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Topic: Opamp setup doubt (Read 10135 times) previous topic - next topic

ChrisTenone

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(Stand alone preamps are expensive specialty items.)   
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Right!? Back in the day, having a separate preamp was a sign that you were a true audiophile. That and a Macintosh (tube) amp. (no relation to the computer.)
What, I need to say something else too?

Marciokoko

Im in Honduras which is why getting these parts is rather troublesome.  It takes me a while to get the exact part down from the US, so I end up trying whatever I can.

I can get an audio amplifier but I want to learn about opamps so I found that youtube tutorial from afrotechmods and tried following it using a TL071 instead of the LM324.  I know its not always going to work (replacing whats required with what I have) but I wanted to give it a shot anyway.  And as it turns out, Im learning from it.

I understand the tl071 may not output enough current as Mike said and I know he is very knowledgeable but I want to be able to discover that from looking at an opamp data sheet by myself.  I still can't find that in the TL071 data sheet.  It must have something to do with Fig 6 he mentioned and how the Max Output Voltage vs load resistance are related, but I really can't connect the dots.  I know ohms law, but that is far from being able to use it properly. 

Grumpy_Mike

Quote
but that link says that it means the voltage can swing anywhere from Vsupply + to -.  So maybe both explanations are the same but Im not understanding it that way
No op amp can go from + supply to - supply. But a rail to rail can get say to just 0.1V short of the supply value, so in effect is is nearly rail to rail. Often it is simpler to say it can get to that value, but it is a simplification, so your confusion is the result of thinking a general simplified explanation some how contradicts my more accurate answer.

Chris is right the fact that nether of those op-amps you have is rail to rail.

Grumpy_Mike

#18
Jul 05, 2016, 12:42 am Last Edit: Jul 05, 2016, 12:43 am by Grumpy_Mike
Quote
It must have something to do with Fig 6 he mentioned and how the Max Output Voltage vs load resistance are related,
Use the graph to see what maximum voltage you can get for a specific resistance. Then voltage / resistance will give you the output current.

You will see the resistance you want to use is not even on that graph so what you are trying to do is not possible with that amp.

In fact you will not find any op-amps that will do what you want. That is a job for power amps.

CrossRoads

Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

Marciokoko

TThanks.  I guess I would have to know what values I would need and I don't even know that.  I just saw the example and wanted to try it.

So I don't even know what Vin and Iin I was working with or what Vout and Iout I needed.  I didn't even know headphones were 32 Ohms.

I guess it would help to know those.  What voltage does a small mic like that generate?  I think I saw 300-400mV?

CrossRoads

Yes, something very small.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

Marciokoko

Ok so if I have 300mV on the I out end, I need to know how much I need on the output end.  So if a headphone set is a 32Ohm load, How do I make "the connection"?

CrossRoads

Your power supply will limit how much you can get on the output end.
Have a 5V supply? Then 0-5V, or +/-2.5V, will be the max you can get.
Power = IV (current x voltage), subbing in V=IR or I=V/R, you get power = V/R * V (V^2/R).
5V x 5V/32 ohm = 0.78W
There is some loss for transistors, so you won't get the full 5V, so power will be somewhat less.
If your supply can't provide enough current, 5V/32ohm = 156mA, then power will be reduced also.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

ChrisTenone

One thing that should be pointed out, since the OP is an op-amp newb: As stated (but not emphasized) in CrossRoads' post above, the power supply need not be +V and -V of the same magnitude. Often you will see op-amps listed as "single supply". This just means that you can make V- equal to zero volts (ground.)
What, I need to say something else too?

Marciokoko

Ok but I have a +/-9V supply made by two 9V batteries as shown in post #1, so I can use P = 18V^2 * 32 = 10,368W, is this correct?




ChrisTenone

Does ten kilowatts from two 9 volt batteries sound reasonable?
What, I need to say something else too?

Marciokoko

#27
Jul 05, 2016, 03:54 pm Last Edit: Jul 05, 2016, 06:47 pm by Marciokoko
No, not at all!  I'm obviously confused.  Oh I see what I did wrong....its 9 * 9/32 = 2.5Watts.

OK I would like to understand from the beginning.  If a small mic can generate a 300mV potential, what do I need to do to that to make it "hearable" on headphones or a speaker?

So basically I need 1 opamp to preamplify the mic sound and then could I use another opamp (or would it have to be a power amp) to amplify the sound for a headphone set?

I found these images from another site:



and



which talk about those 2 stages of amplification.  Does this look like a better circuit to learn from?

Grumpy_Mike

Ok but I have a +/-9V supply made by two 9V batteries as shown in post #1, so I can use P = 18V^2 * 32 = 10,368W, is this correct?
No the , is the delimiter for thousand not a decimal point.

However the problem is the same you can't get 10W from a 9V battery. A voltage of 18V through a 32R resistor gives a current of 18 / 32 = 0.5625 Amps. Those batteries will not supply that current.

Quote
If a small mic can generate a 300mV potential,
Can it? I would say more like 30mV myself.
You have to do two things:-
1) Amplify the voltage
2) Reduce the impedance of the driver.

An op-amp will do 1) but as you have found the op amps you have used will not do 2). Therefore to do 2 you need a power amplifier. That is why there are often two stages of amplification. One to boost the voltage to the right level and the other to reduce the output impedance ( drive current ).

About 600mV will be enough if you can develop it across a 32 R earphone. If you put 10W into an earphone then it is a race to see which melts first your ear or the ear phone.

That last circuit will do 2) and either of those first ones will do 1). You need a voltage gain of about 10 times on the first circuit and a voltage gain of 1 on the second.



Marciokoko

Well that's the kind of thing that's confusing because the 1.  Amplification stage there is done by bjts whereas 2.  Impedance reduction is done by the opamp.

I'm reading a few more sources on opamps and I'll see if I understand the circuit from start to finish.  What I want is to be able to calculate from beginning to end what happens.

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