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Serious Blocking Distortion

Started by Littlewyan, July 05, 2013, 10:54:13 AM

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Roly

Well yes, and no.

The amplitude and phase of a signal are two different properties of the same signal, not two different signals.  {I will now attempt the impossible, explaining AC theory in less than 100 words with no algebra, vectors, or Complex Numbers}.

Consider this circuit;


Initially the capacitor is uncharged and the switch is open.  At time t=0 the switch is closed.

Initially there is no voltage, Vc, on the capacitor so a current, i, flows according to Ohms law, 100 volts and 2k gives 50mA, but as soon as the current starts flowing the cap starts to charge up, the voltage across it rises so the voltage across the resistor, Vr, falls, and therefore so does the current.

Compared to the input, the battery, the voltage rise on the cap takes place delayed from the closing of the switch.  In the DC case the voltage on the cap rises ever more slowly as it comes up to the battery supply, but with an AC sinewave signal instead of the battery the supply starts to reverse before the cap can fully charge up, so it has to discharge again to follow the incoming sine wave, and so is continually chasing the incoming voltage, is effectively delayed from it.

Similarly, because the input voltage is now continuously changing and the voltage on the cap can't catch up, its peak value or amplitude is also smaller compared to the applied signal.

So in terms of the applied AC sinewave the voltage on the cap is also a sinewave but it is both smaller in amplitude and delayed in time, or as we say, phase.

Since a sinewave can be represented as 360 degrees over one whole cycle we express the phase of the output across the cap in terms of the angle between where the input, E, crosses zero and the output, Vc, crosses zero as degrees of phase delay.

With the resistor in series and the capacitor in shunt across the signal path we have a low pass network, and with only a single C and R it's called a first order or single pole network.



If we exchange the positions of C and R we create a first order high pass network, and I will ask you to accept on trust that similar reasoning leads us to the fact that the output signal is phase advanced on the input.



In both the low and high pass cases the attenuation or loss of amplitude from input to output is related to the reactance or effective AC resistance of the capacitor, Xc ohms, which changes with frequency according to;

Xc = 1 / (2 Pi f C)

A rule of thumb to remember is that when the reactance of a capacitor is equal to the resistance in a CR network the amplitude response is half, and the phase angle is 45 degrees.  We normally take the values ten times above and below to be where the phase angle falls to zero or goes to the maximum 90 degrees; similarly where the output amplitude is effectively the same as the input, or effectively zero.


There are a couple of handy graphs for estimating amplitude and phase response here under "Response estimation".
If you say theory and practice don't agree you haven't applied enough theory.

Kaz Kylheku

Quote from: Littlewyan on August 15, 2013, 02:24:21 AM
Haha Roly :P

Ok i've done some reading and am i right in saying the Phase Shift is basically a change in the audio signal? Like a delay perhaps? And that its for mixing two signals together? Which for a guitar would be the original note that was picked and then the harmonics that come after?

Every sinusoidal has phase. Frequency and amplitude are not enough to completely specify a sine wave. You have to know the phase. At a given point in time, a sine wave could start on a crest, or in a trough, or anywhere in between.

A phase shift is not exactly a time delay. Whereas, of course, a time delay will shift the phase, circuit components like amplifiers and filters do not introduce delays (at least not delays that matter in audio): the signals move at close to the speed of light through the circuit.   An amplifier's or filter's phase shift is a frequency-dependent response.  It is not caused by a time delay, but by a response lag (or its opposite, "eagerness", or lead).

For instance, a low-pass RC filter has lag.  Lag does not mean that there is a time delay. When we send, say, a step signal into the circuit's input, the output begins to react instantly, but it does not change instantly: it moves slowly, because the capacitor has to charge through the resistor to achieve the new voltage. It cannot follow the step.  This lag is why the circuit is a low-pass filter; higher frequencies shake back and forth more quickly, and so the lag suppresses them more than it suppresses slow oscillations.

