Use of inductors for sawtooth wave form

[ChrisGreece52]

I bought some 100uH inductors to mess with them and use them to create a sawtooth signal.
Anyway i am trying to do so for an hour but i cant make any progress.
My setup is just a square wave (from arduino)

Tried different freqs ranging from 1 hertz to 1 milihertz.
an led (to see the results)
and inductors (100uH each i used 1 - 5 and 10 of them in series for added inductance)
I even tested the circuit in theory with the Every Circuit app and saw just a litle ripple in the square wave but i need more
I tested it with 1MH inductor and it worked (kinda) but i hope there is another solution
I even put a 2200UF cap to smooth the wave .... (silly but it was just a thought)
Anyway i made a small reasearch and sawtooth wave form and i tried to find some function gens that did not use any  chip just inductors but i had no results.
At school it was pretty clear how to create sawtooth like waves using inductors i dont know where i screw things up....


Thank you.

[Paul Price]

Inductors conduct current in an exponentially increasing way...you need a linear device to create a ramp or a sawtooth. A capacitor driven by a constant current source will create a ramp, an inductor driven by a constant voltage will produce a ramp.

An inductor will produce a kinda ramp in the circuit below with a switching transistor to buffer the output from your a square wave signal generator, a "catch" diode and a resistor for a load on the other side of the inductor.

[ChrisGreece52]

THANK YOU SO MUCH

[ChrisGreece52]

Inductors conduct current in an exponentially increasing way...you need a linear device to create a ramp or a sawtooth. A capacitor driven by a constant current source will create a ramp, an inductor driven by a constant voltage will produce a ramp.

An inductor will produce a kinda ramp in the circuit below with a switching transistor to buffer the output from your a square wave signal generator, a "catch" diode and a resistor for a load on the other side of the inductor.

Sorry for my moronic question but this symbol below the diode and the resistor is a ground right

[c4757p]

Yes.

[ChrisGreece52]

OK Thank you then i will epxeriment with different values to find the perfect one for me
Edit : i tried different values of resistors and inductors but i wont see any major effect on the led .... (maximun inductunce i can get is 1mH )
my setup
Square arduino

Code: [Select]

digitalWrite(led ,HIGH);
delay(100); //milliseconds
digitalWrite(led,LOW);
delay(100); // milliseconds again
300uH inductors (3 x 100 in series)
diode (shall i use something special i use zener ones)
10Kohm resistor from the end of the inductor to ground
and then on the output a red led ....
Did not see any results....

[c4757p]

Dude, you're going to need a lot more than 300uH to do this... L/R (the equivalent of an RC constant for R-L circuits) is 300u/10k = 3ns. This circuit won't do anything interesting until you get near 3ns times.

Go find a spare mains transformer and use the primary as an inductor. That'll get you closer to what you want. (That will be henries, not microhenries.)

And what is the LED supposed to do...?

[IanB]

Sometimes you need to use a little mathematics.

The inductor formula is:

V = L dI/dt

Let's suppose you want the current to rise from 0 to 20 mA in 100 milliseconds.

Then your dI/dt value is:

dI/dt = 20 / 100 = 0.2 A/s

Now, maybe your supply voltage is 5 V. In which case you can work out your required inductance value:

L = V / (dI/dt) = 5 / 0.2 = 25 H

Yes, that is 25 henrys!

The inductor you need is going to be as big as a large transformer and weigh a kg or so.

What you might do is find an actual tranformer, maybe some kind of 12 V 2 A mains transformer, and use its secondary winding as your inductor, leaving the primary unconnected. That will get you a bit closer to the ball park. (Or use the primary as c4757p says, but this will have a higher resistance.)

Oh, and a warning: if you feed a square wave into a large inductance like a transformer primary, the back-EMF may kill anything like an Arduino that is supplying the square wave. Probably best not to actually try this.

[c4757p]

Don't you mean to use the primary? That will have a much higher inductance.

