Low Frequency VCO
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Low Frequency VCO

Description

This circuit provides a low frequency (5kHz to 20kHz) square wave output. The circuit is intended to be driven by a regulated, +12V supply.

Design Details -- How it Works

VCO

The internal Voltage Controlled Oscillator (VCO) of the CD4046 phase-locked loop, U3, is utilized to realize the square wave oscillator. For a frequency range of 5kHz to 20kHZ our desired center frequency, fo, is given by: fo=(20-5)/2+5=12.5kHz. Referring to Figure 4 of the datasheet for the CD4046, at VDD=12V, R1=1M (R6 in the schematic above), and R2=infinity, we should select a 100pF capacitor for C1 (C2 in the schematic above). For the selected components, we desire that the VCO remain linear with respect to the VCOIN control voltage over the desired 5kHZ to 20kHZ range. Since, per the datasheet, fo should occur at VDD/2 or 6V, the upper and lower end of our frequency range should correspond to VCOIN=9V and VCOIN=3V, respectively. (This was confirmed through bench tests. See the table in the schematic above.)

To ensure our device delivers a linear response with respect to our control voltage, we must design VCOIN to remain within the 3V to 9V range. (Slight overlap will be designed  to allow for component tolerances). We desire to control VCOIN via a potentiometer. Although connecting the potentiometer across VDD and ground would ensure availability of the 3V to 9V range required for our desired oscillator frequency, it would also allow the device to operate in a non linear range below about 2.5V and above about 9.5V. To overcome this problem, we will introduce fixed resistors between the potentiometer and ground and VDD. (see Figure below)
It follows that Rh, the 5k pot, and Rl form a voltage divider. Moreover, VCOIN will be at its minimum value when the arm of potentiometer is at Rl (i.e., the resistance from the arm to ground is Rl). Similarly, VCOIN will be at its maximum value when the arm of the potentiometer is at Rh (i.e., the resistance from the arm to +12V is Rh). Given this, we have:

12*Rl/(5+Rl+Rh) = 3V (1)

12*(5+Rl)/(5+Rl+Rh) = 9V (2)

From (1), 12*Rl=3*(5+Rl+Rh) (3)

From (2), 12Rl+60=9*(5+Rl+Rh) (4)

Substitute (3) into (4): 3*(5+Rl+Rh)+60=9*(5+Rl+Rh)

Reducing gives: 10=(5+Rl+Rh) (5)

Substituting (5) into (1): (12*Rl)/10=3 or Rl=30/12 = 2.5

From (5), Rh = 10-5-2.5 = 2.5

Therefore, we select Rl=Rh=2.2k (These are standard values and allow for some overlap. You may plug these values into equations (1) and (2) to confirm a low range of 2.81V and a high range of 9.19V)

From the CD4046 datasheet, at VDD=12V, VCOOUT is limited to about 4.5mA source and sink current. Since this may be insufficient to drive some external circuits, VCOOUT is passed through a single stage of the CD4049 inverter. From the datasheet for the CD4049, it is capable of driving a 45mA load which will provide sufficient drive for most purposes.

 

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Copyright 2006-2009 S Hart
Page Last Modified: 10 January 2009

Fine print: Note that this site is intended as an information and education repository. Readers assume all risks and liabilities associated with the use of any information contained herein. The author assumes no liability of any kind for errors, omissions, or claims against any individual or corporation for use of the information contained herein.