.. accuracy depends on the AD629 difference amplifier and AD8421 instrumentation amplifier, not on the op amp or the external 10 kΩ resistors. In addition, this circuit takes advantage of the precise control that the in-amp has of its output ...
.. be used instead, but a JFET is best for the stand-alone version, where supplies will vary. With the values shown, peaks of ~10 µs will be detected, corresponding to a half-cycle at 20 kHz, giving LED flashes of ~20 50 ms duration ...
.. which will cause the limit value to decrease with rising temperature. For the values shown, this amounts to 8 mA over 10 to 50 C temperature range. The float voltage is determined by the voltage at the U1-GND terminal, which is controlled ...
.. comparator in a servo-controlled charge pump configuration. The input is applied to C1, which is compensated by the 10 µF and 1 µF capacitors to act as an op amp. C1’s output drives the 110k-0.02 µF RC, causing ...
.. A1 during circuit start-up. A V→F of this type is usually restricted to relatively low full-scale frequencies, e.g., 10 kHz to 100 kHz, because of speed limitations in accurately switching the current sink. Additionally, short-term ...
.. CTR at V OUT with a multimeter. Besides serving as the V CC source for the DUT (device under test) circuits, IC5’s 10-V output also provides virtual ground to IC3, so the latter can operate with a single, unipolar power supply. Figure ...
.. Non-inverting input V1 can range from 0 to (2 V2), has a nicely high input impedance ( 1 TΩ) and a low bias current (10 pA). Inverting input V2 has a lower impedance (10 kΩ) but will accept a voltage span from as positive as V1 to as ...
.. half a milliamp. These totals are still rather reasonable prices to pay for accurate and fast conversions. Yes, for a VFC, 10-bit resolution in a millisecond is pretty fast. Note that R1 can be chosen to implement almost any desired V IN ...
.. avoid op-amp output saturation induced by the injected I IN current. The negative feedback resistance is thus equal to 450/10 kΩ, or 45 kΩ. Given the 450 kΩ positive feedback resistor, N for Figure 2 is 0.1, yielding an ...
.. need for current-balancing ballast resistors that compensate for unavoidable mismatch between LED forward voltages. About 10% or more of total available power is typically lost in this way. The circuitry shown avoids this inefficiency by ...