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Topic: Opamp setup doubt (Read 10142 times) previous topic - next topic

Grumpy_Mike

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1.  Amplification stage there is done by bjts whereas 2.  Impedance reduction is done by the opamp.
Why is that confusing? There is no rule that says it can't be like that, and there should be no expectation of it being different.

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What I want is to be able to calculate from beginning to end what happens.
You can't do that because you do not know what voltage you need to feed to your earphone to give what you think is an adequate level. There is also no information as to how much voltage your microphone produces when you give it what you think of an adequate sound input. These are hand waving woolly concepts that stop calculations stone dead.

Marciokoko

The confusion to me is because I would expect the opamp to to the job of amplifying a sound because of its name, amplifier. 

So how does someone go about designing an amplification circuit that takes input from a microphone as small as one taken from a PC and use it to produce sounds on headphones?  Where does one start?

Grumpy_Mike

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The confusion to me is because I would expect the opamp to to the job of amplifying a sound because of its name, amplifier.
It does do the job of amplifying voltage, it does not do the job of impedance matching to a source.

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Where does one start?
One starts by knowing what the inputs and outputs of the system are going to be. That means real numbers, real components and real sound level inputs and outputs values. None of which you seem to know.

Marciokoko

Indeed I don't.  Can you give me an example of how I could figure them out? 

Grumpy_Mike

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Can you give me an example of how I could figure them out? 
You measure the output of your microphone on an oscilloscope.
You feed the earphone with the output from a signal generator and see what output is the "right" level.

I assume that you have not got either of these two pieces of test equipment so your wish to:-
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What I want is to be able to calculate from beginning to end what happens.
Is impossible.

Therefore you have to resort to what most people do and that is use trial and error.

Start with a X10 to X100 preamp followed by an audio amp.


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The confusion to me is because I would expect the opamp to to the job of amplifying a sound because of its name, amplifier.
And what about the op bit? It stands for "operational" and an operational amplifier is not what you want to do all the job. There are lots of different types of amplifier that are wrong as well, like an instrumentation amplifier, thermocouple amplifier and a trans-conductance amplifier, to name but three. They all amplify but they are all not suited to what you are doing.

Marciokoko

Thanks Mike, great explanation.

And I am definitely thinking of getting an oscilloscope

Marciokoko

Btw can you recommend an oscilloscope?

I'm also thinking of a power supply so is there an oscilloscope that might have an added power supply functionality built in?

Otherwise I'm looking in Amazon at Siglent sds1052dl or hantek dso507

Marciokoko

Ok so comparing the LM324 of the original tutorial, my TL071 and the suggested LM386 as well as a true power amp, LM1875:

LM386
-Output 325mW
Small power output, not rail to rail because on Fig Output Voltage vs Supply Voltage it is always about 1-2V less.
So this shouldn't be a good candidate yet it was suggested and it is described in the datasheet as a low voltage power audio amplifier.
Q1/ So does it classify as preamp?  How come it was suggested in post #2 and I saw it in a diagram for a 'power amp' which I referenced in post #27

LM324
Not classified as rail-to-rail
Output 20-40mA, so the current is too small to drive headphones.
Q2/ So it classifies as preamp?

LM1875
Described in datasheet as audio amplifier
Said to deliver 20W to a 8Ω load
Seems to be able to deliver 3-4Amps even at low voltages around 4-5V according to graph.
Not classified as rail-to-rail in datasheet but seems like its more the power amp Mike has been suggesting.
Q3/ Classifies as poweramp?

TL071
Not rail to rail because max output swing 13.5V while 15V supply
Figure 6. Maximum Peak Output Voltage vs Load Resistance shows that for a 32Ω headphone set, Vout would be 0, so...
Q4/ Classifies as preamp?

Opamps do voltage amplification but not impedance matching.  I need to clear up the impedance matching thing because a headphone set is a 32Ω load which will not be "driven" by a preamp such as TL071 because its output current is too small, ok. 

Q5/ But the TL071 Fig 6 says 100Ω/1kΩ/10kΩ load will be driven by it?  So it seems weird the TL071 can drive larger impedance loads but not a 32Ω headset.

Thanks for your time and patience :-)

Grumpy_Mike

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So does it classify as preamp?
A preamp is not a classification. It is a circuit function block you can make. You can make them using, amongst   other things, with an operational amplifier.
An operation amplifier is normally defined as an amplifier with a differential input ( + and - ) and a single ended output. It has an extremely high ( unusable high ) open loop gain and to get the gain to be a usable value you apply negative feedback. The output impedance is normally in the range of 1K to 10K.

 
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I need to clear up the impedance matching thing
Yes it is the source of all your misconceptions.

Ever thought why you can't turn over a car engine with 8 AA batteries? After all they can provide 12V just like your car battery. You know that you can't but why? Well it is a lot to do with the chemistry in a cell but all that, and more, gets bundled up in the concept of output impedance. It is the equivalent of a series resistor in line with the voltage generator. It forms a voltage divider with your load and if this theoretical resistor is several orders of magnitude greater than your load the potential divider action insures that there is "stuff all" voltage across your load.

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So it seems weird the TL071 can drive larger impedance loads but not a 32Ω headset.
Any voltage source can drive a high impedance because as long as the load is comparable with the output impedance it will deliver a voltage to it. When the load is the same as the output impedance you get half the voltage across the load as the voltage generator produces.

This link is about Raspberry Pi GPIO output impedance but the principals apply to any voltage generator, battery, amplifier, output pin, transducer.
http://www.thebox.myzen.co.uk/Raspberry/Understanding_Outputs.html

Marciokoko

#39
Jul 12, 2016, 02:32 am Last Edit: Jul 12, 2016, 02:48 am by Marciokoko Reason: image
Grumpy_Mike,

Im ready to add that transistor to the negative feedback loop now.  

So what you suggest is that I amplify the current through the feedback loop in order to amplify the sound?

How would that work, like this?


Grumpy_Mike

Basically yes, but I would also add a 4K7 resistor from the emitter to ground as well as the load. This makes sure there is always a path to ground even when it is not connected to a load.

Marciokoko


ChrisTenone

4K7 is shorthand for 4.7K. Saves typing a dot. Either would work. 4.7K ohm resistors are common.
What, I need to say something else too?

Marciokoko

#43
Jul 13, 2016, 02:45 am Last Edit: Jul 13, 2016, 02:46 am by Marciokoko Reason: image
Figured.  Here is the final schematic I drew up:



Next Ill see if I can translate this correctly onto a breadboard (couple of breadboards :-)) and then Id like to make the calculations to see what I should expect.

Does that sound right?

Marciokoko

#44
Jul 13, 2016, 04:38 am Last Edit: Jul 13, 2016, 04:45 am by Marciokoko Reason: image
Here it is on the breadboard:

(top part) = Vout to Transistor
(nxt dwn) = Vout-Vin interconnection
(nxt dwn) = opamp
(bottom) = mic entry-cap
On the left +/- rails I have GND with green wires on the black-rail
On the rt +/- rails I have +9V on red-rail and -9V on black-rail



https://www.flickr.com/photos/43666730@N04/shares/55Ds7L

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