??? 02/28/07 01:35 Modified: 02/28/07 02:34 Read: times |
#133902 - It is quite simple... Responding to: ???'s previous message |
Suresh said:
I could not get a clear picture of the current directions at the output of Opamp for a particular type of input and hence could not analyse the possible ways by which a load or any other device could be connected to it thus making the effective use of input. It's quite simple: If the "+" input is more positive than the "-" input, then the output goes high. Why? Because the internal ciruitry "measures" the difference "+" - "-" and amplifies the result by a huge gain (about 100000). And if the "+" input is more positive than the "-" input, the difference is positive and the output goes high. Vice versa: If the input difference voltage is negative, means if the "-" input is more positive than the "+" input, then the output goes low. Due to the internal high gain, an input difference voltage of some millivolts is enough to drive the output into saturation. What the output actually does, when it "goes high" or "goes low" depends on the internal circuitry. Most OPamps provide a voltage output. Means, when the output of a TL072, e.g., goes high, it emits about +14V (without load), when the OPamp is supplied by a +-15V supply. 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 out of the output, means out of the output, through the 2k load and into 0V. This load current is originally flowing from the +15V supply into the Vcc+ terminal of OPamp, by the way. 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. Most OPamps withstand short circuits at their output, because they contain an internal output current limiter, which limits the output current to roughly about +-20mA for the TL072 (both directions). So, when trying to let flow more current out or into of the OPamp's output the output voltage more and more decreases and finally breaks down. Many applications don't like the negative output voltages standard bipolarily supplied (+-15V) OPamps generate. For these applications the industry designed unipolar supplied OPamps, like the LM324, which can have the Vcc- terminal connected to 0V. These OPamps are able to generate output voltages, which are very close to 0V, when the output goes low. A standard TL072 cannot do this, because its output circuitry is different. Another methode to block negative output voltages at the output of OPamp is the use of diodes or similar stuff in series to the output. There are also OPamps, which generate an output current instead of an output voltage, like the CA3080 or LM13700, for instance. These devices are better called OTA (Operational Transconductance Amplifier). But that's another story... Kai |