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| Please need help to calculate gain from circuit. |
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| Audioguru:
I think it will sound awful with a lot of odd harmonics distortion since the output emitter-followers are not inside the negative feedback loop. Some crossover distortion also might be noticeable. Why not make a "normal high fidelity amplifier circuit" instead, or use an LM3886 IC? |
| Zero999:
--- Quote from: b_force on June 24, 2018, 01:25:26 pm --- --- Quote from: Hero999 on June 24, 2018, 01:07:48 pm ---Yes, 25 sounds about right. --- Quote from: FriedMule on June 24, 2018, 12:37:39 pm ---Thanks a lot, so it is about 25Vpp out pr channel? Does that not sound as about 36W out? EDIT: the transformer are 26V-0V-0V-26V on the schematic, it is about 80W out, why? --- End quote --- What's the load impedance? For a sine wave, the RMS power is equal to half the peak power, so with a 4Ohm load: P = V2/(2R) = 252/(2*4) = 625/4 = 78W. A 26V transformer will output 36V, after losses in the rectifier, so 25V peak out seems reasonable. I doubt it'll be able to output full power continuously, as the smoothing capacitors aren't big enough, but the output transistors will probably overheat too, at that power level. It's also not a problem, since peak power occurs typically less than 10% of the time, even at full volume. --- End quote --- Ehm, well better is the use to divide the rail voltage (which is the peak voltage of the sine wave) by the square root of two (= 1.41). Which is obviously the same as the Rms value of the AC voltage (minus some losses), so in this case 26Vrms. (26^2)/8 = 85W @ 8ohm or 170W @ 4ohm (the rail voltage is 26 * 1.41 = 36.7V) However, with Class-AB amplifiers it's not uncommon to give a lot more headroom in the rail voltage. The efficiency of a AB stage is about 60-70% at most, so that means that transistors will dissipate around 50W @ 4ohm. That's quite a lot, so that makes me think, including the low amount of capacitive storage, this amp is suitable for 8ohm or higher. --- End quote --- The original poster said the output voltage is 25V peak, so the output power will be 78W into a 4Ω load or half that into an 8Ω load, irrespective of the supply voltage. |
| b_force:
--- Quote from: Hero999 on June 24, 2018, 11:06:45 pm --- --- Quote from: b_force on June 24, 2018, 01:25:26 pm --- --- Quote from: Hero999 on June 24, 2018, 01:07:48 pm ---Yes, 25 sounds about right. --- Quote from: FriedMule on June 24, 2018, 12:37:39 pm ---Thanks a lot, so it is about 25Vpp out pr channel? Does that not sound as about 36W out? EDIT: the transformer are 26V-0V-0V-26V on the schematic, it is about 80W out, why? --- End quote --- What's the load impedance? For a sine wave, the RMS power is equal to half the peak power, so with a 4Ohm load: P = V2/(2R) = 252/(2*4) = 625/4 = 78W. A 26V transformer will output 36V, after losses in the rectifier, so 25V peak out seems reasonable. I doubt it'll be able to output full power continuously, as the smoothing capacitors aren't big enough, but the output transistors will probably overheat too, at that power level. It's also not a problem, since peak power occurs typically less than 10% of the time, even at full volume. --- End quote --- Ehm, well better is the use to divide the rail voltage (which is the peak voltage of the sine wave) by the square root of two (= 1.41). Which is obviously the same as the Rms value of the AC voltage (minus some losses), so in this case 26Vrms. (26^2)/8 = 85W @ 8ohm or 170W @ 4ohm (the rail voltage is 26 * 1.41 = 36.7V) However, with Class-AB amplifiers it's not uncommon to give a lot more headroom in the rail voltage. The efficiency of a AB stage is about 60-70% at most, so that means that transistors will dissipate around 50W @ 4ohm. That's quite a lot, so that makes me think, including the low amount of capacitive storage, this amp is suitable for 8ohm or higher. --- End quote --- The original poster said the output voltage is 25V peak, so the output power will be 78W into a 4Ω load or half that into an 8Ω load, irrespective of the supply voltage. --- End quote --- Well, he said a 26V transformer. But anyway, it doesn't matter. Nothing is wrong here, I only wanted to point out that it helps to calculate the voltage, instead of dividing the power by two. Both methods obviously work, but if someone is going to bridge an amplifier for example, it shows better that de power goes with a factor of two. |
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