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I been trying to figure out this diagram, and Can't understand what happens from the beginning of the input , how the voltage changes in a way that "shuts off" one transistor. Also cant understand the direction in which it goes in the input signal .

Does the voltage goes off completely on one transistor at a time whle there is a signal ?

How does the signal works in the preamplifier transistor?

What path does the current takes when the signal changes in the input?

(A hobbyist about electricity , learned only the basics)

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  • \$\begingroup\$ Do you understand how a common emitter amplifier works? \$\endgroup\$ – Andy aka Jan 22 '17 at 10:46
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I provide some analysis of a very similar design here on EE.SE regarding this quite similar circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

However, if you read through the text there you will see that there are some measurements you need to take and that the circuit actually needs some tweaking to get it functioning correctly. This isn't something you can do for production. But it's often fine for a hobbyist or builder who wants minimal parts and is willing to tinker.

Still, you can compare the circuit you have presented with the one above, discussed in the above link, to see some slight differences and where I've actually talked about how to adjust it, given arbitrary BJTs, to get it centered and working reasonably as intended.

I also provide a better, more designable approach at that link, as well. It has more parts but it is also doesn't ask the end user (hobbyist) to have to tinker with it to make it work, correctly. The schematic shown at the link above is this one:

schematic

simulate this circuit

Both of the above circuits are for \$5\:\textrm{V}\$, though. And so they would need to be redesigned for \$9\:\textrm{V}\$. But the concepts are similar and I think if you read the above link it will go a long way towards helping you follow the schematic you presented.

It's important for your understanding of your circuit to take note that your circuit includes a large-value bootstrapping capacitor. If you look near the bottom part of my answer to a different question here, you will see some discussion about how that works. It's purpose is to provide a fixed voltage across your \$R_1\$ resistor so that there is a near-constant current flowing through the diodes and to the VAS (your \$Q_3\$.) There are other ways to provide this (and for example the LM380 power amplifier IC I'll mention in a moment uses such different methods.) But these other ideas take extra parts.

Since I just brought it up, I think it would be instructive to you to read another answer of mine about the LM380 power amplifier. Schematic follows here, too:

schematic

simulate this circuit

Here, you will see that they don't use the bootstrapping capacitor trick but instead use a different method of providing a constant current for the two diodes used to separate the bases of power output section. You should be able to notice that they also use an output driver section that uses three BJTs instead of two. However, that's not an important difference, as the basic idea remains quite similar to your circuit (\$Q_{12}\$ is just a helper BJT.) Note that they include the VAS, as well. But that it is driven by a differential amplifier section. This is much better than your case, which... well.... uses NOTHING at all to help out. It's just an exposed VAS, in effect.

A final note worth mentioning, assuming you are willing to take the time to follow those links and what I've said here, is that in the LM380 and in the second schematic I've exposed here, above, there is a negative feedback resistor. In your circuit, this negative feedback resistor is \$R_2\$ and in my first schematic above it is \$R_5\$. In the LM380 and the second schematic above, that's its main purpose. But in your schematic (and my first one above), it also does double-duty in order to bias the VAS. This is part of what complicates such a design.

So, to understand your specific schematic a little better, go to the first link I mentioned at the outset. Read it carefully and it may help somewhat in trying to understand your schematic better.

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It's better to think in terms of each output transistor conducting current.

D1 and D2 set up a voltage drop of 2 diodes across the two bases of the output transistors, just enough to put them both at the edge of conduction.

When Q3 pulls the bases down, or R1 pulls them up, the voltage on the bases shifts slightly so that one output transistor conducts more strongly, to be an emitter follower for the signal, and the other one conducts less strongly, and supplies no current.

This amplifier would sound quite rough, there's a lot more knowhow goes into to getting the output to bias up stably, over temperature, and to reduce the inevitable distortion (crossover distortion) that happens when the load current shifts from one transistor to the other.

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  • \$\begingroup\$ Does c1 makes the voltage drop or rise that changes the affect of Q3 or mostly through r1. Does the input change potential or goes by lower or higher positive voltage? \$\endgroup\$ – user1779374 Jan 22 '17 at 7:43
  • \$\begingroup\$ C1, the output capacitor, charges up to mid rail, so that there is no net DC on the loudspeaker, only the changes in output voltage get through the to speaker. Because of the feedback reisstor R2, it's better to think of the input as a current input. The overall amplifier is inverting, so changes in input current though C2 are met with a change in Q3 voltage, which changes the output voltage, which changes the voltage across R2, until the change in current through R2 matches the input current. There should be a 20k resistor Q3 base to ground to make the input DC bias work properly. \$\endgroup\$ – Neil_UK Jan 22 '17 at 10:05
  • \$\begingroup\$ Thank you for that explaination , also i want to know if when the change in c1 happenes, the purpose of q1 and q2 is to direct the current to the negative side of the voltage input so if it is positive it will go through r2 and if negative it goes through q2? \$\endgroup\$ – user1779374 Jan 22 '17 at 14:07
  • \$\begingroup\$ Q1 and Q2 form what is called a 'push-pull' output stage (look it up). R2 is feedback, it has nothing to do with the output. \$\endgroup\$ – Neil_UK Jan 22 '17 at 15:05
  • \$\begingroup\$ @user1779374 try read this learnabout-electronics.org/Amplifiers/amplifiers55.php and this forum.allaboutcircuits.com/threads/… \$\endgroup\$ – G36 Jan 22 '17 at 15:54

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