| ??? 01/09/06 13:20 Modified: 01/09/06 13:25 Read: times |
#106886 - Misinterpretation Responding to: ???'s previous message |
Farshid said:
My problem is that the Fast Slew rate OP-AMPs are hard to find. Maximum neeeded BandWidth For this project wouldn't be more than 10 Mhz. The reason Bitscope used such a fast OP-Amp was because of the Sub-Sampling technique they used.(This way they can capture up to 1Ghz signals with a 25 MSPS ADC !) Bitscope says, that maximum analog sampling rate is 40Ms/sec. From Shannon-Nyquist theorem this means that a maximum signal frequency of 20MHz can be handled. So, the analog part of the circuit must at least have a bandwidth of 20MHz, which is also stated in their specifications (single-shot bandwidth). A big misinterpretation of sub-sampling technique is, that by sub-sampling a higher bandwidth can be achieved. This is wrong! The only thing that sub-sampling provides is more samples per period. Assume that 20MHz is sampled by 40Ms/sec. Then you get two samples per period with a distance of 25nsec. Sub-sampling technique only adds additionally samples, so that the gap between the two samples is filled by a number of additional samples. The distance between two samples is not 25nsec as with 40Ms/sec sampling rate then, but 1nsec according to a sampling rate of 1Gs/sec. Again, this does not at all mean, that the bandwidth is increased in any way! Why must OPamps be used, which show a much higher bandwidth than 20Mhz for this purpose? A bandwidth of 20MHz means a dampening of 3dB at 20MHz. If you would use OPamps with 20MHz bandwidth, then using four such OPams in series would result in a dampening of 4 x 3dB = 12dB. So, bandwidth of circuit would be much much lower than 20MHz. Because of this, much faster OPamps must be used, to allow that when four of them are put in series the over all bandwidth is 20MHz. Another reason, why much faster OPamps must be used is, that specified bandwidth of OPamp, means "unity gain bandwidth" is only valid for a gain of 1. When a gain greater than 1 is needed then bandwidth becomes drastically lowered. Up to now we have considered the issue in the frequency domain. If you analyze situation in time domain we must have a look at slew rates. A bandwidth of 20MHz corresponds to a signal rise time of 1 / pi / 20MHz = 15.9nsec. Relating to a 5V step this means a slew rate of 314V/µsec. If we had OPamps showing a slew rate of 314V/µec, then putting four of them in series would result in an over all slew rate of SQRT((314V/µsec)^2 / 4) = 157V/µsec. So, again we see that OPams are to be used, which show a slew rate of much more than 314V/µsec. Again, as above, slew rate drastically lowers if a gain is needed which is higher than 1. So, you see, even when only wanting a bandwidth of 20MHz, using OPamps of 300MHz bandwidth and a slew rate of 1100V/µsec is no waste of performance. We have overlooked one part of signal chain up to now: The probe. If a bandwidth of 20MHz is wanted, then a probe providing a bandwidth of 100MHz should be used. Kai |