If we look at your graph, we see that you have a positive phase shift over most of the range, which goes hand in hand with the high-pass-filtering that is going on.  In a high pass filter, the phase shift goes the other way: it leads the original signal. This is because (in the case of a simple RC high pass filter) the output is proportional to the current passing through the capacitor. Suppose we send a step signal into a such an RC filter. The coupling capacitor begins to charge instantly, and so there is an immediate, fast voltage spike on the shunt resistor. The spike then dies down as the current diminishes and ceases. The eagerness of the high pass filter to respond causes it to produce a leading phase shift over sinusoidal signals.

Phase shifts are important in audio, and at the same time they are unimportant.  If you have a stereo (or surround) system, and the signal is inverted in one of the channels, this is bad; it means that every frequency is shifted by 180 degrees. When you sit in the ideal listening position, you will get strange sounding cancelations and comb filtering.   Phase shifts affect bass.  A 90 degree phase shift in a 30 Hz signal represents a time shift. It can throw off  a kick drum in a perceptible way.

Phase shifts are not audible in the higher frequencies (except allegedly in some contrived listening tests with specially crafted material).

All filters have some kind of phase response. For example, the narrow filters in a 31 band graphic equalizer seriously affect the phase coherence of material. Yet, audio engineers and musicians use them anyway.

Phase is the subject of a lot of audiophoolery. Some claim they can hear several degrees of phase shift in a 10,000 Hz signal, and the like, which is nonsense. Snake oil devices exist which supposedly reconstruct the phase coherence of a signal.

Phase is crucially important in the stability of amplifiers which use negative feedback.  All amplifiers have a limited frequency response, and due to multiple parasitic filter poles in their stages, at some frequency they accumulate a 180 degree phase shift. Negative feedback is also 180 degrees, and so this phase shift turns negative feedback into positive feedback.  If the amplifier has a gain of 1 (or, practically speaking, at least one) at this frequency, then it will oscillate.

Phase can tell you things about the response of your circuit. For instance if we look at your graph, the amplitude response looks fairly flat toward the right. Only if we look a little closer do we see that it actually is curved: it reaches a maximum and starts to fall. So there is a frequency response peak hiding there. Where is that peak? The phase response provides a clue: right around the frequency response peak, the phase response drops down to zero degrees, and crosses into the negative (lead turns to lag). That's about the place where high pass switches to low pass.  It is clearer in the phase shift plot; the phase shift clarifies and confirms something to us.




   
   
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Littlewyan

Ah ok thanks guys I think I understand it a bit more now :)

With regards to the amp I have changed all coupling capacitors to 1uF which has improved the sound greatly, I'm sure I can hear a bit more clarity in the distortion now (weird saying I know). I also changed the input capacitor from 100n to 4700pF as JM advised but I may change it back as the sound is a bit trebly for my liking now. However I have also changed the JFET Source Resistor to 5.6K from 4.7K which has lowered the distortion a bit but it is still a tad high, so I'm going to try swapping it for a 6K or even a 6.6K as I get distortion even at low volume! So I'm thinking once I've lowered the distortion the amp may not have as much treble, but we shall see, one step at a time.

As always I've attached an updated schematic, I noticed an error in the schematic where I included an extra capacitor that isn't present in the circuit, apologies.

Littlewyan

I've now changed the Source Resistor to a 6K2 value from 5K6, however I haven't been able to test the amp at high volume (parents and neighbours get annoyed!). I'm hoping to get the amp to a point where I have a clean sound at low volume and distorted at high volume. This should be possible by increasing the Source Resistor shouldn't it? I'm sure I can do that on the 1 watt amp I built earlier this year.

Littlewyan

The other thing is it doesnt really clean up with the guitar volume knob, i'm guessing this is because of the high gain?

Roly

Check the voltages on the drain of the FET and the collector of the transistor.  The operating sweet point ("Q point") we are looking for is that while idle the voltages on these two points should be about half the supply voltage (so that the signal has equal headroom for both +ve and -ve swing).  The ideal point is actually with equal voltages across the device and the load resistor to account for the supply voltage lost across the bias resistor, so a little high on half should be about right.
If you say theory and practice don't agree you haven't applied enough theory.

Littlewyan

The JFET is currently at 19volts and the BJT is at 13 volts. If I change the BJT Collector Resistor from 10K to 6K8 then the collector voltage goes up to 15.7volts which is about right. So I'll just try that first, if it still doesn't sound right then I'll change the JFET resistor from 22K to 27K which will put it at 17 volts. Does this sound ok Roly? If I need a higher value resistor on the JFET then I can get hold of one.

