The Power of Two
Power Amplifier Kits
For builders of electronic audio kits there is a plethora of stock been advertised either from the East or Europe. Whilst I am sure many of these kits are good quality, with a few exceptions, buyers should be cautious of the overzealous power ratings advertised. A case in point would be the ‘rugged’ design using a very popular TO3 NPN transistor, the 2N3055. Compared to modern devices these transistors are still very popular and with careful design can give quite a remarkable sound quality but there are limitations as to what power you can get safely by using only two transistors in a quasi complementary output. This not unlike any other output stage of course but the ratings I have seen advertised go way beyond the norm. Four transistors used in the output stage used to produce about 125W R.M.S. into a 4 Ohm load at 80V (taking supply droop into account). Regulating that supply to 80V DC you are looking at (((80/2 * 0.707)/4) * ((80/2 * 0.707)/4)) * 4, a theoretical output of 200W R.M.S. The formula used is just a play on Ohm’s Law to get Power or I*I*R in R.M.S. value. Four output transistors in a quasi-complementary output stage should suffice, but err on the side of caution and add an additional pair. The amplifiers I have seen advertised can nowhere dissipate that type of power with only two transistors in an output stage, driven in bridge mode into a 4 Ohm load. Design engineers and manufacturers always resort to the transistor specification and the Safe Operating Area or SOA. This will always be given – if not, then do not use the transistor. Of course, stay clear from counterfeits as well, they never do what they are supposed to. As they say, there is nothing like a free lunch, especially when it comes to peace of mind. Good quality transistors are expensive.
R.M.S. or Root Mean Square
R.M.S. power is significant here because all good amplifiers are measured in R.M.S. right? Not really. R.M.S. is merely an alternating current circuit’s equivalent direct current heating capability. In other words, an alternating current circuit with an output measured at 40V peak on the positive going waveform is 0.707 * peak or 28V equivalent D.C. voltage aross a 4 Ohm load (pure resistance, no inductive or capacitive loading) and using Ohm’s Law we get to current V/I or 28/4 = 7A. Power dissipation into this 4 Ohm load is calculated via means of I * I * R or 7 * 7 * 4 = about 200 Watts.
R.M.S. is therefore not Real Mean Sound neither Real Media Sound – it’s a term we love to throw around, that’s all. When measuring amplifier output we also need to know how much distortion is been delivered as well and of course at what frequency. Loudspeakers are really inefficient little blighters. The voice coil is inductive, there is also capacity. In an inductive load current drops as the frequency is raised whilst in the capacitor, the reverse is true. In theory the ideal amplifier output is measured without any distortion at a fixed and common frequency, say 1kHz into a pure resistance of say 8 Ohm. If we do not know what the exact resistance (in a.c. circles we call this impedance) of the loudspeaker at 1 kHz how do we know that the amplifier is in actual fact dissipating the power calculated by measuring the amplitude of the waveform with an oscilloscope and assuming the ‘resistance’ of the loudspeaker is 8 Ohms. So we can safely assume that with a pure resistance of 4 Ohms and measuring the amplitude which may be 40V peak or 28V R.M.S. we have 200W dissipation. A quick methode to calculate what the amplifier ‘could’ put out in terms of power is Vcc squared divide by 8 x load resistance. Therefore if the amplifier supply rails are +40V – 0 – -40V and the load is 4 Ohms = (80 * 80) / 32 = 200W.
Don’t burn your Bridges
Now what happens if we decide to bridge the amplifier. We need two audio amplifiers, the inputs driven 180 degrees out of phase with each other. Sound simple? It is. The problem now is dissipation because we really don’t double the output dissipation because we have two amplifiers feeding one load. In fact if the previous amplitude was 28V R.M.S, this is doubled but not the output dissipation. No, the 56V R.M.S. across a 4 Ohm load means we have V/I or 14A pushed through and in terms of output dissipation, 14 * 14 * 4 or 800W, four times the amplifier rating. If an amplifier is designed to dissipate 200W into a 4 Ohm load, in bridge mode it will tested with an 8 Ohm load, not 4 Ohms. Into 8 Ohms we dissipate 400W, double of the amplifier rating into 4 Ohms. Often modern amplifiers are designed to dissipate large quantities of power into 2 Ohms but on bridging they are designed for a 4 Ohm load.
An amplifier which is designed to drive a 4 Ohm load where the output stage is running just within spec will be damaged if bridged and driving a 4 Ohm load. For that reason rather not bridge and never use a lower impedance that that intended.
Tube and some class D amplifiers on the other hand should never be run without a load. More on this some other time.
Before buying your first or next amplifier kit be sure to know how they did the calculations. The supply rail voltage and transformer VA rating is always a good indication of what power to expect – knowing the specification of the output transistors is sometimes more important.