0
\$\begingroup\$

I take next sample schematic.

enter image description here

This is only an output stage of any real power amplifier. It's clear when it's work in B-class the every BJT amplify only one half of input signal. But what it means if it works in class A. I know that the current always go through BJT if it works in class A. And if signal is too small but enough to open the BJT it will amplify both negative and positive signal swings before the signal level achieves some value above that only part of that signal will be amplified with "N" or "P" stage accordigly. Ok, if both parts of small signal amplified with every parts of a power amplifier, how do they added on load resistance? They both must have opposite direction and must differ from each other and the 0 must be on load. Here is a trick that I can't understand. Any suggestions?

\$\endgroup\$
1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

A class A amplifier also biases the transistor just like the class AB. When the transistor is biased it resembles the load resistor. This is similar to the resistor string to the right. At the bias point, with no input Q1 is adjusted to match R1. A class A amplifier will invert the signal but because of the biasing any signal even a small one will operate the amplifier.

| improve this answer | |
\$\endgroup\$
3
\$\begingroup\$

They both must have opposite direction and must differ from each other and the 0 must be on load.

What makes you think that the output currents of the NPN and PNP transistor have a different direction? You should be careful with the direction of currents and also separate the direction of a DC current (bias current) and the direction of a change in current (due to a small input signal) which we often call AC.

Assume that we have a small but positive input signal, meaning the voltage rises. What will happen:

  • The \$V_{BE}\$ of the NPN will increase

  • The \$I_C\$ and thus \$I_E\$ of the NPN will increase

  • The \$V_{BE}\$ of the PNP will decrease

  • The \$I_C\$ and thus \$I_E\$ of the PNP will decrease

So at the node "Probe1" there's a current increase coming "from above" while at the same time there's also a current decrease coming "from below".

And the current law tells us that the current into load R3 will be:

\$I_{load} = I_{above} - I_{below} = I_{NPN} - I_{PNP}\$

So the currents do not have cancel but actually add up.

But there's a minus sign so they subtract?

Yes but note how \$I_{NPN}\$ increases yet how at the same time \$I_{PNP}\$ decreases. So those changes do add up.

When the NPN supplies more current it is also "helped" by the PNP as the PNP will draw less current.

It is like the NPN is "pushing" more while at the same time the PNP is "pulling" less. So both NPN and PNP contribute to the change in the current through the load.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ A huge glaring hole in this description is what's happening to currents in \$R_1\$ and \$R_2\$. With a rising signal, the voltage drop across \$R_1\$ rapidly diminishes -- right at the time when the NPN needs more base current to support its increasing collector current. At the same time, \$R_2\$ increases its need to sink current. This whole circuit concept is a disaster. Period. \$\endgroup\$ – jonk Feb 2 '18 at 7:38
  • \$\begingroup\$ @jonk I'm talking about small signal behavior meaning small changes in the voltages across \$R_1\$ and \$R_2\$ then what you describe does not happen or at worst, is irrelevant. Also the required increase in base current will be supplied through the AC coupling caps \$C_1\$ and \$C_3\$. Also I disagree with this concept being a disaster. I have seen many Audio amplifier designs which use this exact topology. The trick is to have enough DC biasing voltage across \$R_1\$ and \$R_2\$ or better: replace them with a current source. \$\endgroup\$ – Bimpelrekkie Feb 2 '18 at 7:48
  • \$\begingroup\$ "This topology", stretched out by inserting a current source and a VAS to soak it up, is no longer the same disaster. I agree there. And as it is perhaps it works with small enough swings (or absolutely huge caps, I suppose.) It's just hard to seriously consider the circuit as practical in any way. But I'll +1 your answer. So far as it goes, it is probably what the OP needs at this moment. \$\endgroup\$ – jonk Feb 2 '18 at 7:55
  • \$\begingroup\$ @Bimpelrekkie How do you think may that schematic properly work with R5 and Q2 removed to view only one part of signal swing amplification in output oscillogram (of course for educational purposes only)? \$\endgroup\$ – MaxMil Feb 3 '18 at 14:36
  • \$\begingroup\$ You will not see that when you remove R5 and Q2. When you remove them the DC voltage at the output will rise so the biasing of Q1 will be completely different. Try it in a simulator and see what happens. \$\endgroup\$ – Bimpelrekkie Feb 3 '18 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.