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Clippping audio signals witn diodes

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apakhira

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Well many of the distortion circuits use diodes 2 clip the signal. But apart from saying that "the diode lets the signal 2 go upto a certain amplitude. After that it clips it." But how does i do that? And cud u explain the different things happening when diodes are placed in different ways (eg shunted 2 ground, in the negative feedback loop of an Op-amp, as a half-wave rectifier)?

Thanx
A newbie.
 

evandude

New Member
a diode allows current to flow in only one direction, when the voltage across it is greater than its 'forward voltage' (approximately, it varies slightly with current)

typical diodes have a forward voltage of 0.7v or so, so whenever the voltage of the audio signal goes higher than 0.7v, the diode turns on, allowing current flow (presumably to ground), which causes a voltage drop on the signal source since the output resistance is high... thus holding the signal at the forward voltage of the diode until it drops below again, where the diode turns off and does not affect the signal any more. with 2 diodes facing the right ways, you can clip it on both the positive and negative sides of the signal.

in other applications (feedback loops, half wave rectifier, etc) it again relies on the premise that current can flow in one direction but not the other... in some cases the forward voltage of the diode is used to advantage, other times it is desired for it to be as low as possible, and if the voltage levels are large enough it can often be ignored.
 

apakhira

New Member
So when the signal goes above the forward voltage, only the portion of the signal above the forward voltage is sent to ground? I take it by foward voltage u mean the minimum voltage necessary 2 make the semiconductor work? And how does this voltage change if diodes r connected in series?

Now, could u please explain what happens in case of the -ve feedback loop in an op-amp? I'm attaching a circuit schematic of the Ibanez Tube Screamer TS-9 as an example.

Thanx
 

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audioguru

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The output of the 1st opamp will be limited to a positive-going and negative-going peak voltage that is the forward voltage drop of the diode, plus the peak level of the input signal.

The diodes don't abruptly start conducting when the voltage across them is 0.7V. In this circuit, if the "drive" pot is max, then the diodes begin conducting when the voltage across them exceeds slightly less than 0.4V.

These diodes are in a see-through case. Shine a light on them and they will conduct all the time! If the light is powered from AC then the output will be a buzz.
 

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apakhira

New Member
audioguru said:
The output of the 1st opamp will be limited to a positive-going and negative-going peak voltage that is the forward voltage drop of the diode, plus the peak level of the input signal.

The diodes don't abruptly start conducting when the voltage across them is 0.7V. In this circuit, if the "drive" pot is max, then the diodes begin conducting when the voltage across them exceeds slightly less than 0.4V.
In the -ve feedback loop case, when the diodes r clipping, how do they affect the fuzz/buzz or whatever coming thru the output, since imho(which can be more wrong than right) it is putting the clipped signal int the inverting input, which i think acts as a gain regulator?
 

audioguru

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When a diode starts conducting then it shorts the negative feedback resistors, severely reducing the gain of the opamp so it acts like a follower.
When the fuzz circuit is working, its output is square waves since the peaks of the music are chopped off.
 

apakhira

New Member
When the diodes r working in the feedback loop, they r sending the peaks of the signal into the non-inverting input and letting the signal chopped signal to go thru the output - am i right? And is there any reason 4 making it act as a voltage follower?
 

audioguru

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apakhira said:
When the diodes are working in the feedback loop, they are sending the peaks of the signal into the non-inverting input and letting the chopped signal to go thru the output - am i right?
Correct, but the chopped output level would be modulated by the input signal level. Opamps always try to keep their inputs the same.

And is there any reason 4 making it act as a voltage follower?
Then if the input signal is high enough, the output has some of the dynamics of the input. It is called "soft clipping".
An inverting opamp signal chopper or a resistor feeding a couple of back-to-back diodes to ground would just have ordinary fixed-level square waves at their output.
 

apakhira

New Member
apakhira wrote:
When the diodes are working in the feedback loop, they are sending the peaks of the signal into the non-inverting input and letting the chopped signal to go thru the output - am i right?

Correct, but the chopped output level would be modulated by the input signal level. Opamps always try to keep their inputs the same.
Made a mistake-I think it wud hav been "inverting input". Also, do op-amps try 2 keep their inputs same, or outputs same? :?
An inverting opamp signal chopper or a resistor feeding a couple of back-to-back diodes to ground would just have ordinary fixed-level square waves at their output.
Inverting? I thought these signal choppers were non-inverting...

