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measurement - What is 1. 5%rdg + 4dgt for 15,23 V? - Electrical . . . 0 Where "Rdg"is for reading and "dgt"is for digits So, 15,23 * 0,015 = 0,22845 Rounding to most significant digit it's 0,2 So is this correct? $$15,23 \pm 0,2$$ Since 4dgt is smaller we ignore it Or am i doing this wrong? I need to find uncertainty in the way I did for a physics lab
batteries - Electrical Engineering Stack Exchange I am reading the datasheet of an impedance measurement device Its accuracy is listed as '±0 4% rdg ±8 dgt ' What does 'rdg ' and 'dgt ' mean? And how should I interpret the accuracy? The device is
How do I calculate accuracy marked as ± (4 ppm rdg + 10Digit)? The range is set to 1 volt and has an input of 0 724 DCV With an accuracy specified as ±(4 ppm rdg + 10 LSD), the accuracy is ± ______ DCV This is what I came up with: 10 of 199999 = 0005% 0 724 * 0 000004 + 0 0000005 = 0 000007896 However, it ended up being wrong Just need a little help understanding it
basic - I need explanation on some digital multimeter (DMM . . . And what does tolerance 1% rdg indicate which followed by: 5 dgt mean ? The analog components limit the accuracy to ±1% of the actual value being measured The digital circuitry limits the accuracy to ±5 counts (i e , in the least-significant digit of the display) of the actual value
what the best way to calculate Rg gate driver for Mosfet @ Marko Buršič, Yes it is the maximal gate current I searched and read many Application notes from supplier , but did not find a specific answer or a genaralist formula to calculate the correct value for a Rg resistor for a MOSFET's the right formula must includ the Gate Charge Qg, and the swithing time turn ON - Turn OFF, Or by using the Output voltage slop dVout dt
Finding out the accuracy of the current probes measurement The sensitivity (1 mA) is the smallest current that can be consistently sensed by the instrument, i e it is guaranteed that you will get some reading with anything above 1 mA The gain accuracy tells you your measurement accuracy as a percentage of the measurement itself, so if you are sensing 1 A you can expect +-3%, typically +-1% The figures above are true for a properly calibrated