Op-Amp premature clipping

Discussion in 'General Electronics Discussion' started by boogyman19946, Apr 19, 2012.

1. boogyman19946

38
0
May 2, 2011
Hello again!

Ok, so I decided to put some knowledge to work and wired up a simple amplifier using an Op-Amp. I only had the TL022CP on hand but don't think it should pose all too much of a problem.

The problem I'm facing is that the circuit compresses the waveform instead of amplifying it. What I mean is, the output doesn't swing between the full 0 to 18 volts but about 7 to 10 (from what I can tell from the oscilloscope). I can see that, if it were not cut off, it would be amplified just like I expected, with a gain of about 5 times the difference between the inputs. The input comes from my phone.

Does anyone have an idea of why it clips the signal, as it seems, prematurely?

Here's the schematic

EDIT: the power pins of the TL022CP plug in directly to the 18V and ground busses.

Last edited: Apr 19, 2012
2. Laplace

1,252
184
Apr 4, 2010
The datasheet for the TL022 uses a standard load resistance of 10K. The short circuit output current is a maximum of 6 mA. Whereas the load resistance here is almost 250 ohms so the output would be clipped at +/- 1.5 volts. That might explain why your signal output swings from 7 to 10 volts.

3. timothy48342

218
0
Nov 28, 2011
(Going to show my cluelessness now.)

Is this normal that an OpAmp has such a low short circuit output current?

So what do you do with that output if you want to drive something thast needs more?

Would putting an emitter follower between the OpAmp and the electrolytic in this circuit help?

-tim

4. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
yes

Use something to increase the output current.

Yes, you could do that, but even better may be to add a class B output stage that is inside the feedback loop of the op-amp.

5. timothy48342

218
0
Nov 28, 2011
Thanks (*steve*)!
(*Wow, that site is loaded with good stuff. Going to take a while to digest it.*)

6. duke37

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Jan 9, 2011
Your op-amp has two methods of setting its operating point and these can clash if things are not absolutely identical.
Delete R1 and R2 and let the feedback work.

7. boogyman19946

38
0
May 2, 2011
Ok, I succeeded in adjusting the resistors for the feedback and feeding the output through an emitter follower, but being the noob that I am, I can't seem to figure out how to give the speaker more juice.

Does the B class output stage require a ground point that's between the positive and negative?

8. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
You can do the same thing with the speaker connected to ground and connected to the output via a capacitor. It's not ideal, but it will work.

Be careful about the currents that can flow through the transistors though.

9. TedA

156
16
Sep 26, 2011
boogyman19946,

I wonder what the device labeled "SP" might be. If this is a high impedance headphone, the circuit may work to some degree. If it is a 4 ohm sub woofer, not so much. The impedance and the power requirements are the important parameters, as far as the driving amplifier is concerned.

If you are driving a high impedance device, then the series resistor R6, and the low current capabilities of the TL022 output stage, may still allow a reasonable output level. If "SP" is a low impedance device, it will receive very little power.

The circuit as-drawn has a very low input impedance. At moderately low audio frequencies, this impedance will be mostly determined by C1. The voltage gain is not well defined, but will be quite a bit more than five times. As has been pointed-out, R1 and R2 are not doing much for you.

If you want to provide a reasonable input impedance, and to set a finite voltage gain, you need to add a resistor in series with the input, or else apply the input to the non-inverting input of the op-amp.

The output impedance, as measured at the terminals of device "SP" is about 220 ohms, determined by the value of R6. If R6 is considered to be part of the load impedance, the output impedance is low, and your circuit can be called a "transimpedance" amplifier.

The TL022 may not be the best device for your task. It is a low power, high voltage ( by IC standards ), low speed amplifier, with restricted input and output voltage ranges. The data sheet lists the characteristics with a 30V total supply. It will work at lower voltages, but the data sheet is not particularly helpful about what you should expect.

I have no direct knowledge of this TI part, but the somewhat similar LM358 / LM324 is not a very good audio amplifier, due to its low speed and poor output stage linearity.

That said, if you want to experiment, you can learn a lot messing with what you have at hand.

If you are attempting to drive a low impedance speaker, you will have to add some sort of power output stage to get enough output current.

If you add an emitter follower stage to the op-amp's output, it will need to be a complementary emitter follower, with two transistors. To get reasonable distortion, the transistors will need a bias circuit to set a non-zero idle current.

If you want a bit more power, you might try a complementary pair of transistors driven by the VCC and GND pins of the TL072.

Both sorts of power booster circuits are to be found in common op-amp ap-notes. If you don't find it first, I'll try to link some more info later.

Let us know how your project is doing.

Ted

10. boogyman19946

38
0
May 2, 2011
I've redone the circuit a little bit, this time using two batteries and setting the ground point between -9V and 9V. This way I don't have to put in the voltage dividers.