Now, since i read forward voltage gets doubled if 2 diodes are used in series, what will happen if 2 diodes are connected in parallel in the same orientation- will the forward voltage get halved?
[/b]
 

Nigel Goodwin

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apakhira said:
Now, since i read forward voltage gets doubled if 2 diodes are used in series, what will happen if 2 diodes are connected in parallel in the same orientation- will the forward voltage get halved?

No, they aren't resistors!.

Try thinking of them more like batteries?, if you put two 1.5V cells in series you get 3V, if you put them in parallel you DON'T get 0.75V, you get 1.5V but with twice the current available. Diodes are similar, put then in parallel and you get twice the current handling capacity of a single diode (with certain reservations!).
 

audioguru

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apakhira said:
apakhira wrote:
When the diodes are working in the feedback loop, they are sending the peaks of the signal into the non-inverting input and letting the chopped signal to go thru the output - am i right?
audioguru said:
Correct, but the chopped output level would be modulated by the input signal level. Opamps always try to keep their inputs the same.
Made a mistake-I think it wud hav been "inverting input". Also, do op-amps try 2 keep their inputs same, or outputs same? :?

We both made a mistake. The diodes feedback the output signal to the inverting input of the opamp.
Opamps have a gain of a few hundred thousand so their input voltages are just about the same when negative feedback is added. Make the opamp a follower by connecting its output directly to its inverting input. Then you can see clearly that the inputs are the same.

apakhira said:
audioguru said:
An inverting opamp signal chopper or a resistor feeding a couple of back-to-back diodes to ground would just have ordinary fixed-level square waves at their output.
Inverting? I thought these signal choppers were non-inverting...

This circuit is non-inverting so that it has soft-clipping. I was saying that the signal would have hard-clipping if it was the other ways.
 

apakhira

New Member
OK. I get it. So r u talking about something like this?(See attachment). So how is inverting op-amp increasing lipping?
btw, a dumb question: If op-amps have huge gain, why don't ppl use op-amps 2 make hi-fi amplifiers?
 

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Nigel Goodwin

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apakhira said:
If op-amps have huge gain, why don't ppl use op-amps 2 make hi-fi amplifiers?

They do, almost all transistor HiFi amps are opamps, some even use opamp IC's for low level stages, but the power amplifier itself is almost always a high power discrete opamp (or a high power IC opamp, often a hybrid circuit).
 

Dr.EM

New Member
Don't try connecting a speaker striaght to any old op-amp and set the gain up high though, power amps need to be able to supply far more current than an ordinary op-amp, a 741 or 071 will probably melt if connected to an 8ohm load.

That gain can be useful though, to amplify the tiny levels from a microphone into a more useful level, however, you'll need a very low noise op-amp to do that well.
 

audioguru

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apakhira said:
(See attachment). So how is inverting op-amp increasing clipping?
The diodes across the opamp's negative feedback resistor cause severe hard clipping when the output level of the opamp reaches the conduction voltage of each diode. With a higher input level then the output is square waves with the amplitude of the forward voltage of the diodes.

If op-amps have huge gain, why don't ppl use op-amps 2 make hi-fi amplifiers?
They do, but since most opamps have an absolute max supply voltage of only 36V, the amp's output is only 16W into 8 ohms. Two of these amps could be bridged to provide about 56W into 8 ohms.

Only one opamp that I know has an absolute max supply voltage of 90V. The OPA445AP could be used to make an amplifier with an output of 100W into 8 ohms. Two of these amps could be bridged to provide about 360W into 8 ohms.
 

Nigel Goodwin

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audioguru said:
They do, but since most opamps have an absolute max supply voltage of only 36V, the amp's output is only 16W into 8 ohms. Two of these amps could be bridged to provide about 56W into 8 ohms.

Only one opamp that I know has an absolute max supply voltage of 90V. The OPA445AP could be used to make an amplifier with an output of 100W into 8 ohms. Two of these amps could be bridged to provide about 360W into 8 ohms.

You've not made it very clear Audioguru! - simple opamps WON'T provide 360W, 100W, or even 16W. They only provide very small output power, and generally only into fairly high impedances.

But you could use them, along with suitable output stages, to make high power amplifiers - but it doesn't really simplify things much, so it's not particularly common.
 

audioguru

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Nigel Goodwin said:
You've not made it very clear Audioguru! - simple opamps WON'T provide 360W, 100W, or even 16W. They only provide very small output power, and generally only into fairly high imedances.

But you could use them, along with suitable output stages, to make high power amplifiers - but it doesn't really simplify things much, so it's not particularly common.

OOps! :oops:
Of course I was assuming that the wimpy opamp would drive a high-current complimentary power darlington transistors output stage, like this simple one that uses a 44V-rated opamp. Don't ever short its output or you'll be blinded by the smoke:
 

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