OTA phasers Analysis.

[POTL]

Hello everyone I thought about disassembling the diagram in parts and making it understandable, I understood most of the diagram, but there are still questions that I will need help with solving. As an example, I took the EHX Small Stone schematic, photo is shown below.


1) The input of the circuit is buffered, according to the Datasheet, the OTA input impedance is very low, only 26 kOhm, therefore buffering before each stage is necessary.
2) Between each OTA stage and at the output of the circuit, we see buffers, they are necessary, since the output impedance of the OTA is very high 15MΩ
3) OTA is capable of accepting only a small signal, until distortion appears, usually no more than 50mV, at the input of each stage we see a voltage divider of 27kOhm / 1kOhm, which reduces the 1Vpp input signal to 35mVpp
4) A pair of 27kΩ resistors set the Gain = 1
5) Notch 1 is calculated by the formula tan (45/2) / (2Pi * C * R) Notch 2 is calculated by the formula tan (135/2) / (2Pi * C * R) Where C = 6.8n All R is calculated by gm I could not calculate gm, however according to this article (link below), we can calculate the R values ​​they are 81.5k for the minimum notch range and 13.5k for the maximum notch range.
http://www.dedalofx.com.ar/bioroids/experiments/e03_smallstone.php
6) Pins 1 and 16 are current driven The maximum current that the circuit can accept is 2mA (in real circuits, 1mA is often used). A resistor is selected based on the maximum voltage of the applied LFO. R = U / I = 9V / 2mA = 4.5kΩ, as an example You can use 1 resistor for all OTA stages, can be used for each stage separately. I'm not sure about the calculation for each stage individually.
7) The diode bias inputs (2 and 15) are usually not used, but they also accept no more than 2 mA.

And now there are 2 things that are not disclosed.
1) Correct calculation for filters, how to make it.
2) Individual resistor for each stage OTA = R / 4 (number of OTA)?
3) How many times on the forum I came across information that new OTAs can receive a signal from 100mV to almost 1V without distortion, is there any confirmation? On the one hand, the datasheet does not indicate any differences from the CA3080, on the other hand, there are many reports about the improved characteristics of the LM13700, it is obvious that a decrease in the voltage divider would be great news, because the signal-to-noise ratio will be much better. What is the real headroom for classic 9V power? What is for modern 15-18 volts?

[ElectricDruid]

A quarter of the current goes to each OTA, and you know that the current for each one can't be more than 2mA, so the total current can't be more than 8mA.

But like you said, you could assume 1mA maximum for each OTA, and then size the resistors for that given the LFO voltage, while remembering that the LFO has to provide four times that amount of current (but it's buffered by a darlington pair so it should be fine).

[POTL]

Thank you what about the calculation? I am not trying to figure out how to correctly calculate the filter, I understand that a capacitor and gm are involved in the filter part (or iabc, I'm still confused about this chip). what about the input voltage, the lm13700 is actually capable of accepting almost 1Vpp?

[StephenGiles]

I'm not sure exactly what you are trying to achieve here??

[POTL]

I'm not sure exactly what you are trying to achieve here??

I want to understand how OTA works in this scheme and how to calculate everything correctly.

[ElectricDruid]

Thank you what about the calculation? I am not trying to figure out how to correctly calculate the filter, I understand that a capacitor and gm are involved in the filter part (or iabc, I'm still confused about this chip).

Iabc controls the gain and hence the filter cutoff. In my understanding Gm is the transconductance, the relation between the input current and the output gain. This involves all kinds of physical constant craziness and temperature dependence, but the simplification is that at 25C:

gm ≈ 19.2 x Iabc

(there's a bit more about this on my page: https://electricdruid.net/design-a-eurorack-vintage-vca-with-the-lm13700/ )

what about the input voltage, the lm13700 is actually capable of accepting almost 1Vpp?

You need the graph from the datasheet:


This shows you that you can input somewhere between 150mV and 200mVpp and you'll get 10% distortion. Above that, they don't say, but we can assume it gets even worse. 1Vpp is much too loud for an OTA, but of course, you've got that input divider, so you can adjust the values of that to suit whatever input level you like.

HTH

[StephenGiles]

The OTAs used in that version of the Small Stone were CA3094s surely?

[ElectricDruid]

The OTAs used in that version of the Small Stone were CA3094s surely?

Yeah, but same difference. I haven't got a handy datasheet for that one.

[DrAlx]

If you compare one of those OTA all-pass stages to an op-amp based all-pass stage that uses an LDR as the variable resistive element, then the equivalent variable resistance would be given by

( 1 / gm ) * ( 2 + R/r )

where R and r are the scale down resistors at the OTA input (i.e. 27k and 1k for those middle OTA stages).
So as far as notch location goes, you need to account not only for the capacitance C (6.8 nF), and Iabc (which determines gm as ElectricDruid says) but also the scale down ratio used to minimise distortion.

That first all pass stage is a bit odd because of the mismatch between 30k at input and 27k feedback resistor. Apart from giving less than unity gain, it also puts the pole and zero of that first AP stage at different distances from the origin.  Haven't worked out how much it affects notch location, but the formula I give above for equivalent variable resistance would only be approximate in that case as I assumed matching input and feedback resistors like in the other AP stages.

Regardless of where the notches appear, their frequency locations will always scale in proportion to Iabc. e.g. double Iabc and the notch frequencies will all double.

[ElectricDruid]

Regardless of where the notches appear, their frequency locations will always scale in proportion to Iabc. e.g. double Iabc and the notch frequencies will all double.

This is a good point - the modulation is linear, e.g. V/Hz. This is pretty typical for phasers, but it doesn't have to be that way. It affects the LFO waveform choice because it takes proportionally longer to move the notches in higher octaves. This is why some people swear by "hypertriangular" waveforms (U shaped, roughly) since they spend more time at the bottom and less at the top and compensate for the linear modulation. Another way to think about this is that the hypertriangular waveform is pre-distorted to allow for the linear modulation.

[POTL]

thanks for the answers, I will look at them and try to digest the information. this nkdklya was very busy and I could not fully get to the forum.