Roly

#127
In general when we want to alter the operating, bias, or Q-point of a transistor (or indeed valve) stage the resistance we vary is normally the emitter/source/cathode resistor.

The resistor that is in the collector/drain/anode is called the "load" resistor and normally remains unchanged.  The reason is that the load changes the Q-point alright, but it also changes the stage AC signal gain. Because the emitter/source/cathode resistor is normally bypassed for AC using a capacitor changing this resistor only changes the DC gain, and thus the bias point without changing the AC gain.

Because your FET stage is the more conventional I'll use that as the example (since the simple bias method on the transistor stage has a few gotchas which complicate matters a bit).

Looking at the datasheet for the 2N3819;

http://www.digchip.com/datasheets/parts/datasheet/343/2N3819-pdf.php

...we find that the VGS(off) (Gate-Source Voltage for cutoff) is given as -8 volts, and as we want a linear Class-A stage let us assume our bias target point is half this, or -4 volts.

We will assume an available supply of 30 volts, and will decide (guess) that a suitable device current will be around 1mA.

Your circuit has a 4k7 Source resistor, and 1mA through a 4k7 will give 4.7 volts, pretty close.

Your load resistor is 22k and 1mA through 22k will give a drop of 22 volts.

If we subtract the load resistor drop from the available supply we get an idle Drain voltage of;

30-22 = 8 volts.

This is a bit low on the desired half-supply for best unclipped voltage swing on the Drain, it will bottom out pretty quickly because the FET is already elevated by the Source bias resistor to around 4 volts, so the available negative swing is only 8-4 = 4 volts.

We could reduce the Drain resistor but this would also reduce the stage gain, which we generally don't want to do, but what we can do is increase the bias so that the device draws less current at the idle Q-point.

As a rough calculation we can now find a target Drain voltage of the supply less the bias voltage, 30-4 = 26 volts, divided by 2 to put it in the middle, or 13 volts.

13 volts from the supply is 30-13 = 17 volts drop across the Drain resistor, and the current that will give that is;

I = E/R

17/22 = 0.77272727 (k ohms/volts = mA) or around 0.8mA.

Now we can go back to the Source resistor and work out what value will give the -4 volts bias with this new current;

R = E/I

4/0.8 = 5(k)

So we only need a small increase in the bias resistor to push the Drain voltage up to the desired point and could try changing the 4k7 for a 5k6 (or we could try adding a 330r in series with our 4k7).

Keep in mind that all these devices have a parameter spread, and that FET's in particular often have a very wide spread in the required bias voltage for a given current.

Now if we want to play around with the AC gain of the FET stage we can change the Drain load, increasing for more gain and reducing for less, then resetting the mid point DC conditions using the Source resistor.

We need to keep in mind however that an upper limit is set by the fact that the input impedance of the following stage effectively (for AC) appears to the FET to be in parallel with its Drain load - you can't just make the Drain 10 Megs and get a monumental gain, because the input impedance of the following stage is likely to be somewhere between 100k and 1Meg and this will be the effective upper limit.

This exercise shows that the FET stage isn't too far off the correct bias point ('tho it may be a fair bit lower at around -2.5 volts in reality), and a bit of fiddling in LTSpice shows that this stage doesn't really seem to be much of a problem.  It shows a voltage gain of around x24 which is actually a bit on the high side.  We normally assume that a guitar pickup will output around 200-500mV average with peaks to around a volt, so we would normally be looking for a gain where the stage could cope with that without clipping.  In this particular case reducing the Drain load somewhat could help in both centering the stage Q-point and reducing its gain a bit.

Something that looks good is that this FET stage starts to limit in a fairly nice way, softly at first, and by flattening out the positive peaks a bit.  This would give Hi-Fi-ists the galloping jeebies, but for guitar this sort of distortion gives a "nice" bit of second harmonic, a bit valve-like.

Taking out the bypass cap across the FET Source resistor drops the gain to about x11, but still leaves the following transistor stage clipping badly at 1V in; in fact it still clips at around 500mV in.

