| ??? 03/01/07 05:22 Modified: 03/01/07 05:24 Read: times |
#133987 - Some answers... Responding to: ???'s previous message |
Suresh said:
So if i give say 0.5Vdc at '+' input and 0Vdc at '-' input (excluding DC Characteristics since it should be compensated) without any feedback for gain adjustment and i use a bipolar supply of +/-10V, then i would get an output voltage of +10Vdc (voltage output) that sources some amount of current which a datasheet would say. Is it correct? Yes, but you will not get exactly +10VDC at the output, when using a TL072, or the LF412 of your example. About 1...2V will drop across the internal transistors. But you are right, the output looks like a voltage source, which tries to emit the absolute maximum output voltage possible, just because the input is so heavily overdriven. Suresh said:
My doubt is why do you say "atleast -10V" when the output voltage is going to be -14V (which would fall on 2k)when the '-' input is higher than '+' input? Not exactly clear what you mean. But, the output tries to emit its maximum possible output voltage, which is about -14V typically without load and about 13.5V typically with 10kOhm load. For a load of 2kOhm the output is guaranteed to emit at least -10V. So, it can also be -11.3876543V or 12.99998765V, but that's not guaranteed. Just have a look into the datasheet... Suresh said:
I couldnt analyse how it changes to asymmetric with that voltage source and what would be the voltage drop at R3 for a particular voltage set with that voltage source. But that's quite simple! 
If the lower terminal of R3 is at 5V, for instance, then the upper terminal is at 7.5V, when the output of OPamp emits +10V and at -2.5V, when the output emits -10V. These different threshold voltages mean, that the cap is charged/discharged with differently high currents, resulting in an asymmetric waveform. Kai |
