Evaluation of a Class A Amplifier

This page is to evaluate an ideal amplifier circuit, starting with some theoretical ideals, creating one ideal amplifier and evaluating its characteristics. Theory is bright and sparkling so it's a lot easier to deal with. It's almost not useful, because any real device is not perfect. (If you actually had a transistor that were ideal, it would fit exactly. Real transistors are remarkably close, however.)

The two fundamental properties of an electric amplifier are voltage and current. For instance, if you have a device limited to say, 300V and 100mA---actually, before going on, let's *define* this device very precisely:

We can now define a family of curves for this device:


A graph of voltage vs. current shows a vertical line at Vo = 0V, changing squarely horizontal (ok, I drew a shallow knee, just for clarity), extending to the right (positive Vo) along the specified Io. (That is, at whatever level it is turned on to; fully on, it goes to 100mA; fully off, it remains flat, effectively an open circuit.)

For currents less than 100mA, this thing IS a switch. Think about it: it's not drawing any current in the off state, and voltage drops to zero when on. Limited to 100mA of course, but under that, it's a switch.

As such, you have an absolute maximum of 300 * 0.1 = 30 VA (not watts yet) capacity. You could use it at 300V and only 50mA, but now you're down to 15VA. Ditto for 150V 100mA = 15VA. It's pretty easy to see that 300V, 100mA is the most you can use this thing for.

That's the capacity of the amplifier. You can't just plug this thing into a supply rail and say it's working. If you turn it on, it certainly will dissipate 30 *watts*, but that's not useful power, just heat. One way to get use out of it is to insert a resistor and extract the alternating voltage produced across either the resistor or amplifier. (Since the supply rails are constant at 0 and 300V, it doesn't matter which way you remove it.) For the same ratings, you can immediately tell that the resistor must be 300/0.1 = 3kohms to use the maximum capacity of the amplifier. If it were higher, peak current would be lower (because you can only go up to 300V, no more); if it were lower, current would max out, and voltage would be limited by the constant current characteristic. In either case, power dissipated in the resistor goes down. This can be illustrated by graphing load lines:


Note that, fully off, no current flows through either element, while fully on, no voltage is developed across the amplifier (by definition) and all voltage is placed across the resistor, which thus dissipates 300 * 0.1 = 30 watts. If it takes exactly zero time to switch from off to on states, the amplifier element dissipates exactly zero power, while the resistor dissipates 30W -- 100% efficiency! This is called class D, and transistors may not be ideal, but they are damn close and can easily do 90% efficiency and better.


But anyway, we're concerned with linear amplifiers. For a symmetrical waveform, the solution is clearly to bias the circuit at the midpoint between on and off. For the same resistor and amplifier defined, this is 50mA, 150V (V = I*R, 50mA * 3k = 150V, added to zero or subtracted from 300, same thing). Now if you vary the amplifier from fully off to fully on with an audio signal, varying around this center (idle/bias/quiescent) point, you get such a signal appearing on the resistor. How much? Assuming the AC signal is loaded by the same resistor (V = I*R, so dV/dt = R * dI/dt, a hint of calculus there, d/dt something just means a rate change), which it is, then we can calculate the maximum power output, efficiency and all that goodness.

For zero signal, the change in voltage and current is zero (duh!), so output power is zero and efficiency is zero. (Total power consumption is 300 * 0.05 = 15W, amplifier dissipation is 150 * 0.05 = 7.5W and the load resistor is dissipating same, mind that it isn't dissipating *AC* power, the DC value doesn't count for output *signal*.) We can give the amplifier some signal and an alternating current appears on the output. Since current and voltage are directly related by the resistor (V = I*R at any instant in time), an AC voltage is also produced. We can drive it with more signal until the peaks are just reaching zero voltage and zero current. Let's say the output is a perfect sine wave. The peak-to-peak voltage, then, must be exactly 300V, and current 100mA p-p likewise (I = V/R!). Since a perfect sine wave is designated as having zero distortion, this is the maximum undistorted output that can be produced. How much power is it? Well, the RMS (Root-of-the-Mean-of-the-Square, sort of a Pythagorean theorem for signals) value of a sine wave is sqrt(2)/2 times the peak value, or sqrt(2)/4 times the p-p value, which we have. This comes to 108.066V, approximately. (I would ordinarily say 108V, or even just 110V, but since we're in an ideal world for this piece, we can keep irrationals like square roots in place. Not that it matters.) The signal power dissipated in the resistor is 1082 / 3k = 3.75W, in the ballpark (a bit high because real parts aren't ideal) for such a single-ended tube amplifier using these values.

What about efficiency? Is amplifier dissipation the same 15W it was at idle? Well think about it: I already mentioned that this particular amplifier will dissipate ZERO power if driven to only the ON and OFF states. If you apply an absurd amount of gain to the input signal, you can overdrive the amplifier so hard that it will be pushed into a square wave output. But if you apply less than that, say just a little clipping, or even just use a small signal instead, there simply isn't a point where it just *jumps* from 15W to zero. No, it must be a smooth transition. This tells us that signals *must* be reducing power dissipation, in relation to how much signal there is. How much?

Welllll, I would very much enjoy chugging through the sin(time) definite integral to determine it exactly, but it just so happens, there's an easier way, and it just so happens, it is also exact, and requires no calculus whatsoever! How can that be? Well...

Any time you have a current draw from a DC supply, you have power. Period. There is NO reactive current in a DC supply. Sure, you can send current back into the supply, but if there's more drawn out than put back in, it averages out and you   are   drawing   power! (The observant will note that any such "drawing and replacing of current" is clearly an alternating current phenomenon, and outright doesn't apply to a DC supply! Only the average current is meaningful to power consumption.)

So, we know the thing is +300V. We can stick an ammeter in, and see that it's drawing 50mA. But wait, that was the idle value -- how do we know it's still 50mA? Well, you can stick an ammeter in....... But on paper, you have to calculate what the ammeter sees in the first place! It turns out that, since the amplifier is producing a pure sine wave, the signal is perfectly symmetrical, and a high current draw is balanced in the other half of the cycle with a low current draw. Interestingly, this still holds for an overdriven (distorted) signal, where the peaks of the sine wave get clipped off and flattened. The ultimate clipping distortion, producing a square wave, gives a 50% duty cycle square wave, which simply means half the time, the amplifier is fully ON, the other half, OFF. This too averages to (100 + 0)/2 = 50mA. An amplifier which is always conducting to some extent is called "class A".

That said, we know with certainty that the amplifier will always draw an average 50mA, in all conditions, zero signal or full output or overdriven. Supply is 300V, so power consumption is always 300 * 0.05 = 15W. At idle, voltage is 300/2 = 150V constant, so power is split evenly, 7.5W dissipated in the amplifier and resistor. We know that, at full (undistorted) power output, it makes 3.75W, which must subtract from total power dissipation. But where? The tricky part is I said the 3.75W is dissipated in the resistor, so in addition to the 7.5W it dissipates at idle. Power consumption is constant, so it must come from an equal amount less in the amp. (Again, the extreme case of this, the amp fully on or off, dissipates all 15W of the system in just the resistor, and none in the amp! This should be expected, though it seems remarkable that an amplifier dissipates less when driving harder.)

Now we can calculate efficiency. Total power consumption is 15W, and output is 3.75W, all of 25% efficiency. Incidentially, since this is a theoretical exploration with an ideal amplifier, this happens to be the absolute maximum efficiency you can ever hope to get from an undistorted class A amplifier. Transistors come close, and tubes have a fat chance at this. Against amplifier dissipation (called "plate efficiency" with tubes), dissipation is 7.5 - 3.75 = 3.75W so this efficiency is 50%.

Next: reactive loads such as choke and transformer loads, and maybe some calculus to explore the scenic route to the sine wave RMS value, and power dissipations.

Return to Electronics

Web page maintained by Tim Williams. All rights reserved.