Here I am referring to class B output power amplifier.

enter image description here

This circuit should be easy to build and understand but I'm having problems with biasing since I don't really know how to bias the bases of Q1 and Q2, so that Q1 would conduct only positive polarity signals and Q2 would conduct only negative polarity signals.

It seems that I only managed to properly bias class A amplifier, but not class B.

  • How would I had to bias the upper circuit to achieve class B operation of an amplifier?
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    \$\begingroup\$ There is some discussion related to tweaking the vbias here: 9V battery amplifier. Note that it also discusses bootstrapping, which oldfart refers to in his added comment to you. \$\endgroup\$ – jonk Dec 24 '17 at 20:43

There is a simple known circuit which works as a 'programmable zener'. Below is the principle diagram:


simulate this circuit – Schematic created using CircuitLab

For a real application the variable resistor may be split in three parts to get more accurate control. By varying the resistor you can set the 'zener' voltage between the bases of the two transistors Q1 and Q2 and as such control the quiescent current.

Forgot: Just as a real zener it needs a resistor at the top.

In the good old days that transistor was physically mounted on the heatsink so you also had thermal compensation. Took me a while to find an image on the www but here is one: enter image description here

Post edit
As mentioned in the comment below you have to be careful with this circuit. Before first time use you must make sure the variable resistor set such that the base is at the collector voltage. Thus there is minimal voltage drop. Then you turn the resistor until the bias is 'correct' which normally means you no longer see (scope) hear (ears) the distortion in the output signal. You can turn it a bit further which will increase the quiescent current in output stage. (It will get more the characteristic of a class A amplifier.)

  • \$\begingroup\$ Instead of that Vbias in my circuit, this should replace it? \$\endgroup\$ – Keno Dec 24 '17 at 14:19
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    \$\begingroup\$ Yes, but you need a resistor from V+ as it needs to get current from somewhere. Beware if the zener voltage is set too high the first time you use it, both end stage transistors will be conducting so you have a short from V+ to V-. Make sure the base is connected to the collector! Then slowly turn it down and measure the current in the end stages. \$\endgroup\$ – Oldfart Dec 24 '17 at 14:29

First, understand this is just a double emitter follower using a darlington on each side. The voltage at the output will be pretty much the voltage at the opamp output. The purpose of the emitter followers is to provide current gain.

If each transistor has a gain of 50, for example, then the current the opamp has to source and sink is roughly 50 * 50 = 2,500 times less than what the load draws. For example, if the load is drawing 1 A, then the opamp only needs to source 400 µA.

One problem with a emitter follower is that the output voltage differs from the input voltage by the B-E drop of the transistor. Let's say for sake of example that's about 700 mV when the transistors are operating normally. For a NPN emitter follower, you have to start with 1.7 V in if you want 1 V out. Similarly, for a PNP emitter follower, you have to put -1.7 V in if you want -1 V out.

Due to two transistors being cascaded, this circuit has two 700 mV drops from the opamp to the output. That means to drive the output high, the opamp has to be 1.4 V higher. To drive the output low, the opamp has to be 1.4 V lower.

You wouldn't want the opamp having to suddenly jump 2.8 V when the waveform switches between positive and negative. The opamp can't do that suddenly, so there would be a small dead time at the zero crossing, which would add distortion to the output signal.

The solution used by this circuit is to put a 2.8 V source between the inputs to the high and low side drivers. With 2.8 V difference in drive level, the two output drivers will be just at the edge of being on at 0 output. A little higher input and the top driver will start sourcing significant current. A little lower, and the bottom driver will start sinking significant current.

One problem is getting this offset just right to eliminate the input jump required at zero crossings, but not turn on both drivers so much that they end up driving each other. That would cause useless current to flow and dissipate power that isn't going to the load. Note that 700 mV is just a rough value for the B-E drop. It's reasonably constant, but it does change with current, and also with temperature. Even if you could adjust the 2.8 V source exactly, there isn't a single exact value to adjust it to.

This is what RE1 and RE2 are for. If the 2.8 V offset is a little too high and significant quiescent current starts flowing thru both the top and bottom drivers, then these resistors will have a voltage drop across them. Whatever voltage appears across RE1+RE2 directly subtracts from the 2.8 V offset from the point of view of the two drivers.

Even 100 mV can make a significant difference. That will be caused by 230 mA of quiescent current. Note also that 700 mV is probably on the low side, especially for the power transistors when they carry significant current.

All in all, the 2.8 V source is meant to keep each of the top and bottom drivers "ready", without turning them on enough so that they start fighting each other and dissipating a lot of power.

Of course, everything is a tradeoff. In this case you can trade off more quiescent current for a little less distortion.

Ideally, in class B one side shuts off completely when the other starts taking over. That almost never happens in practice, but this scheme is reasonably close to it.

