Emitter Follower instead of regulator, advice needed

Discussion in 'General Electronics Chat' started by Spence, Mar 24, 2011.

  1. Spence

    Thread Starter Member

    Apr 23, 2010
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    Attached image shows 2 versions.

    Could this method sink enough power to drive the rails of a 50 watts per channel audio amp.

    It simulates OK but I've just read that the emitter follower method is no good for high voltage/ high power.

    To me it's not high voltage or power but I'm concerned at the comment I read, anyone care to offer some opinion.
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    The first circuit is worthless, it does nothing. The second is a primitive voltage regulator. It works, but has a but load of problems. The big one is the voltage at the emitter changes around 0.3V or more with large currents. Modern three terminal IC voltage regulators do not have this problem.

    You have to define how much current and voltage you need, the design accordingly.
     
  3. Spence

    Thread Starter Member

    Apr 23, 2010
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    Well, 50 watts X 2 channels into 8 Ohm speakers is ..... 30v 3.5a, more or less.
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    Maybe for an extremely efficient bridged class-D switching amplifier.

    A normal 50W into 8 ohms class-AB amplifier heats with about 35W so the power supply for two amplifiers will need to be about 63V at 2.7A if the amplifiers are continuously at full output (playing acid rock?).
     
  5. hgmjr

    Moderator

    Jan 28, 2005
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    One of the reasons that the emitter-follower is a poor choice of regulation is the inefficiency of the stage. Looking at your example on the right, you will note that the dc voltage available at the emitter follower's emitter is set by the three 12v zener diodes. That means that the voltage between the collector and emitter of the transistor is approximately 60V-36V+0.7V = 24.7V. That means for every amp drawn by the load there will be a corresponding 24.7*1 Watts dissipated in the transistor. You will need a pretty healthy heatsink to prevent the transistor from being destroyed by this level of heat.

    hgmjr
     
  6. Spence

    Thread Starter Member

    Apr 23, 2010
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    No, nothing like that but thanks for the thought lol
     
  7. Spence

    Thread Starter Member

    Apr 23, 2010
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    Thanks for taking the trouble to look at the problem. So, how is it done then?, I looked at the datasheet for lm7824 which would give me 24v, but it has a 1amp maximum output current. What's the way to go? Drop less current across the transistor and take another approach?

    Incidentally, I'm not short of heatsinks going up to 2+ kilos.
     
  8. Wendy

    Moderator

    Mar 24, 2008
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    One of the reasons digital power supplies and amplifiers are so popular is they are much more efficient. Figure between 80% and 90%, with exceptions. Analog regulators can be much much worse (20% anyone?), but there is no firm number. There is a method to measure the inefficiency, measure the voltage drop across it.

    SMPS (switching mode power supplies) are an example. If a unit is 90% efficient, your 35W to a speaker will take just under 40W from the power supply.

    Class D amps are a way to do audio digitally, with equivalent efficiencies.

    Pulse Width Modulation

    The trade off is noise. Analog regulators are generally much quieter, while switching are much noisier and more efficient.
     
  9. Audioguru

    New Member

    Dec 20, 2007
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    Properly designed audio power amplifiers do not use a regulated power supply.
    Their transformer is powerful enough to have a low resistance so its own voltage regulation is not bad, the main filter capacitors are pretty big so they produce fairly low ripple (hum) and the amplifier circuitry rejects the hum from the supply anyway.

    Low current preamps use a regulated power supply. Their tiny voltage regulators barely get warm.
     
  10. Spence

    Thread Starter Member

    Apr 23, 2010
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    Ah good information. I wanted to make a 50 w.p.c amplifier and gradually work up to something bigger later, I think I should connect it up with the +60v and -60v rails fused at 1.5A and see what happens.
     
  11. Audioguru

    New Member

    Dec 20, 2007
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    50W into 8 ohms makes a sine-wave that is 20V RMS and is 56.56V p-p. It needs a supply that is about 62.56V (plus and minus 31.3V). Your 120V supply that has too much voltage will power an amplifier that produces 203W into 8 ohms if there is enough current.
     
