# Role of Op Amp driver in power amplifier

Discussion in 'General Electronics Chat' started by Himanshoo, Aug 8, 2015.

1. ### Himanshoo Thread Starter Member

Apr 3, 2015
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The text says :

“ As the op amp out put at the junction of D1 and D2 is increased in a positive direction op amp supplies current through D2 and R5 . So the voltage drop across R4 is reduced ,allowing it to pull the Q1 base up to the required level while supplying increased base current to Q1.The Q2 is biased OFF when output voltage is at its positive peak at the junction of diodes.”

But as output goes positive D1 is RB and D2 is FB letting the major current(output current + dc bias from Vcc) flows across D2, resulting in turning Q2 more harder..but the text says that Q2 is OFF during when output is positive….how ?? please justify??????

secondly....

What does the author means by the line "allowing it to pull the Q1 base up to the required level while supplying increased base current to Q1”…Which required level he is talking about???

At node B the voltage is 11.7 V ..and at node A it is 11V but it should be two diode drop below the voltage at node D i.e it should be 10.3 V(because 0.7V voltage is getting dropped at BE junction of Q1 and across D1) please explain why i am having a contradictory idea ???

thnx

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2. ### ericgibbs AAC Fanatic!

Jan 29, 2010
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hi H,
A picture is worth a thousand words.

Note the gradual increase/decrease of the Base currents related to the Vbe voltages due to o1 drive signal.
Look at these LTS plots showing the operation of the Q1 and Q2 transistors, ask if you do not follow OK.

E

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3. ### Himanshoo Thread Starter Member

Apr 3, 2015
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hi eric

from your figure its clear that Q2 conducts when the voltage at node A goes positive...vice versa for Q1...

now the point of my query is that why at node A the voltage is 11V ..it should be 10.3V (ie two diode drop below node B voltage)..???

4. ### ericgibbs AAC Fanatic!

Jan 29, 2010
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hi,
It is Q1 that conducts when Node A [o1] goes positive.

Why do you think Node B should be at two 0.7V drops.?

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5. ### Himanshoo Thread Starter Member

Apr 3, 2015
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At node B the voltage is 11.7 V..and at node A its 11 V ..that what text says....

but now consider the figure...some of the current that flows due to Vcc is going towards Q1 ..and some getting divided and going towards D1 ..
since this current is causing a voltage drop of 0.7 V at Q1 base emitter junction and other part of this current causing another drop of 0.7 V across D1 should leave node A with voltage that is two diode drop below the voltage of node B..i.e 10.3 V ...but this inference of mine is against the text ...

please tell me where I am wrong...

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6. ### ericgibbs AAC Fanatic!

Jan 29, 2010
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Hi H,
The way you are describing Nodes A to B, is that D1 and the diode formed by the transistors Base to Emitter junction is series, that is incorrect.
The D1 diode and Q1 base/emitter are connected in 'parallel' with respect to A and B, R4 current.

But D1 and Q1 Base/Emitter diode are in series with respect to Node A and Vout, ref image.

• ###### AAesp02.gif
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Last edited: Aug 8, 2015
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7. ### AnalogKid Distinguished Member

Aug 1, 2013
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Without getting into how the feedback loop connects everything, consider just Vcc, R4, D1, and Q1 base-emitter junction. With nothing else connected to anything, and only voltmeter input current flowing anywhere, Vcc is 15 V, node B is 15 V (voltmeter input current is very low so voltage drop across R4 is very low), node A is 14.3 V, and Q1 emitter is 14.3 V. The two circuit branches through the two diodes are independent; the diode drops are in parallel, not series, so they do not add. As current in either branch increases, such as by the opamp output pulling diode D1 down, that current flows through R4, increasing its volage drop, and node B goes down. As node B moves up and down, so do both diodes (D1 and Q1 base-emitter). They always move together, in parallel. Whatever voltage is at node A also is at Q1 emitter (withing the Vf matching of the two parts).

The current through R4 is the sum of the currents in the two circuit branches, and those two currents do not have to be equal. If either D1 or Q1 emitter is puled down, R4 current increases and the other component moves down also.

ak

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8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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To be honest I disagree with the "Author". Opamp do not provide any base current for Q1 or Q2. All the base current needed by Q1 transistor during positive cycle is provide be R4. Q1 is controlled by the base current that comes through R4, and diode D1 steals this current that normally will go into Q1 base. But as Opamp output increasing his output voltage the diode D1 current decreasing, so now more current from R4 enters into Q1 base.

