# Oscillator circuit

Discussion in 'Homework Help' started by xxxyyyba, Nov 21, 2015.

1. ### xxxyyyba Thread Starter Member

Aug 7, 2012
249
2

Oscillator circuit is shown.
a) Calculate frequency of oscillations,
b) Calculate minimum value R1 for which circuit still oscillates,
c) Calculate amplitude of voltage VB, for which voltage VA start cuting off

Vz=6.3V, R1=100K, R2=120K, R3=470, R4=2R6=10K, C1=C2=100nF, Vd=0.7V, R4=R5.

I got f_oscillations=232.7Hz and R1_min=250k.
I don't know how to solve task c)
Any idea?

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
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Well, find Vb/Va = gain for Fo and do you know the "cutting off" voltage for this simple diode limiter?
VA start cutting off = "cutting_off voltage" * gain.

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
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Also notice that if R1 is change from 100k to 250k the Fo will also changes.

4. ### xxxyyyba Thread Starter Member

Aug 7, 2012
249
2

$
-I_1-I_2+I_3+I_4=0,
-\frac{V_B}{R_1}-\frac{V_A-V_X}{R_2}+\frac{V_X-V_B}{\frac{1}{sC_2}}+\frac{V_X}{R_3}=0,
\frac{V_B}{R_1}=-\frac{V_X}{\frac{1}{sC_1}}=-sC_1V_X\Rightarrow V_x=-\frac{V_B}{R_1C_1s},
V_X(\frac{1}{R_2}+sC_2+\frac{1}{R_3})-V_B(\frac{1}{R_1}+sC_2)=\frac{V_A}{R_2},
-\frac{V_B}{R_1C_1s}(\frac{1}{R_2}+sC_2+\frac{1}{R_3})-V_B(\frac{1}{R_1}+sC_2)=\frac{V_A}{R_2},
\frac{V_B}{R_1C_1s}(\frac{1}{R_2}+sC_2+\frac{1}{R_3})+V_B(\frac{1}{R_1}+sC_2)=-\frac{V_A}{R_2}$

And we can find now VB/VA. Is it ok?

Last edited: Nov 21, 2015
5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Do you manage to find the gain ??
I got this result :

$\frac{Vb}{Va} = \frac{C1*R1}{(C1+C2)*R2)} = \frac{25}{24} = 1.041\frac{V}{V}$

And Fo is around 147Hz

Also as a side note I immediately recognized this circuit with OP2 alone, as is used in guitar effects
http://www.muzique.com/news/wahantiwah-mod/

6. ### xxxyyyba Thread Starter Member

Aug 7, 2012
249
2
I definitely made mistake somewhere. Your gain Vb/Va doesn't depend on s...

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
But the gain is depend on s. My formula is a valid only at Fo. And if you solve for Vb(s)/Va(s) you already have a voltage gain expression.
All you need is to find gain at Fo frequency.

8. ### xxxyyyba Thread Starter Member

Aug 7, 2012
249
2
Jony30,
I got same result as you in post #5 using analysis in post #4.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Good for you.
And the transfer function (if i do not make any error) in standard form is :
$\frac{V_b}{V_a} =-\frac{ \frac{1}{C2 R2}s } { s^2+\frac{(C1+C2)}{C1C2R1}s\frac{R2+R3}{C1C2R1R2R3}[\tex]$

T(s) = (a1*s)/(s^2 + ωo/Q + ωo^2)

And the gain at Fo is

Av =(a1*Q)/ωo = (C1 R1)/((C1 + C2) R2)