| ??? 02/28/07 07:53 Modified: 02/28/07 09:45 Read: times Msg Score: +1 +1 Informative |
#133915 - Its much clear now. Responding to: ???'s previous message |
Richard said:
The PDF file is highly useful. Thank You. Joe said:
But maybe this will help it make a bit more sense as you study those basics and wonder why. That's my hope anyway. Yes. mainly because of that i.e., to know how those OpAmp functions are practically used for a particular design, i started this thread :) Thank You. Andy said:
Current is said to flow from the more-positive potential to the more-negative terminal. Joe said:
Maybe that's his confusion. Yes. Got confused with that negative voltages. Now got cleared with Kai and Andy's replies. Kai said:
Due to the internal high gain, an input difference voltage of some millivolts is enough to drive the output into saturation 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? Then, Kai said:
When the output of a standard OPamp, like TL072, goes low (means if the "-" input is some millivolts more positive than the "+" input), it emits an output voltage of about -14V (without load), when the OPamp is supplied by a +-15V supply. Again, the output is guaranteed to generate at least -10V across a load of 2k, connected from output to 0V. So, you can let a current of at least 10V / 2kOhm = 5mA flow into the output, means out of 0V, through the 2k load and into the output. This load current is finally flowing through the Vcc- terminal of OPamp to the -15V supply, by the way. Thank You. Its very simple to understand with this Kai. But then, above you have said "at least -10V across a load of 2k,". 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? Finally, i took up a circuit for analysing., 
In the above circuit, iam clear with the Symmetric square wave generation. But for Asymmetric Square wave generation, a voltage source need to added as shown the diagram. 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. Willing to get some information with this. regard's Suresh. |