And the main job for Opamp here is to provide a needed voltage gain additional Opamp helps reduced the crossover distortion.
And the maximum output voltage is equal to
+VL_Max = +ILMax * RL
+ILMax = (Vcc - VbeQ1)/(R6 + RL + R4/(β+1)) = (15V - 0.7V)/(47Ω + 500Ω + 3.3kΩ/(80)) = 24.31mA

+VL_max = 24.31mA * 500Ω = 12.15V

And this equation is only true when opamp is able to provide a large current than Iop > (Vcc - Vee)/(R5).
And because of this limitation it is not a good idea to use opamp as a "driver" for a push-pull output stage.

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9. ### ericgibbs AAC Fanatic!

Jan 29, 2010
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I would say for the 'design' of the output stage of this circuit, with the low output current into the load, that a general purpose OPA would have no problem outputting the required +/-8mA.

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10. ### Himanshoo Thread Starter Member

Apr 3, 2015
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Yes the base current required to ON the transistors is provided by R4 and R5 ....I agree with you on that...But the author also say that "R4 and R5 are there to provide an active pull up for transistor Q1 and Q2..."

what does he mean by it..??

11. ### Himanshoo Thread Starter Member

Apr 3, 2015
238
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Yes bro I got it...as the voltage is same when in parrallel....and diode D1 and Q1 base emitter are in parrallel...hence they will share the same voltage....but different current level...

But why R6(Q1 emitter resistor) is not included in this parrallel combination....since same current flowing across Q1 base emitter junction also flows through R6....ref my drawing...?????

Last edited: Aug 9, 2015

Apr 5, 2008
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Hello,

Your given circuit is limited to the + 15 Volts and - 15 Volts powersupply.
Have a look at figure 32 on the following page of the ESP:
http://sound.westhost.com/articles/dwopa3.htm
It will allow higher powersupply voltages and give more output power.

Bertus

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I do not know. I do not see any "active" pull-up here. Because for sure resistors are not "active" device.
From which book you got this quote ?

14. ### Himanshoo Thread Starter Member

Apr 3, 2015
238
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But what is an active pull up ? Iam not getting it..

By the way the text is from the book.."Electronic devices and circuits" by David Bell ...fifth edition..

15. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I suspect that the author used word "active pull up" to point out that R4 and R5 provide a current for Q1 and Q2.

16. ### Himanshoo Thread Starter Member

Apr 3, 2015
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One thing slipped off my mind to ask is..that ...later in text its written that "the voltage drop across R4 and R5 should be minimum in order to produce the base current.But this minimum voltage requirement keeps the amplifier maximum peak output voltage well below the supply voltage level and thus limits the amplifier efficiency"...
...exactly how???

since if VR4 and VR5 is minimum it will produce large base current which is one of the requirement to drive the transistors in(or near) saturation ...

Last edited: Aug 9, 2015
17. ### Jony130 AAC Fanatic!

Feb 17, 2009
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First note that Q1 is working as an emitter follower and as you should know you cannot saturated the emitter follower until the voltage at Q1 base is larger than Vcc + Vbe. But in this case it is impassible to saturate Q1. This means that transistor always work in the linear region.

The maximum positive output voltage can be find by analysis this situation:

D1 current is 0A. So all the current from R4 will flow into Q1 base. From KVL we have:
Vcc = VR4 + Vbe + Vo = IR4 * R4 + Vbe + IL* RL Also notice that IR4 = IL/(β + 1).
So we end up with this:

IL = (Vcc - Vbe)/( RL + R4/(β + 1))

And

Vo_max = IL * RL (peak value)

As you can see the smaller the value of the R4 resistor the larger IL current will be and vice versa the larger R4 the smaller IL will be.
But in the same time op amp need to be able to provide a current needed to drive D2 and R5 (Iop_max ≈ 2Vcc/R5). If the opamp is unable to provide such a large current the D1 current will never reach 0A, so now the maximum output voltage will now be determined by Iop_max current.
So the new maximum output voltage (peak value) is equal to :
Vo_max' ≈ Iop_max * R4||R5||(β+1)*RL.

Last edited: Aug 10, 2015
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18. ### Himanshoo Thread Starter Member

Apr 3, 2015
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nice explanation jony.....

thnx