  • \$\begingroup\$ Is this the point where switching distortion takes its place? In my book, if I understood it correctly, it is described as both sides (npn and pnp) conduct more than 180 degree of signal? \$\endgroup\$ – Keno Dec 24 '17 at 17:06
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    \$\begingroup\$ @Keno: Crossover distortion can happen both ways. The worst is usually when the high and low side drivers conduct less than half the time. The opamp has to jump over the deadband, which takes finite time. Each conducting for more than half the time does not necessarily cause distortion. It depends on how smoothly they fade in and out relative to each other. Both conduct all the time in class A, for example, and more than half the time in class AB. That's the point of class AB versus class B. Some fade over represents wasted power but not necessarily distortion. A deadband distorts. \$\endgroup\$ – Olin Lathrop Dec 24 '17 at 17:58
  • \$\begingroup\$ I agree with you! But as close as we could get to class B more efficient the amplifier would be, right? \$\endgroup\$ – Keno Dec 24 '17 at 18:01
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    \$\begingroup\$ @Keno: Yes, class B is the optimum efficiency for a linear pass element system. Getting the two sides to switch over exactly right is very difficult. That's why class AB. Allow a little fade over to reduce crossover distortion, at a small price in efficiency. \$\endgroup\$ – Olin Lathrop Dec 24 '17 at 18:04
  • \$\begingroup\$ One more thing. The point/area of conduction where both npn and pnp sides are conducting simultaneously, can this add additional distortion to amplifier or is that simultaneous conduction area not the subject of distortion? \$\endgroup\$ – Keno Dec 24 '17 at 21:04

The difference between class A and class B is the quiescent current through the last stage.

If you make the quiescent current zero then only Q3 or Q4 supplies current when a signal is present. This is class B.

If you make the quiescent current so large that for very large signals (even the largest) both Q3 and Q4 never have an Ic = 0 (are never off), we have class A.

There's also class AB which can be anywhere between class A and class B.

How to set this quiescent current?

That is done by Vbias.

Some examples how Vbias can be implemented:

  • the "Zener" from oldfart's answer

  • a real Zener diode

or this:


simulate this circuit – Schematic created using CircuitLab

The current source can easily be made with a PNP current mirror and a biasinf resistor.

  • \$\begingroup\$ Do you have any ideas how to know for sure, whether the the circuit operates in class A or class B or in between, that is class AB? I scoped the output while changing the bias but all I get is normal sine wave. I could verify the class by measuring the quiescent current through each of transistors, but is there any other way? Maybe with o'scope? \$\endgroup\$ – Keno Dec 24 '17 at 18:03
  • \$\begingroup\$ You can easily measure the current through Q3 and Q4 across the emitter resistors. So apply no signal and measure the current. My guess is that with VBias = 2.8 V this will be a class AB amplifier. Also in class B there will be crossover distortion at the zero crossings. \$\endgroup\$ – Bimpelrekkie Dec 24 '17 at 18:40
  • \$\begingroup\$ @Bimpelrekkie has drawn two examples of a class AB output stage. A small current is always flowing through Q1 and Q2, Q3 and Q4. With sufficient idle current the distortion can be very very low, perhaps .05% or less, but the trade off is that the output stage dissipates a lot of heat. Look up 1,500 watt amplifiers on the web and you will see similar but more elaborate bias designs. \$\endgroup\$ – user105652 Dec 25 '17 at 6:54

You have to understand the output topology well in order to know how to create the biasing for it.

Although someone did mention that your schematic example has the BJTs arranged in Darlington fashion (with added turn-off speed-up resistors), they didn't tell you that such an arrangement almost always has a better topology. So you'd almost never use that topology to begin with. Or, in short, there's no point struggling to understand it in order to bias it.

Why use a Darlington:

  1. High current gain, which is useful in output driver circuits like this because it significantly reduces the biasing circuit's quiescent current and that can be a big help when attempting to slug around large current swings into a small load like this.

Why not to use a Darlington:

  1. Slow turn-off unless a resistor is added (as it is in your circuit example.)
  2. Cannot saturate below approximately one diode drop (plus a little) because of the arrangement. This may mean some added voltage overhead required for the amplifier (which for lower voltage circuits may be unacceptable) and that also may mean some added overall dissipation for the amplifier.
  3. Acts as though it requires two diode drops between the base and emitter, which increases the required biasing voltage span.
  4. Temperature affects both base-emitter junctions, which add in series. So the temperature variation of the biasing voltage span now includes at least four diode drops in series, all of which experience variation over temperature. The complexity of compensation is likely increased, as a result.
  5. There are better alternatives.

The last reason is the main reason about why not to use a Darlington here. If there were no alternatives, then you'd merely be stuck with the idea if you wanted its single advantage.

If you want the high current gain of the Darlington arrangement, then it is almost always better to use the Sziklai arrangement, instead. It looks like this:


simulate this circuit – Schematic created using CircuitLab

This also provides similar high current gain and also cannot saturate below about one diode drop, but also includes the following:

  • Only one base-emitter diode drop per quadrant.
  • \$R_3\$ and \$R_4\$ can be arranged so that \$Q_2\$ and \$Q_4\$ pick up a significant portion of the peak currents (say 25-30%?) This helps stabilize the base-emitter variation of \$Q_1\$ and \$Q_3\$. This option isn't available with the Darlington arrangement.

You already have some comments about how to bias your circuit. Similar ideas may also be used with the Sziklai driver circuit shown above, but you won't require quite as much biasing voltage difference.