Here are the schematics of what I tried:

EDIT: I've reduce R3 from 6875 to 5k to get rid of the clipping distortion. So, R3 is now 5k rather than 6.9k.
EDIT2: Ok, I just realized that the gain for the darlington is MULTIPLIED by the second transistor XD I thought the Betas added together but apparently they are multiplied. That generates a whole lot more current than I expected! That explains why they are frying XD

I changed the op-amp... I'm not actually sure why. I think it was because I wasn't sure if the TL022 would get fried by the battery voltages. The new Op-Amp also has a higher short-circuit current. I guess that doesn't change all too much due to the transistor pair being placed right at it's output. I thought that since the Op-Amp will do about anything to cancel the voltage difference between the inputs, it would overcome the dead-zone of the transistors as described by the article, but the oscilloscope is clearly showing a malformed waveform close to 0V so I'm guessing it's not exactly succeeding. Is this because the Op-Amp is slow? Or can it simply not generate enough juice?

Also, I've attempted two ways of amplifying the output current. The first was feeding it through the same amplifier as the op-amp output but that wasn't really working for me, although I might have wired it wrong or used the wrong values for the resistors.

In the second schematic, I used a Darlington Pair which does ... ok amplifying the current compared to the other ways I tried but the juice running through the transistors will most certainly cook them after a short while.

I haven't been able to figure out how a non-inverting amplifier actually works. I'm using an inverting one because it's the one that I vaguely know how to wire. I'm also still having a little trouble visualizing how capacitors function and finding out the proper values I should use is a whole different story. It's ok when the caps are used alone or in some straightforward filters, but any kind of deviation makes me draw a blank as to what kind of values I should draw.

My questions are mostly the following:

What does it mean exactly to have low input impedance? Does it mean the thevenin equivalent impedance to the "left" of the feedback loop?

If I don't put a voltage divider to bias the input voltage from the phone and into the non-inverting input, won't the amplifier clip the negative portions of the amplified voltages? What I mean is, since in the first schematic ground is the most negative point in the circuit, then if the phone causes the op-amp to have to generate a negative voltage (between output and inverting input) , it seems to me like it wouldn't be able to do it. If the op-amp had a gain of 5, then if the input was 1V, would it be able to generate -5V?

Is it a good solution to use two batteries set up like the above (split voltage)? How does one deal if there is only one voltage source? It seems to me like making the ground point as the midpoint between the highest and lowest voltage would be unfeasible in that situation.

If I stick a resistor in series with the capacitor that couples the phone signal, how do I find the value of the capacitor I need?

And that'll be about all ^.^ I'm sure I'll have more questions as they arise, but for now, that's all I could think off. This is the first circuit I've ever made from scratch and I'm sure the clueless nature of the design shows very clearly. XD Well, one has to start from somewhere.

Last edited: Apr 21, 2012
11. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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The first one is better, however:

1) remove the resistor in series with the speaker.
2) connect the speaker to 0V rather than ground
3) placing a capacitor in series with the speaker is probably a good thing
4a) include the output stage in the feedback loop
or
4b) remove the output stage entirely and just use the driver stage
or
4c) wire up your driver stage like the output stage
5) consider reducing the gain as required.

The output need not swing to the supply rails as you are driving a low impedance output.

Oh, and try to make smaller diagrams. Nothing greater than a typical screen's resolution will ever be needed. For schematics like these, 800x600 would be more than ample.

12. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Yes, that's generally how it's done. Your output won't swing all the way to the supply rails (maybe only to within a couple of volts of it).

In the first circuit you have a push/pull pair with no biasing driving a push/pull pair with biasing. As I said in the post above -- just have one stage and make that a biased one -- AND place it inside the feedback loop.

The op-amp can't do magic. The more gain you're asking for, the less is available to fix up the crossover problem.

Once you remove that 240 ohm resistor you'll find you need far less gain (I hope your speaker is capable of several watts). Note that you'll also want to keep the gain down so that you don't overstress the output transistors (they have a power rating too)

Neither of the ways you show above are correct. See my previous post.

It doesn't really matter. The difference is that for a non-inverting amplifier the output goes more positive as the input goes more positive, and the output goes more negative as the input goes more negative.

For an inverting ampolifier it happens reversed. As the input goes more positive the output goes more negative, and as the input goes more negative the output goes more positive.

In terms of what you hear, there is no difference.

It matters in terms of feedback because positive and negative feedback do really different things. Negative feedback is done via the inverting input and tends to reduce distortion.

Positive feedback is done via the non-inverting input and increases distortion (which is NEVER a good thing with an audio amplifier)

Positive feedback which occurs at a certain frequenct will also cause an amplifier to oscillate at that frequency (which is really bad). For this reason, you should place a small value capacitor across the feedback resistor to reduce gain at high frequencies -- especially when you are placing more stuff in the feedback loop.

That's OK, choosing the correct capacitor value is non-trivial

No, it means that if you consider the output to be a perfect voltage source, it has only a small value of resistance in series with it. Thus it is capable of supplying a high current, and that the voltage varies only a small amount with a small incremental change in load current.

