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You may add a cap in parallel with R to form a 1st order low pass filter.
Your opamp usually has requirements on voltage supply. So using supply as limiters is not a good solution.
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AC analysis is not run point by point in time domain. It's frequency sweep based on transfer function or in freqency domain. DC gain when f=0.
You don't have to get it from AD analysis. You may get from other analysis but you'll find it takes longer time especially when you want to compare...
This is how you calculate:
The input+ has a voltage set by the zener diode=Vz. Input- = Input+ due to feedback. So Vsense=Vz. The current through the resistors=Vz/Rsense
This is a good current sink, better than other designs in this thread.
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If this impedance is so high your signal might be a current, or charge signal. Use charge or current amplifiers.
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Low-cost resistors, capacitors, and LEDs: **broken link removed**
Re: Whats the output of non selected analog multiplexer pins
It's 'high-impedance'.
In other word, theorectically it's 'anything' in term of voltage. If you use a pull-up it's high and if pull-down it's low.
In reality high impedance doesn't mean infinite impedance. And it's high impedance to...
You may get the converter in low cost here:
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because inductors are difficult to manufacture with nice accuracy and Q value. Resistors and caps are easy to control the accuracy and more close to ideal components.
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Re: hi everyone
you need a small DC motor, with a driver or not. Sensing may be an infred transmitter and detector.
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you have to comprise Q if you only have one L and one C: scale L and at the same time scale C. Or you may use a pi sturecture so that you have one extra parameter and one extra degree of freedom. Usually this extra freedom is sufficient
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lcd + cursor
I used buffers too and I felt safer to follow this way.
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constant current regulator
The opamp circuit should be working fine. I guess the heat dissipation is on the transistor? You may check if the transistor is properly biased and the voltage and current are at the levels they are supposed to be at.
Since it's a current source, it tries to...
1:1 Orcad Print out
You may print out the layout with 1:1 in layout if you want to see it physically. or you may measure the dimension with layout tools.
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Power supply
P can be replaced with a higher input impedance voltage sensor. Increasing it to 2K or 10K will be straightforward. Or a V-I-V circuit. Or an opamp will be better
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to answer your question in a system point of view, the two stage circuit form a positive feedback. So instead of keep stabel at a certain bias, the circuit jumps to an extreme stage after an input and stay locked there
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