Something that doesn't look good is that the transistor stage is hard clipping the negative peak early, and need some attention.  This, I think, is the basic cause of your 'distortion at all levels' problem.

At the moment the transistor stage isn't doing anything useful, just distorting early.  I suggest that you re-arrange this stage to be an emitter-follower directly coupled to the FET Source Drain for bias, then at least it will be performing the function of a buffer/impedance transformer to isolate the FET load from the following low impedance tone and volume controls.

Some detail on transistor biasing;
http://www.electronics-tutorials.ws/blog/biasing-transistor-tutorial.html

HTH
If you say theory and practice don't agree you haven't applied enough theory.

Littlewyan

#128
Thanks for that Roly. I've attached a screenshot of what I've put together in LTSpice, these are the changes I will try I have made tonight based on your recommendations. Only thing I didn't do was connect the BJT to the JFET's Source as this seemed a bit bizarre as I won't have any voltage gain at all if I do that? Or am I misinterpreted?

Anyway the amp sounds much better now and cleans up perfectly when rolling back the guitar volume. Only issue now is the tone control doesn't work, as I should have suspected really as its in parallel with the BJT's Emitter Resistor. Going to look into how to get around this another night.

Just had a thought, I could change the Tone Pot configuration to be the same as the volume control, so the middle wiper goes to the output, one side connects to the input and the other then to the capacitor that goes to ground. Would that not work as a tone control?

Roly

Cleaner?  Good.  :tu:

Oops, yes, no, I dun goofed - Drain, not Source.  :duh

aaaaahhhh ... eeeerrrr ... uummmm

A simple top cut control should still work.

Try increasing the emitter resistor, say to around 15k.  This will reduce the current through the transistor (which doesn't need to be that high) and increase the output impedance of the emitter follower a bit which will then give your top cut something to work against.

Or try a Big Muff type single knob control.   8|
If you say theory and practice don't agree you haven't applied enough theory.

Littlewyan

#130
I've got an 18K resistor so I'll install that, that should drop the current down by at least 3ma. Although going by LTSpice this won't make much difference to the tone control, however that Big Muff tone control looks very interesting so I may give that a go :)

Littlewyan

Used the Big Muff Tone Control in LTSpice on my amp and the output went from 5V +/- to 900mV +/-. I'm guessing I'd need another gain stage for this tone control?

stormbringer

You will Always lose signal level with a passive tone stack. So either add a recovery stage, or increase the level Before the tone stack.

Littlewyan

#133
Well the good thing about my old Tone Control was that I didn't lose much signal at all, not compared to this one anyway. When you said Tone Cut Roly did you mean the Tone Control I'm currently using or a different one?

O thats not right, I've lost it  :P

Actually I just checked using LTSpice and whereas the Tone Control did lower the output it would only go from 5v +/- down to 3v +/- which obviously isn't a massive drop compared to the Big Muff Tone Control. Back to the drawing board me thinks.....

I'm wondering if perhaps I should just lose the Tone Control and maybe add a different option in. Perhaps a switch to go from Clean to Distortion? Could switch between Source Bypass capacitors, from 0.68uF for distorted to 0.1uF for clean perhaps?

Roly

The simple cap-in-series-with-pot is called a TOP cut control because that's what it does, it shunts higher frequencies to ground.

Currently your signal level isn't being dropped by your top-cut control, and your control isn't tone controlling; so yep, that's consistent.  The reason it isn't dropping much is because the output impedance of the emitter follower stage is very low, particularly compared to the impedance of your top-cut control, so you could try inserting some series resistance after the emitter-follower before the tone control to give the control something to "work against".  Values to try are between one tenth and one half of the tone pot value; and yes, this will also lead to some overall signal loss.

As @Stormbringer says, the only thing a tone control can do is drop the level; "boost" is a misnomer since "boost" actually means "less cut" than the neutral or flat position.

You can play around with the component values in the Big Muff (or any control for that matter) so that you get less "insertion loss" than 5V down to 0.9V, but at the expense of available "boost".  Or as suggested you could follow up the tonestack with a recovery amp, another transistor or FET stage.
If you say theory and practice don't agree you haven't applied enough theory.