Also, none of the comments about biasing circuits here have addressed themselves to the impacts on your circuit due to temperature variations, while operating. And this may be fairly important to consider. An added collector resistor to the simpler \$V_{BE}\$ multiplier (and tapping off at the collector side of that added resistor, now) can provide a mechanism by which you can make adjustments to match up the behaviors of the multiplier with the variations on the output stage so that the quiescent current is relatively stable over temperature. (Assuming you thermally couple the multiplier BJT to the output BJTs.) And it also can add compensation for the Early Effect, as well.

Just as a rough model, the schematic might now look like:


simulate this circuit

You'd adjust \$R_7\$ and \$R_8\$ and \$R_9\$ in order to set the biasing voltage difference required (arranged so that when quiescent the voltage drop across \$R_1\$ and \$R_2\$ would be about \$50\:\textrm{mV}\$ each -- after you decide on how to size them in the first place -- not discussed here, yet.) You'd also adjust \$R_7\$ itself (and, as a consequence of that, also perhaps \$R_8\$) in order to match up the thermal variation behavior to maintain that voltage drop across \$R_1\$ and \$R_2\$ when you use a hair-dryer or some other heat source on this entire output stage. (I've assumed that you've thermally coupled the BJTs together on a single heat sink.) \$C_1\$ provides some useful bootstrapping and \$C_3\$ provides an AC bypass across the \$V_{BE}\$ multiplier for the bases into the two output Sziklai quadrants.

\$C_2\$ provides Miller compensation for the VAS (\$Q_6\$), although this isn't the only way to drive the circuit -- an opamp might be used instead (so, no \$Q_6\$ in that case.)

The above assumes that you really do have bipolar supply rails and a grounded, DC coupled load. I've also not shown the negative feedback that is probably going to be required, eventually. Things would be somewhat different if the load is AC coupled and you have only a single supply rail to work with.

  • \$\begingroup\$ Nice! But why is the C3 connected to the collector of Q5? And C1 which is considered to "bootstrap" something (?) - I still don't get its function, although have read few of the posts you recommended to me so far. \$\endgroup\$ – Keno Dec 27 '17 at 18:49
  • \$\begingroup\$ @Keno For now, just ignore the capacitors. Note that \$R_7\$ is usually a very small value (about \$50\:\Omega\$ or so) you could just as well have hooked the collector end of \$C_3\$ to the node joining \$R_6\$ and \$R_7\$ (without moving \$Q_2\$'s base connection to the collector, though.) The function of \$C_1\$ is to raise the effective resistance of \$R_6\$ and thereby raise the loop gain that results from it being at \$Q_6\$'s collector load (applied to little-re, \$r_e=\frac{k T}{q I_{C_6}}\$, to approximate this gain.) \$\endgroup\$ – jonk Dec 27 '17 at 20:24
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    \$\begingroup\$ @Keno You have stuff to learn. I think one of the main points here is that designing a good output stage from discrete parts takes a certain level and breadth of knowledge about various effects. Temperature being one of the more important ones, if it is to be a good power driver. You often don't find detailed treatments of discrete designs (though you do see the schematics) because with the advent of good, cheap ICs there is very little need, anymore. Except to learn. Old books are more often the only place you find this info, sadly. \$\endgroup\$ – jonk Dec 27 '17 at 20:27

Actually the class B amplifier doesn't has a base bias. The bias occurs at AB class. But you can bias the base in many ways.

If you're using an op amp just like in the image, you could just use feedback. It makes the output equals to the input, just like a buffer but with a power stage.


simulate this circuit – Schematic created using CircuitLab

You could also use two voltages sources.


simulate this circuit

You could use diodes and a constant current source.


simulate this circuit

And at last but not least the Vbe multiplier. It takes the @oldfart's idea. The current the resistors R1, R2 and R3 is given by approx $$ I_r=\frac{V_{be2}}{R_3} $$ And, $$ V_{BB} = I_r(R_1+R_2+R_3) = V_{be2}(\frac{R1+R2+R3}{R3}) $$.


simulate this circuit

NOTE: The R2 resistor is for fine adjust.

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    \$\begingroup\$ Not having any emitter resistors on the final output transistors is a bad idea except in your first circuit. Even if you adjust the voltage offset between the bases to not cause much quiescent output current, you are still asking for thermal runaway. As the output transistors get hotter, their B-E drops go down. This causes more quiescent current with the same input bias offset. That causes more heating, which causes lower B-E drops... etc. \$\endgroup\$ – Olin Lathrop Dec 24 '17 at 15:37
  • \$\begingroup\$ You're right. I answered it theoretically because the second and the third circuit is almost never used. The last circuit you can couple thermally the Q1, Q2 and Q3 and it solves the thermal runaway. \$\endgroup\$ – Francisco Gomes Dec 24 '17 at 16:03

class B is defined as 180 deg conduction angle - so class B is biased to the point of conduction - otherwise its really class C (especially for small signals). The emitter resistors are key to both biasing stability and to allowing each device to turn off during the opposite half cycle.

class AB is when the conduction angle is between 180 and 360


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