Power supplies are slow

[hanhan]

Hi,
I am reading about "Ceramic Capacitors 0.1uF - 25 Pack" in the page: https://www.atomsindustries.com/Ceramic-Capacitors-01uF--25-Pack_p_58.html
And here is the description of it:
0.1uF Ceramic Capacitors. 25 of them. You need these in your circuit...you just do ok! Well here is the general reason why. Power supplies are slow and when you increase your current draw suddenly by say changing the High/Low state of some pins on an microcontroller the voltage will droop from your power supply. Putting one of these between your Vcc and GND of your load (like your chip) will work like a little reservoir of electrons to pull from while the power supply catches up to what your circuit is doing. The closer to the load the better. They will also filter out spikes and droops caused by other sources as well. So to keep your circuit acting as expected, using decoupling capacitors in just the logical thing to do.

Could you help me explain the bold sentences? What do they mean by saying that "power supply are slow"?

 

[ericgibbs]

hi anhnha,

Consider that a MCU pin is driving an LED with say 20mA.

Initially if the LED is OFF [unlit] the current drawn form the PSU by the pin will be zero, now if the MCU pin to the LED is set High the 20mA LED current will be drawn from the PSU.

IF the PSU has NO decoupling capacitors on the Vdd to the PIC its possible that the PSU regulator will be unable to respond fast enough to maintain the Vdd, so it will momentarily drop.

Also consider the Inductance of the wiring from the PSU to the MCU, any sudden demand for the LED current would limited by the inductance of the wiring and the Vdd at the MCU would momentarily drop.

The decoupling capacitors will supply the current demand for a short time, hopefully holding the Vdd constant

E

 

[hanhan]

Thank you Eric!

IF the PSU has NO decoupling capacitors on the Vdd to the PIC its possible that the PSU regulator will be unable to respond fasy enough to maintain the Vdd, so it will momentarily drop.

Is this because the parasitic inductance in PSU and wire?

The decoupling capacitors will supply the current demand for a short time, hopefully holding the Vdd constant

I am wondering how this capacitor isn't affected by parasitic inductance?
And in order to reduce inductance we have to put the capacitor as close the led as possible, right?

 

[ericgibbs]

Thank you Eric!

Is this because the parasitic inductance in PSU and wire?

I am wondering how this capacitor isn't affected by parasitic inductance?
And in order to reduce inductance we have to put the capacitor as close the led as possible, right?

hi,
The decoupling capacitors should be as close as possible to the MCU in order to minimise losses, resistive and inductive, in the connecting wires of the Vdd supply.

Usually the decoupling capacitors are of two types, larger value electrolytic's that are able to supply higher current demands in the circuit and lower value ceramic types that can be located close to the MCU.

Generally electrolytic's tend have have an higher impedance at the higher frequencies whereas ceramic's have better high frequency performance.

Typical values are 100uF thru 1000uf for electrolytic and 10nF thru 100nF for ceramic.

E

 

[ronsimpson]

I am sitting in my 'library' reading your posts on my smart phone. As I look at the sink, shower and toilet I can describe bypass caps by water pipe theory.
If I get a drink, the water comes from a 1/4" pipe that runs runs from the far end of the house. Then from a 1" pipe that runs 1/4 mile to the road, turns and runs 1/2 mile to a larger pipe for 3 miles to a large tank then 5 miles to a pumping station.....This is good for getting a drink of water.

When I flush the toilet I need a rush of water that will fill a 2 inch pipe for 4 seconds. I can not get this amount of water from a 1/4 inch pipe. Right behind me is a 2 gallon tank with enough water to fill the 2 inch pipe and do the job. Then slowly the tank will fill. If I flush 10,000 times the big tank miles away will empty and the pumps will start up.

You micro computer (load) might switch from little power to full power in 1nS. 1/1,000,000,000 The power supply can not respond to that. It needs 1/1,000 second. The wires have a delay. Some where you need to store energy to fill in the difference in time. Use a small capacitor at each load point and a large capacitor at (in) the power supply.

 

[hanhan]

Thanks ronsimpson,
Much appreciated and your imagination is very impressive!