect on calculations.

Simulating the circuit as shown in Fig. 1 with the Linear Technology LTC6244 rail-to-rail 5MHZ op-amp we have the following results:

Quite symmetrical band pass curve on a log of frequency vs voltage db scale with

Fo = 2KHZ

F low at -3db = 1.24KHZ approx.

F high at -3db = 3.3KHZ approx.

-20db at 20KHZ approx.
-20db at 200HZ approx.

Phase shift at Fo is -90 degrees.

With an input signal of 0.62893, the output signal is 1 volt as predicted from the gain K calculation.

Note that

Fo = ( F low * F high )^0.5 = ( 1240 * 3300)^0.5 = 2023HZ , very close to Fo = 2KHZ.

You will note also that the band pass shape for the rates of attenuation is proportional to a percentage of the frequency, not an absolute value of frequency delta from Fo.

2ND ORDER LOW OR HIGH PASS FILTER

Fig. 1 shows a circuit that can be configured as either a high pass or low pass filter depending on which components are populated. The values shown are for a low pass filter with a cut-off frequency of approximately 200HZ.

In the case shown R2 and R3 are set to 8.1K, and C1 and C4 are set to 0.1UFD. The component parts parallel to these are set to parasitic values (not populated) but would be populated for the high pass version and R2, R3, C1, and C4 would not be populated for the high pass configuration.

We can calculate the cut-off frequency value for either the high pass or low pass filter from the equation

Fo = 1 /( 2 * Pi * ( R2 * R3 * C1 * C4 ) ^ 0.5 ) = 196.48 HZ

R2 and R3 are usually set equal to each other and C1 an C4 are usually set equal to each other. We can then simplify the formula to:

Fo = 1 / ( 2 * Pi * R * C)

which we should recognize as the equation for a simple RC filter, but in this case we have a 2nd filter instead of a 1st order filter with the circuit of Fig. 1.

So if you know the frequency cut-off you want you can set the capacitor values and calculate the R values. For example, say you want a cut-off of 500 HZ and you will set the C values to 0.1UFD as before. The R values are then

R = 1 / ( 2 * Pi * Fo * C ) = 3.183 K

Or if we know the R values and the cut-off frequency, we can calculate the capacitor values;

C = 1 / ( 2 * Pi * Fo * R ) = 1.000 * 10^-7

or 0.1UFD.

The circuit in Fig. 1 was simulated with the components shown using +5V and -5V power supplies on a SPICE simulator and verified to have the calculated cut-off frequency.

What is a class T amplifier?

A class T amplifier is a type of class D amplifier except that it has some characteristics based on advanced technology in switching modulation and feedback techniques, such as very high switching frequencies, sigma-delta type pulse width modulation, and other usually proprietary features that are not published as generally available information. What might be some of the advantages of this type of design? We can expect that the higher switching frequency would help to reduce distortion and lower noise. Sigma-delta modulation spreads out the switching frequency over a range of frequencies which helps to reduce EMI testing problems.

What would we expect to be the disadvantages of the class T system? The variable switching frequency has some disadvantages over a fixed frequency system, especially in testing, and filtering out the switching frequency in the final output signal. The circuitry is usually more complicated and harder to design than the classical fixed frequency ramp generated pulse width modulation.

Should you buy a class T system? This is a question of personal taste, but I can tell you that the class T system has been very well accepted by large companies that are including the system in TV’s and home theatre systems. If the system sounds good and you do not want to spend a lot of money on high end equipment, class T may be good for you. If you are looking for a high end system you may want to shop carefully and do your research first. You may still find that a class T system is great and just right for you. Good hunting and good listening.

what are differences of full bridge vs. half-bridge class d

What are the advantages of full-bridge vs. half-bridge Class D amplifiers?

The following is a quick summary:

Half-Bridge
PWM Two-level switching (high and low logic levels)
Current Limited to the rating of each mosfet, assuming 1 mosfet high side, and 1 low.
No. of mosfets 2 minimum
Gate drive One dual gate driver possible
DC offset Critical -- must be virtually zero
Supply voltages Need positive and negative voltages
Distorsion Less than 1% possible
Bandwidth 20-20KHZ possible depending on design
Power regulation A pumping effect makes regulation more difficult

Full bridge
PWM Three-level possible
Current capacity Twice the half-bridge rating for the same mosfet types
No. of mosfets 4 minimum
Gate drive More complex than half-bridge
DC offset Self-canceling
Supply voltages One voltage can be used. Also can work from positive and negative supplies
Distorsion Much less than 1% possible
Bandwidth 20-20KHZ practical
Power regulation Easier than half-bridge

One comment on the gate drive system for the full-bridge circuit is that the logic is straight-forward but it may be necessary to add gate drive buffers to increase the strength of the gate drives.