What is the purpose of this capacitor? - Page 1
Forum section: electronics / beginners

author: jj5
author_rank: Unknown
timestamp: February 09, 2025, 11:56:11 pm
content: Hi there. I am building theJaycar and Silicon Chip Mini Project #003 Thermal Fan Control by Tim Blythmanand I don't understand the purpose of the 100nF capacitor in theXC4494 Temperature Sensor Module. Can someone explain that to me? Thanks!

author: calzap
author_rank: Frequent Contributor
timestamp: February 10, 2025, 01:34:09 am
content: I assume it's for smoothing ripple or spikes. Mike

author: jj5
author_rank: Unknown
timestamp: February 10, 2025, 01:56:24 am
content: Quote from: calzap on February 10, 2025, 01:34:09 amI guess I should clarify that I want to know "what it does" but also "how it does it". I expect there is something fundamental for me to understand here...

author: themadhippy
author_rank: Super Contributor
timestamp: February 10, 2025, 02:15:20 am
content: Have a read up on decoupling capacitor , it should explain everything

author: jj5
author_rank: Unknown
timestamp: February 10, 2025, 02:41:07 am
content: Quote from: themadhippy on February 10, 2025, 02:15:20 amOh, the decoupling capacitor is the other 100nF capacitor in this circuit. I am asking about the other 100nF capacitor, the one in the temperature sensor module. (But I should read up on decoupling capacitors too!)

author: D2236
author_rank: Regular Contributor
timestamp: February 10, 2025, 03:42:23 am
content: It may be to force the fan to start in off state when the circuit is powered up. The -ve input of the opamp will be pulled up to supply for a few ms, turning off the MOSFET until the cap is charged up. May also be added to keep +V noise out of the module.

author: forrestc
author_rank: Unknown
timestamp: February 10, 2025, 03:46:39 am
content: Quote from: jj5 on February 09, 2025, 11:56:11 pmIt appears to be there to form a low-pass filter.The 4.7k resistor and 10K thermistor form a "voltage divider" which results in the voltage at the connection between the two (which is connected to S) varying based on temperature.What the capacitor will do form a "low pass filter". If you look up passive low-pass filter you'll find that they consist of a resistor in series followed by a capacitor in parallel. This arrangement will cause signals over a certain frequency to be attentuated. So if it was audio, you could set up a filter like this so you'd have the bass sound but not the treble.In this case, if I stick 4.7K and 100nF into a low pass filter calculator, I find that the cutoff frequency is around 330 Hz. (I'm too lazy to figure out what the actual effective resistance range is). But... the point here is that if the temperature sensor puts out noise over 330Hz, it will not exist in the output S pin.

author: jj5
author_rank: Unknown
timestamp: February 10, 2025, 04:32:10 am
content: So the 4.7kΩ resistor is part of a voltage divider and also a part of a low pass filter?

author: forrestc
author_rank: Unknown
timestamp: February 10, 2025, 05:47:25 am
content: Quote from: jj5 on February 10, 2025, 04:32:10 amTechnically, both the 4.7k resistor and the thermistor are part of the low-pass filter.To oversimplify and, as a result, be a bit inaccurate: When you have a voltage (or signal) source and a resistor, you have an effective source impedance. That impedance is what determines the R part of the low-pass filter.Imagine your circuit without the bottom 4.7K resistor. In this case, the impedance would be the thermistor value. That impedance would then be used to determine the cutoff frequency.But with both halves of the voltage divider present, determining the effective impedance becomes a bit more complex. I'm guessing it's the parallel resistor formula:$$ r=\frac{1}{\frac{1}{r_1}+\frac{1}{r_2}} $$where r1 and r2 are the two resistor values - but I wouldn't take any bets on that being correct. I'd probably shove it into a simulator to double check my wild guess.

author: PGPG
author_rank: Frequent Contributor
timestamp: February 10, 2025, 10:54:11 am
content: Quote from: forrestc on February 10, 2025, 05:47:25 amYou are correct. But I don't remember when I sow that formula such written. Since I remember I use R1*R2/(R1+R2).Search for Thevenin's equivalent voltage source.