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Simple transistor headphone amp

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audioguru

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I got the distortion down to less than 0.1% by simply adding another transistor for more open-loop voltage gain.
I reduced the frequency because the number of cycles shown kept being too many.

Go Advanced has changed and made me think my attachment was not going to show.
 

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crutschow

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I will expand my use of the word "feedback" to the engineering definition to include things that are not actually feedback but which mimic the effect of actual feedback.
But an emitter resistor does provide "actual" feedback. It reduces (subtracts from) the effective input voltage across the base-emitter junction with a voltage proportional to the collector output voltage, similar to what a resistor from collector to the input would do. The difference is that the feedback is applied to the emitter side of the base-emitter junction, not the base side.
 

Mr RB

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But an emitter resistor does provide "actual" feedback. It reduces (subtracts from) the effective input voltage across the base-emitter junction with a voltage proportional to the collector output voltage,
...
Hi Crutschow, I understood what Re does and I'm not arguing with you about it's ability to help make Ic more linear to changes in Vb.

My issue was with the earlier description of the AMP having negative feedback. Please consider these example schematics I just scribbled;

Figure A is an opamp with negative feedback. When a load is applied the resistor Rfb applies negative feedback and the amp raises its output voltage to help combat the load.

Figure B is a common emitter amp with negative feedback. When a load is applied Vc tends to drop, and resistor Rfb applies negative feedback to the base turning Q1 more off and the result is that Vc rises to help combat the load.

Figure C is a common emitter amp with no negative feedback. When a load is applied Re does basically nothing but maintain the same current into the collector, so Vc drops.

Re makes the common emitter amp more linear in it's open loop response, and in that way it gives a similar effect to negative feedback under perfect conditions (when there is no load). But an amp that only has the correct output voltage when there is no load is a good definition of open-loop... ie; "without negative feedback".

So my view is that just because Re makes the common emitter amp better behaved (more linear) in it's open-loop behaviour does not make the AMP best described as "a common emitter amp with negative feedback". Personally I would reserve that term for figure B which actually has negative feedback.

(edit) Nice distortion figures AG. :)
 

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audioguru

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My issue was with the earlier description of the AMP having negative feedback. Please consider these example schematics I just scribbled;
Your circuit B does not have a series input resistor like all of our similar amps have so its amount of negative feedback
is determined by the output impedance of the input source.

In a transistor with an emitter rersistor when the collector has a load then its current and the emitter current increases. The increased emitter current has opposite phase to the emitter current provided by the input so it is negative feedback.
 

carbonzit

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Update: AG's amp ahead by a length

So following the latest results from this other thread (about using FFT in LTspice to measure THD), it appears that Audio McDuck's latest design is the clear winner in the distortion department, with a THD (measured at 1KHz) of only 0.06%.

However, it turns out that my humble original 3-transistor design (with a couple of modifications) isn't too shabby either, with a THD of only about 0.7%. Which makes it quite usable as a real headphone amplifier, if one wants to go the non-op amp, non-IC route. (Both .asc files attached below.)

I ended up adding a small (100Ω) emitter resistor, which although it raised the THD marginally, reduced the quiescent current substantially, important for battery-operated devices.

(Regarding the LTspice simulation, alert readers will notice that I've parameterized the frequency used both for the input voltage source and the FFT analysis, so one need only change one number (on the .param freq nnnn statement) to change the frequency. Hat tip to Alec_T for this.)
 

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audioguru

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CarbonZit,
Adding the 100 ohm emitter resistor to your circuit reduced the gain to 0.1 so that the input is 200mV peak and the output is 20mV peak which is almost nothing. My circuit has a gain of 2.8 and an output of 560mV peak.
 

Mr RB

Well-Known Member
...
In a transistor with an emitter rersistor when the collector has a load then its current and the emitter current increases. The increased emitter current has opposite phase to the emitter current provided by the input so it is negative feedback.
:eek: How does loading the collector increase the emitter current?

The way I see it, if the base signal is at Vb 1.6v, Ve is 1.0v, Re 1k is 1mA, so collector current is (approx) 1mA. It is acting as a constant current sink of 1mA into the collector (Ic).

Now if the collector (amp output) is loaded by the load (some current to ground), there is no change to Ic sunk which is still 1mA, so the voltage Vc drops according to load as Rc has more curren tthrough it.

There is also no change to the voltage on Re which is still 1v (as there is no negative feedback from output to emitter) and there is no change either to the 1.6v Vb (as there is no negative feedback from output to base).

So under a specific situation (amp with no load) Re performs a function like negative feedback making change in Ic more linear to change in Vb, but the common emitter AMP itself has no negative feedback.
 
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audioguru

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When a load causes the collector current to to have a higher current swing then the emitter current also has the same increased current swing.
When the NPN transistor's base has a signal then without an emitter resistor the transistor has a very high voltage gain of about 180 so a very small input makes it produce a large distorted output. But with an emitter resistor the emitter voltage follows the base voltage so that the actual base-to-emitter signal voltage and current are much less due to the negative feedback from the emitter resistor which reduces the distortion and reduces the gain.
 

carbonzit

Active Member
Really bad amp!

You want to see a truly atrocious 3-transistor amplifier? Check this out:



I got this off the web (here) and simulated it in LTspice (attached below).

80% THD!

(And this is the author's "improved" version!) No doubt one reason for the horrible performance is the fact that it incorporates positive feedback (?!?!?).
 

