need to make a Sawtooth VCO with variable gain output?

MichaelWeaser

Joined Aug 25, 2021
4
I need to figure out how to make a sawtooth VCO that the output matches up to this graph. I only need frequencies up to around 225 hz .

The only idea I have for right now, is using a regular linear output VCO , but add a voltage controlled amplifier , so whatever voltage is set using the vco, use the same voltage amount to set the gain of the VCA to make it the same as the graph. But really I just want a single voltage input VCO, that can change the output based on frequency, without the need of additional circuitry.

Obviously lower frequencies have a higher output , then higher frequencies . The frequency range I am using the output would only need to be 40 to 45 dB based on the graph. I really don't think the output needs to be the same output as the graph, I just think the frequency ranges need to have the same curve as the graph.

LowQCab

Joined Nov 6, 2012
2,881
What are You building ?
There might be may ways to achieve the same end result.
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AnalogKid

Joined Aug 1, 2013
10,213
Please post a link to the site or source of the graph. It appears to be related to noise generatiion.

If you convert the graph to a logarithmic x-axis, I think you will see that it is a straight line with a slope of approx. -3 dB per octave. This is the power spectrum of a pink noise source. A very common way of producing this is to start with a white noise source and pass it through a 3 dB per octave lowpass filter. Search for 'pink noise schematic' to see many examples of this approach.

What is it you are trying to achieve? And, why is there a requirement to use a sawtooth signal? Is this school work?

ak

Ian0

Joined Aug 7, 2020
6,974
Relative Output power? Relative to what?

MichaelWeaser

Joined Aug 25, 2021
4
Please post a link to the site or source of the graph. It appears to be related to noise generatiion.

If you convert the graph to a logarithmic x-axis, I think you will see that it is a straight line with a slope of approx. -3 dB per octave. This is the power spectrum of a pink noise source. A very common way of producing this is to start with a white noise source and pass it through a 3 dB per octave lowpass filter. Search for 'pink noise schematic' to see many examples of this approach.

What is it you are trying to achieve? And, why is there a requirement to use a sawtooth signal? Is this school work?

ak
I am making a clone of the Bell Labs Voder from 1939 ( speech synthesizer that is controlled using a keyboard )

Here is a link of where the graph comes from : https://drive.google.com/file/d/1FplhII4b_Iy59Y4CfwqMO8MZi6GcP-H4/view?usp=sharing

This is a document written by the engineers that designed the original Voder , on how it works , and this document has information on its circuitry , and does have the actual circuits used for the white noise and the sawtooth VCO they used ( known as relaxation oscillator in the document )

I am sure this graph has nothing do with pink noise, and its actually the output graph of the sawtooth generator vs the white noise generator

Ian0

Joined Aug 7, 2020
6,974
What it's telling you is the spectrum of the waveform of the oscillator for a fundamental frequency of 100Hz.
If you make an oscillator with a similarly shaped waveform (vaguely a sawtooth) then it will have a spectrum as shown in Figure 7. I'm assuming you don't have a thyratron to hand, so you might just have to use a 555, and the waveform on pin 2 and 6 will have a similar shape (similar enough) and will give you that spectrum.

MichaelWeaser

Joined Aug 25, 2021
4
What it's telling you is the spectrum of the waveform of the oscillator for a fundamental frequency of 100Hz.
If you make an oscillator with a similarly shaped waveform (vaguely a sawtooth) then it will have a spectrum as shown in Figure 7. I'm assuming you don't have a thyratron to hand, so you might just have to use a 555, and the waveform on pin 2 and 6 will have a similar shape (similar enough) and will give you that spectrum.
Now I'm really confused, I was told by a different person that the output changes based on the frequency. I found a guy who designs VCOs and other electronics using tubes, so I sent him the image of the thyratron circuit without the graph because I wanted to ask if this circuit is basically a sawtooth VCO. I was told this circuit works similarly to a VCO, but really isn't a VCO at all because the output voltage changes based on the frequency selected. Obviously what I was told was that lower frequencies have a higher gain output , then higher frequencies.

So obviously I thought that graph must be the output for that thyratron based circuit based on what the frequency is chosen.

Ian0

Joined Aug 7, 2020
6,974
Seems clear to me.

MichaelWeaser

Joined Aug 25, 2021
4
Just realized you are completely right, due to what it says " Such a wave, as is well known, consists of a fundamental and numerous upper harmonics." then it talks about fig. 7 . which is the graph. so it is graphing the harmonics of the sawtooth like wave at 100 hz.

Ian0

Joined Aug 7, 2020
6,974
Just realized you are completely right, due to what it says " Such a wave, as is well known, consists of a fundamental and numerous upper harmonics." then it talks about fig. 7 . which is the graph. so it is graphing the harmonics of the sawtooth like wave at 100 hz.
Yes - badly - because the graph implies that there is harmonic content at, say 160Hz, which there isn't. It should be a series of vertical lines, at 200Hz, 300Hz, 400Hz etc.
I love these old papers and patents from the early days of electronics - I have copies of the first op-amp, Schmitt's original trigger etc.