I have decided to construct an amplifier using single chip components, so-called 'chip amps', and I thought it would be an opportunity to show those without the knowledge or experience the steps involved in constructing such an amplifier. First, we must decide on a few things, like power output, power supply, components, general construction and layout. What are the needs?:
I need an amplifier to power my computer loudspeakers, and my lab loudspeakers. So what are the power requirements of these loudspeakers? The computer L/S have a power handling of 30 watts each drive unit, so they will handle 60 watts, although at 4 ohms! The tweeter will handle 50 watts of music power, whatever that is. The lab. speakers will handle 50 watts for the bass/mid range, and 60 watts for the tweeter, a little more if I add a heat sink!. So 50 watts seems to be a good figure to aim at.
Looking at a few specs has shown that the TDA7294 will provide the necessary power.
The spec sheet is here: spec sheet TDA7294
Depending on power supply voltages, the output is above 50 watts into 4 - 8 ohms, at less than 0.1% harmonic distortion. It will suffice. What we need next to decide on is the power supply. We need to consider the voltage on the power rails, and the current they need to supply.
Here, we need Ohm's Law, voltage = current x resistance.
How much voltage and current does an 8 ohm loudspeaker require?
Well, we have three terms, but two unknowns, but we also know that power is in watts,
and watts is just voltage x current.
So power = 50 watts = voltage x current.
But from Ohm's Law, we know that voltage is current x resistance, so we can re-write
power = current x current x resistance, often quoted as I²R, just as watts is sometimes referred to as VA.
so, power output = 50 watts or I²R, and we know resistance is roughly 8 ohms, so
power = 50 watts = I² x 8, so that I² = 50/8 , so that the square root of 50/8 = the current required = 2.5 amps.
So our 50 watt amplifier needs a supply of 2,5 amps into 8 ohms, which, from Ohm's Law, we get the voltage
V=IxR = 2.5 x 8 = 20 volts, RMS (where did that come from?) RMS stands for root mean square, and is a way of expressing power in an alternating current (AC) circuit, as an amplifier for music is. To get the peak voltage, we simply multiply by the SQRT2, or 1.414, to get 28 volts, or there abouts. So we need a voltage rail or 28 volts to get 50 watts RMS? Not quite, but nearly. The output transistors can drop about a couple of volts, so 30 volts would be nearer the mark.
So what size power supply do we need? Surely its just 30 volts at 2.5 amps? which is 75 watts or VA. Well, no. The amplifier will generate a lot of heat, because it is not very efficient. Look at the spec sheet, and look at the power dissipation for a given output power, fig14., so for our ±30 volt supply, we are looking at a dissipation max of 24 watts when driving 8 ohms. So we will need 100 watts for a single amplifier, for two, that's 200 watts, or VA, as transformers are often quoted in.The output from the amplifier will be about 75 watts peak, or 50 watts RMS.
Now what about the heat sink requirements? We have seen that the amplifier will generate a maximum of 24 watts of heat for the 50 watt RMS output we require, so we will need to get rid of that heat, if the amplifier is survive longer than a few minutes. We can do that by using an appropriate heat sink, of sufficient dimensions. Heat sinks are usually quoted in °C/W, that is, they will rise in temperature x°C for every watt dissipated. We have 24 watts to dissipate.
The other thing to consider is what is the maximum temperature do we want the heat sink to reach without burning ourselves, or shutting down the amplifier, as it has auto over temp shut down.
Lets say the heat sink should not be more than a hand could stand, this is about 60°.
This is including ambient temperature, of say 25°C, so the rise is 60-25 = 35°C and we have 24 watts to dissipate, so it is 35/24 °C/watt, or about 1.5°C/W. However, this assumes a constant output of 50 watts into 8 ohms, which is unlikely, unless we listen to very loud rock music with significant clipping. In normal operation, the power dissipation is likely to be far less. However, the computer 'speaker is essentially a 4 ohm design, so the dissipation will be higher.
These are the components I've decided to use.
Here you see the two built amplifier boards, one of the two heat sinks, and a common power supply, a 200VA transformer, a 35 amp bridge rectifier (the square thing in the middle), and a couple of smoothing capacitors. I'm constructing what I've termed a 'pseudo-active' amplifier system, a separate amplifier in each loudspeaker (together with crossover components), and a common power supply.
The transformer, bought from ebay for little coin, was from a B&O unit, giving 23-0-23 volts (and a couple others) at 200VA, just what is needed.
When the AC 23 volts is rectified by the diode bridge, it will give 23 x sqrt(2) volts, DC, which is 32.5 volts, or there abouts, as it will vary a little with the current drawn. All that is really required to complete the main part of the power supply are a couple of smoothing capacitors, hung between the 30 volt rails and earth. The amplifier board has more capacitors of lower capacitance, including a couple of 0.1 µF polyester ones, to earth very high frequencies.
to be continued.........