Hello all,
For my current project that I am working on, I am in need of a positively and negatively ramping triangular wave voltage signal. This signal is to be symmetrical, sweep between +/- 2 V at a frequency around 500 mHz (slow).
Ideally, I would like to use a single coin cell battery to power this signal. Is there a simple way to split say a 5 V battery into two +/- 2.5 V supplies?
Here is the standard circuit that I am using to build the wave:
**broken link removed**
I understand that the frequency can be found as:
[latex]f = (\frac{1}{4*R_{t}*C})\frac{R_{2}}{R_{1}}[/latex]
I am just not sure of how to interpret the voltage peaks for my triangular wave.
Design:
I decided on values for the components based on the above equation:
R1 = 1 MΩ
R2 = 2 MΩ
Rt = 1 MΩ
C = 1 uF
Vss = +5 V
Vee = -5 V
Op-amp used = LF412CN (dual)
Therefore, using the equation above: f = 0.5 Hz
Results:
After wiring this up on my breadboard and checking the scope, I noticed the follow:
f = 460 mHz
Vpeak(+) = 1.68 V
Vpeak(-) = -2.16 V
Appears to be symmetrical but is offset.
My question is, what is the relation between the +/- supplies and my peak voltages, also how do I get rid of this offset?
***EDIT***
[latex]|V_{pk}| = (\frac{R1}{R2})|V_{sat}|[/latex]
Thanks,
JP
For my current project that I am working on, I am in need of a positively and negatively ramping triangular wave voltage signal. This signal is to be symmetrical, sweep between +/- 2 V at a frequency around 500 mHz (slow).
Ideally, I would like to use a single coin cell battery to power this signal. Is there a simple way to split say a 5 V battery into two +/- 2.5 V supplies?
Here is the standard circuit that I am using to build the wave:
**broken link removed**
I understand that the frequency can be found as:
[latex]f = (\frac{1}{4*R_{t}*C})\frac{R_{2}}{R_{1}}[/latex]
I am just not sure of how to interpret the voltage peaks for my triangular wave.
Design:
I decided on values for the components based on the above equation:
R1 = 1 MΩ
R2 = 2 MΩ
Rt = 1 MΩ
C = 1 uF
Vss = +5 V
Vee = -5 V
Op-amp used = LF412CN (dual)
Therefore, using the equation above: f = 0.5 Hz
Results:
After wiring this up on my breadboard and checking the scope, I noticed the follow:
f = 460 mHz
Vpeak(+) = 1.68 V
Vpeak(-) = -2.16 V
Appears to be symmetrical but is offset.
My question is, what is the relation between the +/- supplies and my peak voltages, also how do I get rid of this offset?
***EDIT***
[latex]|V_{pk}| = (\frac{R1}{R2})|V_{sat}|[/latex]
Thanks,
JP
Last edited: