I asked somebody this once and he said that it's the parallel of the 2 resistors because I should consider the circuit with the voltage source being ideal and short it to ground.
If I consider it a series RC circuit with a load and how to account for the fact that it will take longer to charge the capacitor particularly as the maximum voltage would never get applied to it and it would discharge quite quickly.
To put some solid numbers to it, suppose you have a 15V supply, and your voltage divider is a 10K resistor on top, and a 20K resistor on the bottom, for an output of 10V. The equivalent resistance is 10K in parallel with 20K, or about 6.7K. So the capacitor will behave exactly the same as if you were charging it from a 10V supply through a 6.7K resistor.
Okay fine but I'm also looking at the discharge time. As I said before the bottom resistor is a temperature sensor. So the vault is in the middle is going to vary. What I'm trying to ascertain is how fast will it keep up with a given capacitor in parallel. I think I'm going to get the signal generator out, the oscilloscope out and some parts.
Okay fine but I'm also looking at the discharge time. As I said before the bottom resistor is a temperature sensor.
So this is why I wanted to analyse what is effectively a voltage divider with a capacitor in parallel with one resistor and what it's frequency response would be given that is not just a simple RC circuit.
It is in effect a circuit in which I that you consider the resistor changing rapidly or the voltage applied changing albeit by not very much.
It's not the sensor response I'm looking at. What I am looking at is if this was just a resistor divider with capacitor what would the frequency response be to say for example injected noise.
So what I'm actually planning to do it in my hopes of over sampling on the ADC was to inject some of the PWM output from my controller back into the input pin via a resistor. Obviously I don't want that input to change instantly so I put a capacitor in so that to some degree the voltage will swing around about the measurement point. The hope is to gain a bit more resolution (it's purely an experiment and I don't strictly need the resolution) and ultimately there is going to be noise to filter out anyway so my sampling algorithm will filter the signal. Whether or not I get the extra resolution is another matter and beside the point there is another thread on that. What I was wondering was what value of capacitor to use so that I produce a reasonable amount of noise rejection but don't over filter for my intentionally injected noise which would come via a high value resistor. So in this case the sensor and its pull-up are at around 470 ohms each. If I were to use a 470 kilo Ohm resistor from my output to my sensor input I would want to look at the change of voltage that resistor being connected first to 5 V and then to ground causes. The effect of immediately jumping the voltage up or down by a few millivolts will be delayed by the capacitor. This would allow the voltage to be moved around its actual voltage and hopefully squeeze a bit of resolution out.
I have just carried out an experiment and the voltage jumped about by 5 to 10 mV with a 5 V input to the divide and using the 470 Kilo Ohm resistor to inject a 5 V square wave onto the capacitor.
Instead of putting a capacitor directly across the thermistor, why not put a separate RC filter on the output of the voltage divider? Then you can calculate the filter properties independently of the thermistor circuit. If you make the R in the RC significantly higher than the themistor resistance it will not affect the thermistor voltage response.
If you worked out the Thevenin equivalent you would see that it is a simple RC - first order low pass with extra attenuation due to the resistive divider.
I think you're chasing a nonexistent rabbit into hole with this extra resolution idea. But if you are going to dither the input in order to eke out a few more bits at least use a noise source that is not correlated with your signal.
Not really possible as the resistance of the sensor is fixed and the allowed input impedance is fixed so I'd never get orders of magnitude apart. It's going to be a bit of a fudge either way and I was getting curious on a theoretical point of view regardless of how I fundge it together and see what comes out.
Not really possible as the resistance of the sensor is fixed and the allowed input impedance is fixed so I'd never get orders of magnitude apart.
if the main voltage source has gone dead
if the main voltage source has gone dead
Dead ? Or to zero volts ? If the source is still connected, even at zero volts, it is still part of the circuit and the "top" resistor is still active.
Add a resistor in series, from the divider, to the ADC. Make this resistor value several times larger than the thermistor's highest value. Place the cap-to-GND after this resistor.
Add a resistor in series, from the divider, to the ADC. Make this resistor value several times larger than the thermistor's highest value. Place the cap-to-GND after this resistor.
That's what I said in Reply #13. I'm glad my suggestion wasn't totally crazy.
I don't understand this. The input to the ADC has to have a high impedance. It is not likely to influence the readings from your voltage sensor.
The ADC is optimized for analog signals with an output impedance of approximately 10 k? or
less. If such a source is used, the sampling time will be negligible. If a source with higher impedance
is used, the sampling time will depend on how long time the source needs to charge the
S/H capacitor, with can vary widely. The user is recommended to only use low impedance
sources with slowly varying signals, since this minimizes the required charge transfer to the S/H
capacitor.