The input should be relative to your 0V, i.e. half way between +9V and -9V. Your batteries are a voltage divider too.

yes

With a voltage divider (or some other means of biasing the input voltage above ground.

Nevertheless, that's what you might do. It doesn't always have to be exactly half way either. This is especially true in class A stages.

Technically... You choose a value that gives you an appropriate 3db point for low frequencies.. Typically, you pick one that's big enough.

13. boogyman19946

38
0
May 2, 2011
EDIT: Sorry about the huge diagrams. XD The viewing programs always scale the pictures to fit on the screen so I had no idea how big they actually were.

Woot! It's alive!

I removed the driver stage and coupled the signal with a capacitor directly through the speaker to 0V. Now it's really starting to sound like it should! Here's what I got:

EDIT: THE GROUND POINT IS ALSO CONNECTED TO THE JUNCTION OF THE TWO BATTERIES.

I plugged the speaker between the batteries for a dual rail set up. In the schematic with a Class B driver, the ground point is the 0V point so I wasn't sure what you meant, so I just plugged it at 0V. The speaker I'm using is marked at 0.5W.

I'm not sure if I have done it correctly in the circuit above. I put a voltage divider in the push/pull pair right after the output of the amplifier. I'm guessing it'll ease the op-amp's job not having to account for the push/pull's dead-zone. I'm not sure if that's the correct way to do it, but it looks like the wave produced features less distortion (although there is still quite a bit of it).

I haven't tried reducing the gain yet but I'll do it eventually. If I wanted to put in volume control, I figure it would be best to put it inside the feedback loop right?

Is there a particular reason why positive feedback increases distortion? I was trying to figure out why the circuit for my guitar amplifier had capacitors in the feedback loop and while I realized that it would reduce the gain for high frequencies, I wasn't sure why it would want to do that.

I thought getting minimal variation in supply voltage with respect to load current is a good thing. Does this mean I should aim for low input impedance?

So in other words, just overshoot the value? Sounds like a good deal to me!

Last edited: Apr 22, 2012
14. boogyman19946

38
0
May 2, 2011
I've downloaded a tone generator on my phone to see what happens to the signal at various frequencies and it turns out, that the sine wave... doesn't really look like a sine wave by the time it leaves the amplifier. Here's the shape (as best as I could remember it) of the waveform generated as the output at 3047Hz:

I drew it by hand in paint.

EDIT: I think the capacitors are the ones that distort the wave but I'm not sure why.

15. TedA

156
16
Sep 26, 2011
Phone? Is that a 600 ohm device?

What are you connecting to your input? You want something with a fairly high output impedance to feed your current input. Don't try to feed a voltage in there.

Try reducing the signal level and see if the distortion decreases at some point. You need an input signal around 1 mA peak, or less.

Ted

16. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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You show a ground at your input, but you don't show the ground between the batteries. The junction of the 2 batteries is (or should be) your ground

17. boogyman19946

38
0
May 2, 2011
It's a phone that I'm connecting to the input, yes. I'm not sure how much impedance it sports though.

I've reduced the signal from the phone (changed the volume while it was on) there was a point where the signal wasn't distorted at the speaker, but, at the same level, the signal right at the high-pass filter (right at the input) wasn't a signal at all: it was just noise. Also, at that level, the speaker didn't really produce any sound, it just kind of hummed. One notch higher, and the speaker would start playing the signal and at the same time, the noise at the high pass filter turned into an actual waveform, albeit distorted. Does this mean that it's the input that is being distorted and the op-amp amplifies it?

You're right, there should be a ground symbol between the two batteries, I've somehow forgot to add the symbol to the schematic, but that's where the ground is connected XD

18. duke37

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Jan 9, 2011
The circuit that you show three or four posts back, is a virtual earth amplifier. If you have plenty of gain, there should be negligible voltage at the input. As explained previously, the amplifier is controlled by current input.
The gain will be R3/input impedance. Try a 1k in series with the input and turn up the volume.
R2 is not necessary but will do little harm.
C1 in series with 1k will make a high pass filter, passing frequencies above about 2kHz, put a 10uF capacitor here.

19. boogyman19946

38
0
May 2, 2011
I've done what you said duke but the waveform hasn't changed (or maybe it has but the jagged waveform was still present). On the other hand, I'm curious as to why the amplifier is controlled by the input current as opposed to the voltage. I was under the impression that it's trying to set the voltage at both its inputs to be the same and I'm having a hard time understanding why it's the current that does the job in this case.

There is something else I realized though. I've take my phone and tried testing the signal outside of the circuit and the it turns out, that the "sine wave" generated by my phone looks like so:

EDIT: sorry about the size of this one

Can this cause the strangeness in the output signal? Do the capacitors behave differently at the more level hills than they do while the signal changes?

Last edited: Apr 23, 2012
20. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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What is the amplitude of your input signal?