Class-AB Efficiency vs. Output Power: Related to Dynamic Resistance (Re) of BJT?

[TimNJ]

Hi all,

I've been studying some Class-AB IC amplifier datasheets and I've noticed that a typical efficiency vs. output power curve usually looks like attachment #1.

Notice that the efficiency is quite terrible at low output powers and gradually increases with power, essentially logarithmically.

Here's how I've been trying to explain this to myself: A small current into the base (Ib) means a small collector current (Ic). A small Ic derived from a relatively high voltage supply (Vcc) means the transistor has to present a high dynamic resistance (Re) to limit the current through the load. If we freeze the time-varying signal in an instant of time, or perhaps take its RMS value, we can think of the transistor and the load impedance (RL) as the top and bottom halves of a voltage divider. Under these conditions, suppose the transistor "looks like 24 ohms" and the load is an 8-ohm speaker, then only 25% of the power delivered by the supply will be applied to the speaker, and the rest 75% will be burned up as heat in the transistor.

Now, if we apply a larger base current (crank up the volume), then the dynamic resistance goes down, but the load impedance, of course, remains the same. Therefore, a greater percentage of the supplied power is applied to the speaker. For instance, if the transistor now presents a dynamic resistance of 8 ohms, then the efficiency is 50%.

It seems that the typical efficiency curve of a Class-B/AB amplifier matches up with Ic vs Vbe curve of a BJT. Near cutoff, the dynamic resistance is quite high, but gets lower and lower with greater Vbe (or Ib). Is this the main driving force behind why these amplifiers have poor efficiency at light load?

Thanks!

[Benta]

A somewhat tortuous analysis and missing the point. You could do the same analysis for a class AB power amp using MOSFETs and where is your V_BE/I_C relationship then?

Forget the whole transistor characteristics relationships and simply model an ideal amplifier with zero bias current (ideal class B), a little bias current (class B), a lot of bias current (class AB) and full bias current (class A). Model the power supplied to the speaker and the power dissipated. This should give you the right answer.

[TimNJ]

Thanks. I appreciate the suggestion and it sounds like that will clear it up once I run through those examples.

But regarding the Vbe/Ic equivalent MOSFET relationship. Does not the Vgs/Id relationship look quite similar?

Thanks again.

[Benta]

No.
BJTs are current controlled, which means I_BE vs I_C is relevant.

Still, you should look at audio power amps as voltage sources no matter which type of power transistors are used.

[T3sl4co1l]

Well... BJTs are voltage controlled, but nevermind that.

Better question: what possible relationship, other than coincidence, would r_e have with output efficiency?  What happens for different load resistances?  Does the relationship hold?  What about different supply voltage or current, or different transistors (the FET equivalent to r_e is the small-signal source impedance, 1/Gm)?  What about a theoretical device with constant Gm, or negative incremental Gm (Gm decreases at high current)?  (We can approximate this by adding series resistance to the transistor's common terminal.)

Incidentally, negative incremental gm defines the class D amplifier.  That's a roundabout way of saying, the transistor is operated in saturation, where increased Vgs (or whatever) causes no increase in Id (because, in this case, Vds is so low there's no room left, so of course gm is trash).  So that would seem evidence that the properties are independent.

Tim

[TimNJ]

Well... BJTs are voltage controlled, but nevermind that.

Better question: what possible relationship, other than coincidence, would r_e have with output efficiency?  What happens for different load resistances?  Does the relationship hold?  What about different supply voltage or current, or different transistors (the FET equivalent to r_e is the small-signal source impedance, 1/Gm)?  What about a theoretical device with constant Gm, or negative incremental Gm (Gm decreases at high current)?  (We can approximate this by adding series resistance to the transistor's common terminal.)

Incidentally, negative incremental gm defines the class D amplifier.  That's a roundabout way of saying, the transistor is operated in saturation, where increased Vgs (or whatever) causes no increase in Id (because, in this case, Vds is so low there's no room left, so of course gm is trash).  So that would seem evidence that the properties are independent.

Tim

Hey. Sorry for the late reply. I've read it a few times and I've tried to formulate a sensible answer with good questions, but honestly, it seems that I don't know as much about discrete transistor amplifiers (or transistors, in general) as I thought I did..

From what I can gather, the theoretical efficiency of a Class B amplifier (for instance) depends on the percentage of maximum output swing (of the total available Vcc). My research points me to a theoretical efficiency curve which is equal to n = a*(pi/4), where a is the percentage of maximum swing between 0 and 1. That's 78.5% maximum for Class B.

As far as the relationship between r_e and output efficiency, I'm still thinking about that one. For one, I think r_e works to reduce the gain of the amplifier. Assuming constant Gm, and thus a singular value of r_e, then I would say it would decrease the efficiency proportionally(?) That is, efficiency will be reduced across the board by a factor of RL/(r_e + RL)...So if r_e is 1 ohm and the load is 8 ohms, then the maximum theoretical efficiency will be 8/9 * 1 * (pi/4) = 69.8%. Different transistors will incur more loss than others, it seems. At the expense of?

Anyway, thank you very much for the good pointers.

[T3sl4co1l]

Note that you only ever use gm OR r_e, never both.  That would double count it!  r_e is the hybrid-pi model equivalent of gm.  That's all.

Perhaps a counterpoint would help: what if the device takes 20V input drive to make 10V out?  How about 100V in?  What does it matter?

If the input current is approximately zero (many times smaller than the output current, anyway), there is very little power required to drive the outputs, and efficiency only depends on the output voltage range (in terms of total supply voltage). 

Think about what an amplifying device is, as an ideal transistor or what have you.  It is a device which draws some current through an output circuit, dependent on some input (voltage or current).

Suppose it's linear, so Ic = Vbe * gm, and if you like, Ib = Ic / hFE (where hFE is constant).  Which means Vbe/Ib == Rb, a constant base resistance (for small signal purposes, in the hybrid-pi model, this is called r_π).  Suppose there's no Early effect, so the collector output curves are flat horizontal.  Then r_e = 1/gm, as well.*  With no limit to voltage or current (it's a linear device with no bounds), you can pick any arbitrary area on the output curve, draw a load line across it, and drive that load with power equal to the area under the line.  Now we can look at power output and device dissipation -- efficiency -- for a general case, with no quirks of an underlying device to worry about!

*Alternately, we might choose vertical curves, so that Vce = Vbe * mu (i.e., a voltage gain mu), or an intermediate case where the curves have a finite slope, or a variable resistance ("triode region").  For the finite slope case, there is the relationship: gm * h_oe = mu (h_oe is the slope of the output curve, the output resistance -- normally given by Early effect).  If you like, the constant-current case has infinite mu, while the constant-voltage case has infinite gm.  Note that the constant-current family of curves with parallel output resistance, is equivalent to the constant-voltage family of curves with series output resistance, by the Thevenin-Norton source theorem.

Load lines -- for further reading, google the term.  It's a geometric technique that's as old as, well, tubes.   It's not important -- not taught -- today, because transistor curves are less reliable in some respects (e.g., Early effect, hFE variation), more reliable in others (Ic = Is * e^(Vb/Vth)), and much more flexible in ratings (whereas a given tube is a huge waste if operated much below its ratings), but at the same time... teaching a method in the abstract isn't suitable for all readers, so it's nice to check out.

Tim