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  12. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    One use of a emitter follower regulator is a ripple reducer by connecting a capacitor between base & ground in first figure [Q 1]of thumb nail.
     
  13. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Half a volt to a couple of volts of ripple etc is lost in 20volts RMS at the output.

    However the input at 2millivolts to 20 to 200 millivolts is another matter.

    This is one reason why the early stages of an amplifier may need to be fed from a stabilised supply, even if the later ones do not.

    This is no biggie as the early stages are low power, low current.

    Since you are considering building your own amps be aware that most designs separate the voltage and current amplification, doing the voltage first and the current in the output stages.

    go well
     
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  14. Spence

    Thread Starter Member

    Apr 23, 2010
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    I decided to fry the head instead of the transistors lol, so I went back to the simulation and I don't understand why the circuit draws the same power if I put 20v rails or 60v rails, just can't get my head round it.

    If I double the output transistors (parallel) I can see the power double, and I suppose it would keep increasing until I exhaust the power supply.

    But I don't understand why changing the rail voltage doesn't automatically change the output (although, I can see that the driver and output transistors still have the same base and emitter voltages)

    The attached simulation shows a sine wave simulating the preamp.
     
  15. studiot

    AAC Fanatic!

    Nov 9, 2007
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    If you have the same load and the same gain, once you have enough supply voltage to accomodate input x gain you will not increase the output.
     
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  16. Audioguru

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    Dec 20, 2007
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    The voltage of the "rails" do not affect the output level because your simple amplifier has a voltage gain of about 0.95.

    But the voltage of the rails do affect how high the output level will go if you increase the input signal level.

    It is wrong for the output power to increase much when you add additional transistors in parallel unless it is because the original transistors were overloaded.
     
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  17. Spence

    Thread Starter Member

    Apr 23, 2010
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    From my point of view (a persistent simpleton) it does but it's probably my misconception.

    Q1 and Q2 both pass/draw 750 m.A. and I see 1.5 A across the speaker, if I connect Q3 and Q4, they too draw 750 m.A. each and I see 3A across the speaker.

    I hear what you say, could it be that the simulation is too perfect and hides the limitations of the transistors and power supply?

    Another point, If I change the rail voltages from 20 to 60, the output transistors will then drop 50+ volts and discharge the extra power as heat (no?). Yet the supply V4 does not pass more current, that I still don't understand.

    Thanks for the help.
     
  18. Audioguru

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    Dec 20, 2007
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    When you add output transistors in parallel they share the current that single output transistors would have. Therefore each output transistor in parallel carries half the current of a single output transistor. Then they also share the heating.

    Adding parallel output transistors allows slightly less base-emitter voltage loss and slightly less voltage loss across the 0.22 ohm emitter resistors.
    Input= +12V:
    With one output transistor and a 0.22 ohm emitter resistor the BD139 has a base-emmiter voltage drop of 0.7V, the TIP31 output transistor has a base-emitter voltage drop of 0.9V and the voltage at the series 0.22 ohm resistor and 8 ohm load becomes +10.4V with a current of 1.27A.

    With two output transistors with emitter resistors the BD139 has a base-emitter voltage drop of 0.68V, the TIP31 output transistors have a base-emitter voltage drop of 0.85V and the voltage at the parallel 0.22 ohm resistors in series with the 8 ohm load becomes +10.47V with a current of 1.29A.

    The current in the 8 ohm load has increased from 1.27A to 1.29A which is no where near double like you had.
     
  19. studiot

    AAC Fanatic!

    Nov 9, 2007
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    You never see amps across anything - that would be volts.

    You see amps through something.

    So if you have connected an ammeter across the speaker in your simulation or reality you will not be reading the value you think you are.
     
  20. Spence

    Thread Starter Member

    Apr 23, 2010
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    Thanks Audioguru, tremendous help and information you have provided.

    The Amplifier is almost finished, works well and sounds good. The biggest problem was installing fans, running from a seperate transformer tapping and power supply.

    The last amplifier I made suffered from mains related pops and clicks, this one doesn't, however, there's a small switch on thump, I don't know what to do about it.
     
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