- #1
Exulus
- 50
- 0
Hi guys, I need a bit of help with this. I've got an op-amp and the standard formula:
[itex]V_{out} = -\frac{1}{RC}\int V_{in} dt[/itex]
And i need to integrate a square wave from it in order to determine some capacitor/resistor values to get an output amplitude of 5V and freq 200Hz (triangle wave). Every time i try to do it i end up getting 0 as an answer (which is kind of logical, considering one half of the cycle is the negative of the other). For example:
[itex] V(t) = V[/itex] for [itex] 0 \leq t \leq 0.0025[/itex]
[itex] V(t) = -V[/itex] for [itex]0.0025 \leq t \leq 0.005[/itex]
[itex]V_{out} = -\frac{1}{RC} [ \int_{0}^{0.0025}Vdt - \int_{0.0025}^{0.005}Vdt][/itex]
[itex]V_{out} = -\frac{1}{RC} (0.0025V - (0.005V - 0.0025V)) = 0[/itex]
How can i get around this problem? Cheers.
[itex]V_{out} = -\frac{1}{RC}\int V_{in} dt[/itex]
And i need to integrate a square wave from it in order to determine some capacitor/resistor values to get an output amplitude of 5V and freq 200Hz (triangle wave). Every time i try to do it i end up getting 0 as an answer (which is kind of logical, considering one half of the cycle is the negative of the other). For example:
[itex] V(t) = V[/itex] for [itex] 0 \leq t \leq 0.0025[/itex]
[itex] V(t) = -V[/itex] for [itex]0.0025 \leq t \leq 0.005[/itex]
[itex]V_{out} = -\frac{1}{RC} [ \int_{0}^{0.0025}Vdt - \int_{0.0025}^{0.005}Vdt][/itex]
[itex]V_{out} = -\frac{1}{RC} (0.0025V - (0.005V - 0.0025V)) = 0[/itex]
How can i get around this problem? Cheers.