Opamp setup doubt

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.

How much current do you need? You can find op amps with pretty high output current/channel:
http://www.digikey.com/product-search/en/integrated-circuits-ics/linear-amplifiers-instrumentation-op-amps-buffer-amps/2556125?k=op%20amp

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?

Yes, something very small.

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"?

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.

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.)

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?

Does ten kilowatts from two 9 volt batteries sound reasonable?

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?

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?

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.

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.

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.

  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.

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.

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?

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.

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.

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

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:-

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.

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.

Thanks Mike, great explanation.

And I am definitely thinking of getting an oscilloscope

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

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 :slight_smile: