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LM4810 датащи(PDF) 13 Page - National Semiconductor (TI) |
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LM4810 датащи(HTML) 13 Page - National Semiconductor (TI) |
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13 / 18 page ![]() Application Information (Continued) the magnitude of “clicks and pops”. Increasing the value of C B reduces the magnitude of turn-on pops. However, this presents a tradeoff: as the size of C B increases, the turn-on time increases. There is a linear relationship between the size of C B and the turn-on time. Here are some typical turn-on times for various values of C B: C B T ON 0.1µF 80ms 0.22µF 170ms 0.33µF 270ms 0.47µF 370ms 0.68µF 490ms 1.0µF 920ms 2.2µF 1.8sec 3.3µF 2.8sec 4.7µF 3.4sec 10µF 7.7sec In order eliminate “clicks and pops”, all capacitors must be discharged before turn-on. Rapidly switching V DD may not allow the capacitors to fully discharge, which may cause “clicks and pops”. In a single-ended configuration, the output is coupled to the load by C O. This capacitor usually has a high value. C O discharges through internal 20k Ω resistors. Depending on the size of C O, the discharge time constant can be relatively large. To reduce transients in single-ended mode, an external 1k Ω–5kΩ resistor can be placed in par- allel with the internal 20k Ω resistor. The tradeoff for using this resistor is increased quiescent current. AUDIO POWER AMPLIFIER DESIGN Design a Dual 70mW/32 Ω Audio Amplifier Given: Power Output 70 mW Load Impedance 32 Ω Input Level 1 Vrms (max) Input Impedance 20k Ω Bandwidth 100 Hz–20 kHz ± 0.50dB The design begins by specifying the minimum supply voltage necessary to obtain the specified output power. One way to find the minimum supply voltage is to use the Output Power vs Supply Voltage curve in the Typical Performance Char- acteristics section. Another way, using Equation (5), is to calculate the peak output voltage necessary to achieve the desired output power for a given load impedance. To ac- count for the amplifier’s dropout voltage, two additional volt- ages, based on the Dropout Voltage vs Supply Voltage in the Typical Performance Characteristics curves, must be added to the result obtained by Equation (5). For a single-ended application, the result is Equation (6). (5) V DD ≥ (2V OPEAK +(VODTOP +VODBOT)) (6) The Output Power vs Supply Voltage graph for a 32 Ω load indicates a minimum supply voltage of 4.8V. This is easily met by the commonly used 5V supply voltage. The additional voltage creates the benefit of headroom, allowing the LM4810 to produce peak output power in excess of 70mW without clipping or other audible distortion. The choice of supply voltage must also not create a situation that violates maximum power dissipation as explained above in the Power Dissipation section. Remember that the maximum power dissipation point from Equation (1) must be multiplied by two since there are two independent amplifiers inside the package. Once the power dissipation equations have been addressed, the required gain can be determined from Equa- tion (7). (7) Thus, a minimum gain of 1.497 allows the LM4810 to reach full output swing and maintain low noise and THD+N perfro- mance. For this example, let A V=1.5. The amplifiers overall gain is set using the input (R i ) and feedback (R f ) resistors. With the desired input impedance set at 20k Ω, the feedback resistor is found using Equation (8). A V =Rf/Ri (8) The value of R f is 30k Ω. The last step in this design is setting the amplifier’s −3db frequency bandwidth. To achieve the desired ±0.25dB pass band magnitude variation limit, the low frequency response must extend to at lease one−fifth the lower bandwidth limit and the high frequency response must extend to at least five times the upper bandwidth limit. The gain variation for both response limits is 0.17dB, well within the ±0.25dB desired limit. The results are an f L = 100Hz/5 = 20Hz (9) and a f H = 20kHz * 5 = 100kHz (10) As stated in the External Components section, both R i in conjunction with C i, and Co with RL, create first order high- pass filters. Thus to obtain the desired low frequency re- sponse of 100Hz within ±0.5dB, both poles must be taken into consideration. The combination of two single order filters at the same frequency forms a second order response. This results in a signal which is down 0.34dB at five times away from the single order filter −3dB point. Thus, a frequency of 20Hz is used in the following equations to ensure that the response is better than 0.5dB down at 100Hz. C i ≥ 1/(2π * 20kΩ * 20Hz) = 0.397µF; use 0.39µF.(11) C o ≥ 1/(2π *32Ω * 20Hz) = 249µF; use 330µF. (12) The high frequency pole is determined by the product of the desired high frequency pole, f H, and the closed-loop gain, www.national.com 13 |
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