| ??? 03/31/06 03:33 Modified: 03/31/06 03:37 Read: times |
#113452 - Answers Responding to: ???'s previous message |
Suresh said:
Voltage at non inverting i/p = 1.75V
Inverting i/p: freq of mod voltage = 1111HZ 100uF with 2.2K series resistor maintains the freq below 3dB. It blocks the DC component too. not quite sure about the 10nF cap - seem to be a low pass filter. the 10K with Rf wuld be used for gain adjustment. the cap accross the Rf makes the circuit works as a integrator at high frequencies. So the difference between two i/p's is amplified with the above setup. so the difference voltage amplitude is equal to 1.75 - .85 = .9 and 1.75 - .51 = 1.24 (without gain) but is it not that, we only need to get .85 and .51 at the o/p? Modulation frequency is 1 / (900µsec + 900µsec) = 555.6Hz 100µF with 2k2 and 10k series resistors forms the corner frequecny of high pass filter. 2k2 resistor and 10nF cap provide a low pass filter, needed to keep away steep edges from a "slowly" OPamp. 2k2 plus 10k resistors are used for gain adjustment. The cap across the Rf keeps the feedback stable. The difference voltage is 0.85V - 0.51V = 0.34Vss and the additional gain of 8 helps to increase the relative resolution and allows you to use a simple 8bit ADC for the application. Kai |