Edit: I see your edit!  But in most cases, the secondary is going to be much lower than what is needed here. Even on the primary, you'd be lucky to find 25H.

[IanB]

Don't you mean to use the primary? That will have a much higher inductance.

Edit: I see your edit!  But in most cases, the secondary is going to be much lower than what is needed here. Even on the primary, you'd be lucky to find 25H.

I just added a second edit too. Transformer primaries are dangerous!

[c4757p]

Oh, and a warning: if you feed a square wave into a large inductance like a transformer primary, the back-EMF may kill anything like an Arduino that is supplying the square wave. Probably best not to actually try this.

Good point. Chris, the "sawtooth" you will get from doing this will be almost identical to the one you'd get by applying the same square wave through a resistor to a capacitor. I'd do that instead.

And if you want a properly linear one, charge it through a constant current circuit instead. Two transistors, two resistors, and you're set. If you want it to be symmetric (triangle, not sawtooth), you'll need two of them.

[c4757p]

I just added a second edit too. Transformer primaries are dangerous!

Possibly an even more important safety warning would be that if you're sticking a waveform up the secondary, you may not want to touch the primary...

[ChrisGreece52]

Oh, and a warning: if you feed a square wave into a large inductance like a transformer primary, the back-EMF may kill anything like an Arduino that is supplying the square wave. Probably best not to actually try this.

Good point. Chris, the "sawtooth" you will get from doing this will be almost identical to the one you'd get by applying the same square wave through a resistor to a capacitor. I'd do that instead.

And if you want a properly linear one, charge it through a constant current circuit instead. Two transistors, two resistors, and you're set. If you want it to be symmetric (triangle, not sawtooth), you'll need two of them.

Ok ill try that instead

[ChrisGreece52]

Let's suppose you want the current to rise from 0 to 20 mA in 100 milliseconds.

Then your dI/dt value is:

dI/dt = 20 / 100 = 0.2 A/s

I think you did a mistake there.
with the time you devided 20 with 100 instead of 0.1 (100 milliseconds = 0.1 second)
I only say that because your result is 0.2 A/s not 0.2 A/ms

[ChrisGreece52]

Dude, you're going to need a lot more than 300uH to do this... L/R (the equivalent of an RC constant for R-L circuits) is 300u/10k = 3ns. This circuit won't do anything interesting until you get near 3ns times.

Go find a spare mains transformer and use the primary as an inductor. That'll get you closer to what you want. (That will be henries, not microhenries.)

And what is the LED supposed to do...?

Nothing i just wanted to watch the results through the led so i can use the sawtooth wave from later on any project.
But it could work with the things i have i can put 10 inductors in series and use a 100 ohm resistor instead of a 10k so 1000/100 10 seconds (or i am terribly wrong right here :/ )

[baljemmett]

Let's suppose you want the current to rise from 0 to 20 mA in 100 milliseconds.

Then your dI/dt value is:

dI/dt = 20 / 100 = 0.2 A/s

I think you did a mistake there.
with the time you devided 20 with 100 instead of 0.1 (100 milliseconds = 0.1 second)
I only say that because your result is 0.2 A/s not 0.2 A/ms

IanB is correct; the millis in 20mA / 100ms cancel, leaving 20A / 100s = 0.2 A/s.   Or if you prefer not cancelling the prefixes, 0.02A/0.1s = 0.2A/s.

[ChrisGreece52]

Let's suppose you want the current to rise from 0 to 20 mA in 100 milliseconds.

Then your dI/dt value is:

dI/dt = 20 / 100 = 0.2 A/s

I think you did a mistake there.
with the time you devided 20 with 100 instead of 0.1 (100 milliseconds = 0.1 second)
I only say that because your result is 0.2 A/s not 0.2 A/ms

IanB is correct; the millis in 20mA / 100ms cancel, leaving 20A / 100s = 0.2 A/s.   Or if you prefer not cancelling the prefixes, 0.02A/0.1s = 0.2A/s.

OOHHHH ok thank you