What does this resistor do in a OTA Phaser?

[Chugs]

What effect on the response of the phaser does the 1.8K resistor across the positive and negative inputs of the LM13700 have? What does changing the value do? The Small Stone has 1K in the same position.

[iainpunk]

its a resistive devider, AKA a fixed volume reducer.
the OTA has a relatively high minimum gain, to account for that, the level is reduced before the gain stage.

cheers, Iain

[Mark Hammer]

its a resistive devider, AKA a fixed volume reducer.
the OTA has a relatively high minimum gain, to account for that, the level is reduced before the gain stage.

cheers, Iain

Um, did you mean to say "a relatively low maximum gain"?

[PRR]

> Um, did you mean to say "a relatively low maximum gain"?

No.

It might be more-correct to note that the maximum clean input directly at an OTA is 20mV-50mV, which is much less than a hard-strummed guitar. So we have to knock-down the signal to get clean phasing.

[iainpunk]

its a resistive devider, AKA a fixed volume reducer.
the OTA has a relatively high minimum gain, to account for that, the level is reduced before the gain stage.

cheers, Iain

Um, did you mean to say "a relatively low maximum gain"?

well, if you have an OTA stable at unity gain, and put a guitar in to it, it has a lot of distortion.
to account for that, we reduce the input signal strength and up the gain.
i have also never gotten an OTA to be stable at gains under 5x, using NFB, but that could be on me tho.

cheers, Iain

[Chugs]

So the lower the value of that resistor the more the signal is reduced at each input to less the chance of overdriving that stage?

[iainpunk]

So the lower the value of that resistor the more the signal is reduced at each input to less the chance of overdriving that stage?

yes, that's basically their function. over driven OTA's sound quite owkay so if you want a phase-fuzz, you could take it out of the last phase stage. bit its an odd/weird/hard to use fuzz sound due to the phaser action going on

cheers, Iain

[Chugs]

Thanks! Makes sense.

Different question, same phaser: R27 appears to be reducing the value of the Regen pot from 500k to approx 25k. Is that the case?

If so, could R27 be removed and the Regen pot replaced with a 25KC pot for the same range of Regen?

[iainpunk]

no, not really, when the pot is at full regen, there is a 500k resistance to ground, not 25k.
i dont think it matters to much, but in my experience, 500k is cheaper than 25k, since the local guitar store sells 250k and 500k for really good prices (2euro for Alpha pots!), but for other values, i need to go to the (way more expensive) electronics store in the inner city. if you order on line, i think the prices are very similar.

cheers, Iain

[Chugs]

Yes, I am guessing they originally use 500KC as that is the same value as the Rate control.

If I am seeing this correctly: The Regen pot controls how much signal is fed back into IC2b through the 1uf cap and 27k resistor. It does this by acting as a voltage divider, dumping signal to ground. As such, presumably other value pots would not alter the Regen response very much if at all? 

[iainpunk]

Yes, I am guessing they originally use 500KC as that is the same value as the Rate control.

If I am seeing this correctly: The Regen pot controls how much single is fed back into IC2b through the 1uf cap and 27k resistor. It does this by acting as a voltage divider, dumping signal to ground. As such, presumably other value pots would not alter the Regen response very much if at all?

i'm like 95% sure you are correct.

cheers, Iain

[ElectricDruid]

> Um, did you mean to say "a relatively low maximum gain"?

No.

It might be more-correct to note that the maximum clean input directly at an OTA is 20mV-50mV, which is much less than a hard-strummed guitar. So we have to knock-down the signal to get clean phasing.

+1 agree with this. The problem is not the gain of the OTA, but the input level that it can stand. If you look at the datasheet for the LM13700 (for example) you'll see that distortion goes up to somewhere between 1% and 10% when the signal is only 100mV (depending on whether you use the linearisation diodes or not:


If you want anything that's going to sound "clean", you need to be keeping the input signal down below 50mVpp or so. Hence the heavy input dividers.

[PRR]

....... input signal down below 50mVpp or so. .....

OK, say 45mV p-p which is 15mV RMS. Divider is 16:1. This accepts 240mV RMS input cleanly. While gitar "can" do more than a quarter Volt, most of the time not, and the player may choose to play less violently for clean (or stronger for BJT-pair distortion).

[ElectricDruid]

....... input signal down below 50mVpp or so. .....

OK, say 45mV p-p which is 15mV RMS. Divider is 16:1. This accepts 240mV RMS input cleanly. While gitar "can" do more than a quarter Volt, most of the time not, and the player may choose to play less violently for clean (or stronger for BJT-pair distortion).

