how and why power amps clip

[hpupo]

Hello...

I am in need of some explanation. I understand that guitar tube amps may have their power amps designed to clip. I do understand that it will clip if one require the power amp to put out more power than the power supply can provide. But what exactly does "more power than the power supply can supply" means? Does it mean that current will continue to grow and voltage will drop (P = V*I)? Or does it mean it will reach a point which neither current or voltage can be increased?
I have more questions about this issue, but one thing at a time...

thanks

 

[sakellogg]

maybe im wrong... but i thought they distorted not clipped. distortion from clipping is diffrent, right?

 

[sofaspud]

Consider this simplified example: Say you have an amp that is powered by 12 volts. And it is fed a signal of 2 volts that will be amplified 8 times. That is 16 volts. The output will go to 12 volts and remain there for a time (determined by the wavelength of the signal) because it can't go any higher. Neither current nor voltage can be increased above that point. The power out cannot exceed the power in. So it clips.

All clipping is distortion, but not all distortion is clipping.

 

[TheGimp]

If you look at the final output stage in a push pull amp, there is a center tapped transformer connected to two output tubes (or multiples thereof), and the center tap to B+.

For simplification assume that the output voltage is 400V, and the voltage gain of the output stage is 40. Therefore an input signal of 10V peak sinewave (20V p-p) will cause one of the output tubes to shut off as it approaches 400V and the opposite tube to saturate as it apporaches 0V. This is as much as the stage can put out. Any increase in the input voltage will casue the output to reach it's limit earlier in the waveform phase and come out of max later in the waveform phase.

This is clipping. The top and bottom of the waveform as viewed on an oscilloscope looks like they were cut off or clipped.

 

[TheGimp]

Clipping produces distortion as the clipped waveform is a series of frequencies combined together (fourier series). Second harmonic is produced first, then third, fourth, fifth, etc. The percent of each harmonic is dependent on the tube, circuit topology, and other factors. These are harmonic distortion.

Clipping also introduces intermodulation distortion (TIM) where the different frequencies modulate each other.

Phase distortion may be introduced as well.

As the tube clips, it will not shut off/saturate instantaniously, but rather gradually transition. This gradual transition can be represented by a series of even harmonics, as oposed to a square wave which is a series of odd harmonics.

In addition, the non-linearity (output current vs input voltage) of the tube increases as one approaches cutoff/saturation. This non linearity introduces predominantly second order harmonics.

 

[hpupo]

Thanks for the replies. So correct if im wrong, but can i still think of a power amp as a voltage amplifier???

 

[sofaspud]

Not necessarily. A power amp is a power amp. P=I*V, so you can have a voltage amplifier, a current amplifier, or both. With audio, I'd say it's usually both. But all three amplify the power.

 

[TheGimp]

In order to make power, one has to have both current and voltage. The input stages of the tube amplifier are predoiminantly voltage amplifier stages. There may be current gain in latter stages, but early stages are typically 12AX7s which are used to progressivly increase the voltage of the guitar output (up to 400mV?) up to the drive level needed by the grid of the output tubes (20Vp-p to 70vp-p typically).

The output tubes because of their greater input capacitance will require more grid drive current than smaller voltage gain tubes so current gain is required in the tubes driving the grids of the output tubes.

Furthermore, the output tubes are used for current gain as they drive inductive transformers. Voltage without current could not transfer power to drive the speaker (and would imply very high internal impedance of the output tubes) .

 

[hpupo]

i think i got. Now to something else i do not understand. I learned from my studies on power amps, that because of the transformer (i still do not fully comprehend this effect) the voltage will rise to more than the supply voltage (something to do with collapsing of the magnetic field in the transformer). My questions is, if the transformer allows the voltage to go higher, how can it clip? Or is plate current that is being clipped in this case?

thanks

 

[sofaspud]

A transformer has limits for voltage, current, and frequency same as (virtually, if not literally) all other components. It can step up the power supply voltage, true, but it is still vulnerable to the limits described in the previous responses. The transformer voltage output is directly proportional to the voltage input.

 

[TheGimp]

In some limited cases, the voltage across an inductor can reach twice the supply voltage. In a single ended amplifier stage, as long as the load impedance stays uniform, one can drive an inductive load in such a manner that once one reaches the limit of the plate supply and current starts to collapse, the inductor voltage continues to increase. This is not true for a push pull output circuit as the voltage across the primary can be seen as twice the supply voltage. This is the result of one side is going positive while the opposite side is heading to 0.

The exception to this is removal of the load from the secondary of the transformer. In this case the high impedance seen buy the output tubes will result in very rapid di/dt and several thousand volts across the primary. The unloaded output is a major failure mode in tube outputs.

 

[hpupo]

Ok, but again, considering what TheGimp just explained (thanks by the way), and if i understood it correctly, if the plate voltage goes higher than PS voltage, how can it clip?

 

[quikie22]

1) clipping distortion need not necessarily occur at the output (with a loud screech), it can be occur at the pre-stage, or at the first drive stage... and then the distorted signal fed to the output stage.. voila you still get clipping distortion

2) True the voltage can go higher than the plate with an inductive load. But will it have a linear relationship to the signal? eg at low levels, assuming a gain of 8x, maybe at 1V, you get 8V out, but at 2V, you can get a max signal of 12V due to limitations of the output stage... Clipping is when the input and output gain relationship becomes non-linear due to either insufficient current or voltage that can be supplied by the output stage.

