Simple amplifier schematics wrong?

Discussion in 'Analog & Mixed-Signal Design' started by upand_at_them, Sep 17, 2016.

  1. upand_at_them

    Thread Starter Active Member

    May 15, 2010
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    I happened to be looking through one of my electronics books the other day and found the attached circuit. Looks wrong to me, so I went to a popular beginner electronics site and found two similar circuits:

    [​IMG]
    [​IMG]

    All three of these seem wrong to me. How can there be any collector current in the PNP without a bipolar supply? Am I missing something?

    The two images from www.electronics-tutorials.ws might just be mislabeled. But the one from my book is clearly suggesting it as an alternative to having a bipolar supply. Where is the PNP collector current coming from?
     
  2. blocco a spirale

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    Jun 18, 2008
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    Think about it from the transistor's point of view, it doesn't "know" what the power-supply configuration is, it can only "see" relative potentials on its terminals. The first two schematics are not practical examples of complete output stages but simplified diagrams to demonstrate how biasing works.

    The final schematic in fig 13.6 shows something that looks more like a practical amplifier where the output capacitor is used to remove the DC offset from the loudspeaker terminals. When the amplifier is first switched on there will be a loud pop through the loudspeaker as current passes through the voice coil and charges the capacitor. Once the capacitor is charged, the PNP transistor can sink current just as it would if this were a split-rail power-supply arrangement.
     
    Last edited: Sep 17, 2016
  3. upand_at_them

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    May 15, 2010
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    There needs to be a voltage potential for current to flow. The NPN's have it, and collector current will flow in the NPN's when turned on. But I don't see that any current will flow through the PNP collectors.
     
  4. upand_at_them

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    May 15, 2010
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    I edited the circuit to show what I mean.
     
  5. hp1729

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    Nov 23, 2015
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    I agree with you. Those examples need a -V instead of ground.
     
  6. blocco a spirale

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    If we're talking about fig 13.6, the potential is stored on the output capacitor where its LHS is at 1/2 vcc and the PNP emitter is also at around 1/2vcc and its collector is at 0V.

    In reference to your latest post #4, don't forget that there is 1/2vcc across the capacitor and therefore when the PNP transistor is on, the LHS of the capacitor is pulled down to 0V and the RHS of the capacitor is now at -1/2vcc (not 0V as you have labelled it). The capacitor in series with the loudspeaker voice coil provides the current that passes through the PNP transistor.
     
    Last edited: Sep 17, 2016
  7. upand_at_them

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    How can the LHS of the output cap be at 1/2 Vcc when the NPN is off?
     
  8. AnalogKid

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    Aug 1, 2013
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    0 V on the left side of the output coupling capacitor is incorrect. In a well designed circuit of this type, that point is V+ (positive power rail)/2.

    Disconnect both collectors and look at the resulting series string. From V+ you have R, base-emitter diode, R, R, base-emitter diode, R, GND. In round numbers, the NPN operates between V+ and V+/2, and the PNP operates between V+/2 and GND. The only thing split supplies gets you is eliminating the output coupling capacitor, because then the resting point between the two emitters is at 0 V, assumedly the return potential of the load. If the two supplies are, relative to whatever V+ is in your drawing, V+/2 and -V+/2, he same voltage ranges are across the same transistors, just shifted.

    ak
     
  9. blocco a spirale

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    If the NPN transistor was never on in the first place then you are correct, everything would be at 0V but in a real amplifier this is not the case. In a real amplifier, in the quiescent state, the point between the output transistors (and therefore, the LHS of the output capacitor) is around 1/2vcc.

    Even though the supply is single-rail, the output capacitor makes it possible for the + terminal of the loudspeaker to swing between (nearly) +1/2vcc and -1/2vcc.
     
    Last edited: Sep 17, 2016
  10. upand_at_them

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    May 15, 2010
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    I think I realize my confusion.

    This PNP won't turn on at all, right?...

    [​IMG]

    In the other circuits I assumed that the NPN was off during the negative excursion of the input signal. So those other circuits are AB (not B) amplifiers...And the NPN's provide the current to the PNP collectors?
     
  11. crutschow

    Expert

    Mar 14, 2008
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    For that circuit (post #10) to work properly the input needs to be biased at 1/2 the supply voltage and the output needs to be capacitively coupled.
     
  12. AnalogKid

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    Aug 1, 2013
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    Wrong.
    Nope. Either the load provides the current for the PNP collector, or the NPN emitter does. For example, if the load is a resistor from Vout to 0v, thenthe PNP never conducts and the NPN acts as a simple emitter follower that never turns off.
     
  13. blocco a spirale

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    Your confusion arises because you are looking at a misleading and incomplete output stage diagram, the purpose of which, is to demonstrate the effect of biasing and is not to be taken as an example of a practical amplifier.

    As I said before, the diagram in fig 13.6 is the one to use.
     
  14. hp1729

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    Nov 23, 2015
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    Re: 13.6
    I must be missing something then.
    Maybe the non-specific voltage levels are throwing me.
    I see an AC waveform coming in. When the input goes negative how does the PNP transistor turn on? How does the output go negative?

    input as an sine wave riding on a DC level, unspecified.
    When the input is at midpoint, going down, I can see the base of the PNP decreasing, but what voltage is the emitter at?
     
  15. Veracohr

    Well-Known Member

    Jan 3, 2011
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    You're assuming the midpoint is 0V. There's no reason it has to be, and this circuit won't work if it is. Figure 13.6 is just an output stage, not a full amplifier. You wouldn't put a bipolar input signal directly into this circuit.
     
  16. DGElder

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    Apr 3, 2016
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    Assume you power up 13.6 with Vcc at 10V. No input signal. Both transistors will turn on due to the diode biasing. The impedance of the series transistor and resistor in the top half is the same as the impedance of the series resistor and transistor in the bottom half, i.e. an equal voltage divider. After the inrush current settles down the voltage across the speaker will be zero and the capacitor will be fully charged. The voltage on LH plate of the capacitor will be 1/2 Vcc (5V) and the RH plate will be at 0V. So the capacitor voltage (Vc) is 5V. That capacitor is large enough that it will hold close to 5V even after the AC input signal is driving the circuit.

    So when the input signal (capacitivly coupled so that it will apply an AC signal on top of the established 5V DC bias) goes positive the NPN is turned on, the PNP is turned off and the LH capacitor plate is raised up toward 10V. At the peak of the positive input signal the speaker will see Vcc - Vce - Vc: 10V -1V -5V = +4V (Vce is just an approximation for the NPN minimum collector emitter voltage for illustration). When the input signal goes negative the NPN turns off and the PNP turns on and draws the LH capacitor plate down toward 0V so the speaker now sees 0V -Vce - Vc: 0V +1V -5V = -4V.

    The speaker is driven to plus and minus 4V.
     
    Last edited: Sep 19, 2016
  17. upand_at_them

    Thread Starter Active Member

    May 15, 2010
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    Can I use 2N3904 and 2N3906 transistors to test this?
     
  18. Sinus23

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    Sep 7, 2013
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    I do not see why not and besides you should be fairly quick to breadboard it to test.
     
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