Agree. You choose what amount of distortion counts as "acceptable" or "desirable" and you adjust to taste. Clearly the guitar you play into the front end of this makes a huge difference since signal levels out of guitars vary so widely. I generally use my friend's basic Strat as a "typical" guitar (after all, what's more typical that a fairly cheap/half-decent Strat copy?!). I got that guitar on the oscilloscope and it put out between 100mVpp and 1Vpp depending on where you played and whether you were blasting big chords or plucking single strings.
1Vpp down to 50mVpp implies 20:1 divider, so the 16:1 is fine if you can cope with a little distortion when you dig in (probably most players can, and even *expect* that).
But like I said...the designer can choose their own level. There isn't an objectively "best" level of distortion. It simply depends what sound you're after and what your aims are. That's why this is such an interesting game to play. If it was pure engineering, it wouldn't be half so much fun. As it is, the engineering follows the art, which for me is entirely the way round I'd want it to be!

[merlinb]

If you look at the datasheet for the LM13700 (for example) you'll see that distortion goes up to somewhere between 1% and 10% when the signal is only 100mV

FYI, 10% distortion on a guitar signal is basically inaudible. At most, the guitarist will think it is adding 'color' or slightly 'fattening' the tone; it doesn't sound like the distortion knob on your dirt box. 100mVpp is fine in a guitar OTA circuit.

[ElectricDruid]

If you look at the datasheet for the LM13700 (for example) you'll see that distortion goes up to somewhere between 1% and 10% when the signal is only 100mV

FYI, 10% distortion on a guitar signal is basically inaudible. At most, the guitarist will think it is adding 'color' or slightly 'fattening' the tone; it doesn't sound like the distortion knob on your dirt box. 100mVpp is fine in a guitar OTA circuit.

Fair enough. Thanks!

100mVpp still isn't exactly a huge level, which is why OTA circuits tend to get noisy. Noise at a few mV becomes a significant percentage of the signal pretty easily.

[POTL]

Hi very useful information about the resistor as old circuits are misleading. what about calculating notch filters? if I understand the circuit correctly, then a pair of 27k resistors sets the gain = 1, like a pair of resistors in other phasers on optics and transistors. Apparently 27k + 6.8n sets the cutoff frequency, but how do you calculate the changes caused by the operation of the LFO? If 27k / 1.2k is a voltage divider, then we are losing more than 20dB, 27k + 27k sets unity, where do we return the gain after the divider?

[DrAlx]

Hi very useful information about the resistor as old circuits are misleading. what about calculating notch filters? if I understand the circuit correctly, then a pair of 27k resistors sets the gain = 1, like a pair of resistors in other phasers on optics and transistors. Apparently 27k + 6.8n sets the cutoff frequency, but how do you calculate the changes caused by the operation of the LFO? If 27k / 1.2k is a voltage divider, then we are losing more than 20dB, 27k + 27k sets unity, where do we return the gain after the divider?

If you take a regular opamp based all-pass filter (APF) stage there are sort of two parts to it. A "unity inverting" part (from 2 equal resistors at the inverting input), and a "frequency dependent part" (a simple LPF or HPF) at the non-inverting input. These parts act together and make the overall gain 1 (or minus 1) at DC depending on whether you use LPF (or HPF) in the frequency dependent part.
Many FET based phasers will have an HPF as the frequency dependent part and so tend to invert (i.e. gain = -1) low frequency signals and let through high frequency ones without inversion (i.e. gain = +1) . You can think of that as 180 degrees phase shift for low frequencies, and zero phase shift for high frequencies.  The exact phase shift for a frequency depends on the RC time constant of the HPF. An op-amp based phaser will typically vary R somehow using an LFO.

With this OTA based APF there are similar parts. A pair of equal 27k resistors set inverting unity gain. But the frequency dependent part (the RC) no longer has a resistor R. Instead of R you have 1/g where g is the transconductance of the OTA (which is determined by the control current supplied from the LFO).   Don't think of the voltage divider at the OTA input as affecting the gain of the "unity inverting part". It is more to do with making sure the OTA behaves nicely.
For example, imagine you have high frequency audio input. The capacitor in the APF then acts like a short circuit, so the input goes straight to the Darlington buffer, and so to the output.  So for high frequencies, the gain is 1 and phase shift close to zero. So no signal loss due to the voltage divider at the input.