 

[cbdb]

Audio power amps are voltage amps, they take the input voltage and multiply it an exact amount, like 30 times. They also put out current but depends on the load. (if you have no load, you get no current)

The transformer can multiply/divide the voltage but the current changes to keep the output power the same as the input. This dosnt prevent clipping.

Clipping in amps (or any circuit ) is caused by running out of voltage. If your amps rails are +/- 30 volts, thats the max voltage it can put out. The amp will clip at around 30 volts. (the voltage rises to 30 then flattens out.) The harder the clipping (the flatter the voltage) the harsher the sound (more distortion). Clipping distortion is usually unpleasant so guit amps are designed to distort in other ways.

The intentional distortion in guitar amps is a combination of other distortions that can be tailored to make the amp sound the way you want (a Marshal sounds different than a Fender). And adjustable so you can change the sound of a specific amp.

There was a great paper posted somewhere here at diyaudio on designing a tube guitar amp that explained the different ways to get a tube amp stage to distort. You will have to search for it.

 

[TheGimp]

Merlin Blencowe has a nice site, and even better books that do a good job of explaining both tube theory and guitar amp theory.

http://www.freewebs.com/valvewizard/

 

[hpupo]

thanks again, but so far i believe no one was able to explain (or i was not able to understand) how you can clip the output stage (let use a SE for simplicity) being that the inductive load with allow it to go higher the PS voltage. Let's use the following example (based on a real fact): if i plot a load line (assuming a fixed load impedance, which i know it is not real) for a EL84 tube and set the bias point midway the load line and plot the signal gain (sin wave, with fixed frequency), one will see the plate voltage can swing to more than the power supply voltage. Everywhere i read, it was said this was due to current collapsing on the inductive load. But it such stage was to clip, where it would clip, since the plate voltage has already surpassed the PS voltage?

 

[quikie22]

thanks again, but so far i believe no one was able to explain (or i was not able to understand) how you can clip the output stage (let use a SE for simplicity) being that the inductive load with allow it to go higher the PS voltage. Let's use the following example (based on a real fact): if i plot a load line (assuming a fixed load impedance, which i know it is not real) for a EL84 tube and set the bias point midway the load line and plot the signal gain (sin wave, with fixed frequency), one will see the plate voltage can swing to more than the power supply voltage. Everywhere i read, it was said this was due to current collapsing on the inductive load. But it such stage was to clip, where it would clip, since the plate voltage has already surpassed the PS voltage?

yes.. agreed that the plate voltage has surpassed, but will the voltage be in linear relationship to the signal!! Clipping is non-linear relationship.

 

[jjman]

I see my guitar amp output stages clip if/as/when the grid reaches zero volts during the signal's swing.

 

[wakibaki]

thanks again, but so far i believe no one was able to explain (or i was not able to understand) how you can clip the output stage (let use a SE for simplicity) being that the inductive load with allow it to go higher the PS voltage. Let's use the following example (based on a real fact): if i plot a load line (assuming a fixed load impedance, which i know it is not real) for a EL84 tube and set the bias point midway the load line and plot the signal gain (sin wave, with fixed frequency), one will see the plate voltage can swing to more than the power supply voltage. Everywhere i read, it was said this was due to current collapsing on the inductive load. But it such stage was to clip, where it would clip, since the plate voltage has already surpassed the PS voltage?

It will clip at the point where the input voltage times the stage gain exceeds twice the PSU voltage. The situation is no different than if there was no additional voltage available due to the collapsing field in the inductor (transformer) other than that the limiting output voltage is twice what it would otherwise be. In fact twice is an idealised output, it won't reach twice the PSU voltage in practise.

Even if the stage gain is <1, as in the case of a cathode follower where the gain is close to unity, the stage can be made to clip if the voltage swing applied to the grid is sufficient, i.e. (in that case) greater than the supply voltage to the overdriven stage. Bear in mind as well that a stage may not have a symmetrical swing available at the output. In your example it is biassed to the midpoint, but this may not be the case, causing it to clip on the positive- or negative-going side before the other side.

w

 

[Printer2]

thanks again, but so far i believe no one was able to explain (or i was not able to understand) how you can clip the output stage (let use a SE for simplicity) being that the inductive load with allow it to go higher the PS voltage. Let's use the following example (based on a real fact): if i plot a load line (assuming a fixed load impedance, which i know it is not real) for a EL84 tube and set the bias point midway the load line and plot the signal gain (sin wave, with fixed frequency), one will see the plate voltage can swing to more than the power supply voltage. Everywhere i read, it was said this was due to current collapsing on the inductive load. But it such stage was to clip, where it would clip, since the plate voltage has already surpassed the PS voltage?

Just to lead things in a certain direction I will ask a few questions. The magnetic field is collapsing. Why? What is the tube doing to cause the magnetic field to collapse? Does the increase in voltage help or hinder the output tube? Does the collapse cause power to flow out to the load or into the tube?