Below you can see the circuit we’re currently talking about.
The resistor and capacitor on the left of the inverter make up an RC delay element:
But how does the current flow in this circuit?
If you try to follow the current flow you might end up scratching your head.
The inverter actually has connections to plus and minus. To simplify, these are usually hidden in circuit diagrams. You just need to know that they’re there.
This means it’s not as easy as “what goes in, must come out” since the current can flow into and out from the plus and minus connections too.
The easiest way to think about current flow when you have logic gates is to assume no current flows into the input. And you can think of the output as either connected directly to plus (if it’s 1) or minus (if it’s 0).
This means we can look at the left and the right side of the inverter individually.
Current flow on the right side
The right side of the inverter is simple:
If the inverter outputs a 1, currents flows down from the inverter output, through the resistor and LED, down to minus.
If the inverter output is 0, no current flows on the right side.
Current flow on the left side
The left side of the inverter takes some more concentration.
But remember our rule:
Current always flows from a higher to a lower voltage – if there is a path for the current to take.
When you flip the switch from 0 to 1, the left side of the switch is connected to plus. The capacitor is currently discharged, so it has 0 volts across it. Which means there is a path from plus through the resistor and capacitor, down to minus – therefore current will flow.
When the capacitor is fully charged, it doesn’t allow any more current to flow through it. So the current flow stops.
“Fully charged” means the voltage across it has reached the voltage it is being charged with.
When you flip the switch back again from 1 to 0, do you see where the current flows?
Yes – it looks like it’s flowing backwards..!
But, as you now realize, current simply flows from higher voltage to lower voltage. If voltages change – and they often do – the current redirects from the new ‘high’ to the new ‘low’.
So you now have a positive voltage on the capacitor. The switch is connected to 0 volts. And there is a path from the capacitor, through the resistor and down to minus through the switch. So current will flow here.
The current flows until the capacitor is fully discharged (i.e. no more voltage across the capacitor) making it the same voltage (ie zero) as the minus of the battery.
Sadru says
April 10, 2019 at 9:49 pmFascinating!
admin says
April 11, 2019 at 4:45 pmGlad you like it!
David H says
April 10, 2019 at 10:34 pmVery interesting and easy to understand thx
admin says
April 11, 2019 at 4:40 pmGlad to hear!
C. Dixon says
April 11, 2019 at 1:53 amI love the way you teach!
admin says
April 11, 2019 at 4:43 pmThat’s very cool to hear =)
Zac says
April 12, 2019 at 12:18 pmwhat is the equation to figure out time delay? and how do you do it?
George says
April 18, 2019 at 7:52 amThe capacity of the capacitor in Farads X the resistance of the resistor in Ohms = the delay in seconds
George says
April 18, 2019 at 7:55 amWhat happens to the capacitor in the circuit is now clear. But I don’t see any description what happens to the resistor + LED part… When the switch is at 0V and the capacitor starts discharging, there is a path to the plus, but what happens to the LED at that moment?
admin says
April 23, 2019 at 2:27 pmOn the LED side of the circuit, all depends on if the output is HIGH or LOW. When HIGH, the LED is on. When LOW, the LED is off.
The inverter has a threshold voltage that decides exactly when to switch the output on or off. In this circuit you need to use a Schmitt-triggered inverter where the threshold voltage for going from low to high is a bit higher than the one for going from high to low.
George says
April 18, 2019 at 8:01 amCan we say that when the switch is “1”, the capacitor gets charged till it is fully charged AND the input of the inverter is high, which makes the output low (which means the LED is off).
And when the switch is set to “0” (and the capacitor was already charged), current flows from the positive capacitor to the swtich to ground AND it keeps the input of the invertor still high as long as the capacitor is not fully discharged (which means the LED stays off till the capacitor is fully discharged)?
admin says
April 23, 2019 at 2:28 pmYes, that’s pretty much it =)
Clive says
May 25, 2019 at 1:23 pmHi
I’m trying this using a 741 inverter. 9v supply to pins 7&4 input to pin 2 output to pin 6.. When connecting to pos supply , the led glows dimly , and then when connect to negative it glows brightly.
This I can see this is the effect of high input low output and low input high output of the inverter.
But I am puzzled as to where the voltage is comming from, to make the led glow although dimly, when there is a high input, eg connected to pos supply.
There should be zero output?
Like your emails, and easy explanations ,will probably join your site after the summer.
admin says
May 27, 2019 at 3:41 pmHi Clive,
The 741 is an opamp, not an inverter, so it will amplify the difference between what is on the input pins 2 and 3. Are you sure you’ve set it up as an inverter?
Clive taylor says
May 28, 2019 at 10:24 amHi.
Ah. I seem to be missing the point, and as my lack of knowledge , has shown ,just because the pins are called inverting and non-inverting , does not mean it’s an inverter. I found a flashing light diagram on the web, also with feed back loop, and this worked. But as I say it flashed , but I guess was not inverting.
Ok lesson learnt .
As I’m learning , what numbers are inverters, i might have one in kit of bits I have.
I’ll look out for the next email.
Thanks
Muhammad Hamwi says
June 20, 2019 at 2:24 ami must admit i’m a little confused, i’m trying hard to understand the capacitor, when you say;
“And there is a path from the capacitor, through the resistor and down to minus through the switch. So current will flow here.”
would not the current take the less resistive path and flow from its ground connection ? it may be that i’ve completely miss understood capacitors.
when a capacitor is being charged a negative charge builds up on its negative plate & a positive charge builds up on its positive plate, and once its charge equals that of the supplies, no more current flows, when you flip the switch from 1 to 0, thats what i don’t understand, is the 0 terminal connected to the negative plate of the capacitor ? very sorry if i sound confusing, if someone could please point out where i’m getting mixed up that’d be great, thank you
admin says
June 20, 2019 at 3:16 pmHi Muhammad,
“would not the current take the less resistive path and flow from its ground connection?”
Current always from flow a high potential (Volt) to a low potential (Volt). Since ground is the lowest potential in this circuit, you won’t have current flowing from ground and back into the circuit.
Muhammad Hamwi says
June 20, 2019 at 9:31 pmah yes thank you, also when you say;
‘Which means there is a path from plus through the resistor and capacitor, down to minus – therefore current will flow.’
does the current literally flow through the capacitor ?
admin says
June 24, 2019 at 9:32 amYes, current flows through the capacitor unless it’s fully charged.
Electronics guy says
July 9, 2019 at 12:43 pmWhy does the current go backwards instead of going to the inverter and from there to neutral?And also what type of inverter is the inverter (nmos,pmos etc.)?
admin says
July 9, 2019 at 1:47 pmThe inverter only cares about voltage. It looks at the voltage on its input and sets the output to the opposite.
No current flows into the input because of the way the inverter is built (ok, a tiny bit of current does flow when the input is high, but usually so little that you can ignore it)
The inverter type for this circuit to work needs to be a schmitt-triggered inverter. It doesn’t matter what transistor types are used to build it.
Benjamin Dorion says
July 12, 2019 at 8:47 pmPut more simply, is the voltage that crosses the inverter at ’14’ and ‘7’ meant only to ‘activate’ the inverter, or am I completely missing something?
admin says
July 15, 2019 at 11:49 amYes, that’s correct – it only activates the inverter.