Continue to Site

# Sinusoidal Synthesizer

Status
Not open for further replies.

#### zeronen

##### New Member
Hi !

Here is my assignment to build the following circuit :

The goal of this circuit is to so as the output signal is as good sinusoidal as possible. I have no idea about its operational mechanics ... I could select the components and build successfully the synthesizer but I don't understand what is going inside ? Anyone could help me ..

The component values :

Rs = 220 ohm
R1 = 1K ohm
R2 = 680 ohm
R3 = 330 ohm

(Like R3 = 3R1 and R2 = 2R1)

And this is my result,

Thank a lot,

Try this circuit...

#### Attachments

• triangletosine.GIF
2.3 KB · Views: 491
Basic theory

OK.
Simplify the circuit so you only consider Rs, R1 and its series diode...

Under what conditions will it conduct ?
What effect will this have on the total circuit (think about resistor ratio ?)

Suppose the voltage at the nodes is Vo

The first branch conducts when the Vo is greater than 0.7 V and the voltage across the R1 is (Vo - 0.7) so the current is (Vo - 0.7)/R1

The second one conducts when the Vo is greater than 1.4 V

The third one conducts when the Vo is greater than 2.1 V

Suppose the Vo is from 0.7V to 1.4V -> there will be a current through the Rs and the R1 and the diode and the Vcc = Vo + I.Rs or

-> Vo = Vcc - I.Rs

Because Vcc is a linear changed linearly and Rs is a constant but the current I, based on the graph of the current versus the voltage through a diode -> It changes exponentially -> So therefore Vo also changes exponentially -> Is that the reason makes the linear voltage a little bit bent into curve, isn't it ?

Sori ... I just think another explaination ...

Suppose the Vcc is from 0.7V to 1.4V (or maybe more) ...

The voltage across diode is relatively constant 0.7V then the voltage across Rs and R1 is (Vcc - 0.7) V ... By the voltage division, the voltage acroos R1 is

(Vcc - 0.7) * R1 / (R1 + Rs)

So the voltage output has the form of y = ax + b

Vout = 0.7 + (Vin - 0.7)*R1/(R1 + Rs) -> the factor a is less than the orginal factor -> That is the reason the output voltage is little bit bent, isn't it ?

Here is my resultant measurement, like the above circuit but there is only the first branch ... That means only 2 resistors Rs and R1 and the diode.

I think the others just make the waveform more like the sinusoidal.

Spot on!

That's correct - the diode is a non-linear resistance- here it is operating as a voltage controlled switch - at 0.7v (some would argue 0.6v - makes no difference to the operation of the circuit).
Below this voltage the equivalent circuit is very different ... effectively just Rs.

Expand this logic and you have the whole circuit sussed - and could suggest improvements ?

Status
Not open for further replies.

Replies
9
Views
3K
Replies
5
Views
2K
Replies
6
Views
4K
Replies
5
Views
2K
Replies
6
Views
2K