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The Electrician

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So under a specific situation (amp with no load) Re performs a function like negative feedback making change in Ic more linear to change in Vb, but the common emitter AMP itself has no negative feedback.
It's not just a function like negative feedback; it is negative feedback.

See for example: http://mysite.du.edu/~etuttle/electron/elect7.htm

Note near the top of this page, the figure showing the four ideal feedback configurations.

Down near the bottom, there's a figure labeled "Voltage amplifier". To the left of that figure in the text, you'll find: "We observe that the output current is sampled, by the 1k emitter resistor, and the voltage across this resistor is fed back to the input (the emitter is the "inverting input" of the transistor) with the proper phase to give negative feedback. Therefore, we have series-series feedback in this case."

Also, have a look at page 7 of this:
http://www.electro-tech-online.com/...TransistorFeedbackCircuits_SW_ED4_PR2_web.pdf

There is much more on the web to be found by searching for "transistor series-series feedback". For example, a slide from a power point presentation is attached.
 

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carbonzit

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It's not just a function like negative feedback; it is negative feedback.

See for example: http://mysite.du.edu/~etuttle/electron/elect7.htm
Thank you! Bookmarked--and printed. I love this guy's stuff. (He has a lot of great material on vacuum tubes, not to mention seemingly every other aspect of electronics.)

Note near the top of this page, the figure showing the four ideal feedback configurations.
Right. He uses wordier descriptions ("series mixing, shunt sensing") than my textbook, which simply calls them SS/SP/PS/PP, but same difference.
 

audioguru

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You have the amplifier you found fed from a source impedance of zero which causes the output to clip its head off because the negative feedback is shorted.
I added a 10k series input resistor then the output was not clipped. The distortion is 0.75%.
 

audioguru

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EDIT: You should stretch out the sinewaves so you can see if they are clipping by changing the term "0.95" in startup to "0.995".
This is what a your clipped sine-wave output looks like:
 

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carbonzit

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I added a 10k series input resistor then the output was not clipped. The distortion is 0.75%.
Huh? I tried what you suggested, but got a different result (0.81%). I noticed you edited your message between first posting it and my reply. You originally claimed 0.5%. Where are you getting these figures?

If you're going to make claims pertaining to a SPICE simulation, could you please post your file? Mine's attached below. (You'll notice that I decreased the timestep to 1uS; at the default, the THD figure was over 2%.)
 

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carbonzit

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EDIT: You should stretch out the sinewaves so you can see if they are clipping by changing the term "0.95" in startup to "0.995".
That's totally not necessary; no matter what parameters you use on the .tran statement, you can always zoom in to displayed waveforms at will.
 

carbonzit

Active Member
Thank you! Bookmarked--and printed. I love this guy's stuff. (He has a lot of great material on vacuum tubes, not to mention seemingly every other aspect of electronics.)
This guy's first name is Elizabeth: http://mysite.du.edu/~etuttle/ :D
Thank you again, and I submit myself for 25 lashes with a wet noodle for ASS-U-Ming that the author was male, an assumption I see far too often in places like this, judging from the pronouns used.

I love her writing, and her inquisitive, experimentative nature.
 

Mr RB

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When a load causes the collector current to to have a higher current swing then the emitter current also has the same increased current swing.

...
Sorry AG but you are talking complete nonsense!

Given a Vin of 1.6v there is a fixed Vb of 1.6v, which will NOT change as there is no negative feedback from the output of the amp back to the base.

With a fixed Vb of 1.6v you get a Ve of 1.0v due to the open loop properties of the B-E junction dropping approx 0.6v. With Ve of 1.0v and 1k Re you get a fixed Ic of approx 1mA.
This is *fixed* at 1mA as there is no negative feedback from the output of the amp to either the base or to the emitter.

So the amp functions as an open-loop constant current sink where current sunk into the collector (Ic) is fairly linearly proportional to changes in Vb.

BUT Vb is fixed, and it in turn fixes Ve and Ic, and there is NO negative feedback of the amp output (Vc) to either base or emitter so when the collector is loaded its voltage drops accordingly, and Vb and Ve and Ic all remain fixed and unchanged!

I have no idea where you get "increased load current causes increased emitter current". :eek:



MrAl said:
It's not just a function like negative feedback; it is negative feedback.
...
It is negative feedback of Ic, but NOT negative feedback of the AMP. See what I said to AG above. The AMP runs very much as open-loop.

MrAl said:
...
Down near the bottom, there's a figure labeled "Voltage amplifier". To the left of that figure in the text, you'll find: "We observe that the output current is sampled, by the 1k emitter resistor, and the voltage across this resistor is fed back to the input (the emitter is the "inverting input" of the transistor) with the proper phase to give negative feedback. Therefore, we have series-series feedback in this case."
Not right. The amp's "output current" is not sampled at ALL. That is the problem! The current that is "sampled" is the current sunk into the collector. That is NOT the output current of the amp. Output current of the amp is current drawn into the load, and is entirely separate to Ic and not will not affect Ic (as Ic is a constant current sink).

(Please see circuit C in my post #43 above)
As I said to AG, if you draw current FROM the amp output, the emitter current remains fixed and the amp output voltage Vc drops in response with load. It doesn't get any more open loop than that.

Given a load, there is no negative feedback from Vc to either Ve or Vb in response to the load.
 
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