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Audio Amplifier Analysis

  1. Aug 17, 2015 #1
    Hi. I'm trying to analyse an audio amplifier circuit (DC, frequency response, stability, etc.).

    I began with the DC bias analysis but I'm having troubles finding where to start. I've attached the full circuit and a DC simplified version. I've placed an 8 ohm resistor as the load (8 ohm speaker), is this correct?

    All I could come up with is (I'm assuming collector currents equal emitter currents):
    ##I_{C1}=V_{EB4} /R_6##

    Should I start by doing the two-port equivalent of the feedback network (C3, R4, R5)? Somewhere else... ?
    I'm pretty much lost.

    Thank you in advance.

    PS: I also found that ##I_{R17}=V_{BE9}/R_{17}## but that got me nowhere.

    Attached Files:

    Last edited: Aug 17, 2015
  2. jcsd
  3. Aug 18, 2015 #2
    Without going into nonlinear equations and numerical/iterative solution we are force to use to use a simplified solution based on some assumption.
    Because this circuit is nothing more then a opamp based/style non-inverting amplifier with the gain of one (for DC voltage this circuit work just like voltage follower, but AC closed loop gain is Av ≈ 1+ R5/R4 ≈ 22V/V ).
    So Vin = 0V this means that also is equal 0V ( we ignore Q1 and Q2 the base current).

    Ic1 ≈ Ic2 ≈ I1/2 ≈ 1.18mA
    Ic4 ≈ (Vcc - VbeQ6)/( R9 + R10) ≈ 5.2mA
    Ic9 ≈ Ic4 - VbeQ9/R17 = 5.2mA - 0.38mA ≈ 4.82mA

    The output stage current is undetermined because we do not know the exact value for a Vbe5 and Q6.

    The simulation show as this result
    Ic1 = 0.856mA
    Ic2 = 1.51mA
    Ic4 = 5.25mA
    Ic9 = 4.81mA
    Ic5 = 4.65mA
    Ic6 = 3.8mA
    Ic7 = 208mA
    Ic8 = 209mA
  4. Aug 18, 2015 #3
    First of all, thanks for replying!

    I've attached the results of the simulation that I ran.

    For the differential amplifier I found the following (which more or less matches the simulation):
    ##I_{C1}=V_{EB4} /R_6=0.47 V/560 Ω=839.29 μA##
    ##I_{C2}=I_1-I_{C1}=1.54 mA##
    ##V_{B4}=V_{CC}-V_{EB4}=35 V-0.47 V=34.53 V##
    ##V_{B1}=-I_{C1}/β_{1}.(R_2+R_3)-V_{BE1}=-839.29 μA/325 . (22 kΩ+1 kΩ)-0.66 V=-0.72 V##
    ##V_{CE1}=V_{B4}-V_{B1}=34.53 V+0.72 V=35.25 V##
    ##V_{CE2}=V_{CC}-V_{B1}=35 V+0.72 V=35.72 V##

    I understood this perfectly.

    However I don't understand this equation. I get that ##I_{C4}=I_{R9}=I_{R10}## but I'm not seeing how the voltage drop across R10+R9 equals Vcc-Vbe6.

    Attached Files:

  5. Aug 18, 2015 #4
    On the speaker front:

    Speakers have frequency dependent impedances. They vary between about ½ the nominal (8Ω) to perhaps 5 times the nominal depending on the box they are in. (The box affects the power transfer depending on frequency.)
  6. Aug 19, 2015 #5


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    Actually, it equals Vcc - Vr14 -Vbe6

    If you assume the output voltage is 0V, then the voltage at the top of R9 is Vbe6+Vr14 below 0. But if you assume Ir14 == 0 then that means Vbe6 could be very low or zero.

    On the other hand, If it is operating class A, then there is a drop across R14 even if Vrl == 0.
  7. Aug 19, 2015 #6
    Are you sure about VB1 ?? Notice that Q1 is a NPN transistor so the base current will flow into the Q1 base and this is why VB1 = Ib1*(R2 + R3) ≈ 60mV and VE1 = VB1 - Vbe1. Also to be more precise Ic1 is equal to Vbe4/R6 + IB4

    Notice that the voltage at Q6 emitter is equal to 0V (as we assumed) and if we ignore R14 voltage drop the voltage at Q6 base is Vbe6 below the 0V. So this means the voltage across r9 and R10 is Vcc - Vbe6.
  8. Aug 22, 2015 #7
    Thanks, will look into that.

    Now I got it, thanks!

    You are right, I had some troubles with the sign of IB1; and it should be VE1 in my equation, not VB1.

    ##V_{E1}=I_{C1}/β_{1}.(R_2+R_3)-V_{BE1}=839.29 μA/325 . (22 kΩ+1 kΩ)-0.66 V=-0.60 V##

    In my last post I forgot to upload one simulation result (attached here).

    Thanks everyone!

    Attached Files:

  9. Aug 22, 2015 #8
    After reading your last post I notice that I also made a small mistake in finding VB1.
    As I said earlier the base current enters into the Q1 base. So this means that the current will flow from GND through R2 and R1 and will flow into the base.
    So the voltage at Q1 base must be negative VB1 = -Ib1*(R2 + R3) ≈ -60mV and Ve1 = VB1 - Vbe1 = -60mV - 0.61V = -0.67V
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