Clock edge detector using capacitor and resistor problem

Thread Starter

korgong123

Joined May 1, 2022
1
I have watched recently this video:
At the end of it he describe a way to make edge detector by using capacitor and resistor. The thing is i don't see it being discharged in any way. Would't it make it impossible to charge again and in consequence make current flow through the resistor and make voltage peak? Could someone explain how it works? I am beginner at building electronic circuits and this is my first post so sorry if i asked stupid question or made mistakes writing it.
 

olphart

Joined Sep 22, 2012
107
A> Forget a big capacitor (only limits frequency)
A small (.01 - .1uF) is good for noise, but frequency dependent.
B> Use a resistive voltage divider to get it into range (if necessary).
C> Either an ADC (if value matters) or a comparator (for yes/no) to get desired signal.
G. H. <<<)))
 

eetech00

Joined Jun 8, 2013
3,149
I have watched recently this video:
At the end of it he describe a way to make edge detector by using capacitor and resistor. The thing is i don't see it being discharged in any way. Would't it make it impossible to charge again and in consequence make current flow through the resistor and make voltage peak? Could someone explain how it works? I am beginner at building electronic circuits and this is my first post so sorry if i asked stupid question or made mistakes writing it.
Initially, there is an assumption that the capacitor is discharged, and the +side of the cap at zero volts.
So the cap represents a very high resistance to DC. Therfore the R/C junction is at 0v since the other end of the resistor is connected to ground. The moment +V is applied to cap, the cap becomes a very low resistance for a few milliseconds, and the R/C junction goes to +V. As the cap begins to charge up, its DC resistance slowly increases. When it is fully charged (5 time constants) , it represents an open circuit to DC, so the junction is now at 0v again. The end result is a momentary positive pulse at the R/C junction.

Hope that helps...
 

dl324

Joined Mar 30, 2015
14,459
Welcome to AAC!
Could someone explain how it works?
Can you give time index? I'm not going to watch a 16 minute video.

Better yet, just draw a schematic of the circuit in question.
I am beginner at building electronic circuits and this is my first post so sorry if i asked stupid question or made mistakes writing it.
Your first mistake is relying on YouTube. There are a lot of stupid so called experts there. Go for a text book, school material, or a peer reviewed site like AAC.
 

AnalogKid

Joined Aug 1, 2013
10,054
The guy's explanation of the circuit was incomplete. This graphic shows *both* input transitions.

Note that without additional parts, the negative-going input transition causes an output that goes *below* GND. This is useful in some power conversion circuits, but can harm the input stage of some logic families. A common fix is to put a small signal diode in parallel with R (anode to GND). The diode conducts when the input is at -0.6 V, "catching" the spike before it is large enough to hurt something.

ak

1651494523108.png
 
Last edited:

dl324

Joined Mar 30, 2015
14,459
The thing is i don't see it being discharged in any way. [snip] Could someone explain how it works?
The circuit is called a differentiator. Since the voltage across a capacitor can't change instantaneously, on the rising edge of the input, both side of the capacitor follow that rise. When it reaches its maximum, the side of the capacitor connected to the resistor starts discharging towards ground through the resistor.
\( \large V_f = V_ie^{-\frac{t}{RC}} \)

You could debate whether it should be called charging or discharging, but the waveform follows capacitor discharge behavior.

From RCA ICAN-6346:
1651505388338.png1651505424865.png
 
Last edited:

RPLaJeunesse

Joined Jul 29, 2018
213
AnalogKid stated "A common fix is to put a small signal diode in parallel with R (anode to GND). The diode conducts when the input is at -0.6 V, "catching" the spike before it is large enough to hurt something." Actually, most logic families (not all!) have built-in diodes to clamp an input pin to nominally the supply rails (plus a diode drop, of course). The thing to remember is that those diodes usually don't want to see much over a few milliamps of current. (Research "Latch-up" if you want to know why.) So there should be in series with the capacitor a resistor that is very much smaller than the timing resistor to the rail, so as to not affect the output swing. For a 5V supply and 2.5mA the series resistor should be 2K Ohms. That would mean the timing resistor should be up past 50K Ohms (Note: as low as 20K would probably work fine for driving 4000 series CMOS gate).
 

MrChips

Joined Oct 2, 2009
26,087
Another way of looking at this, a C-R circuit is a differentiator, i.e. it takes the derivative of the input signal.

Thus it responds to fast transitions of the input signal. A fast rising edge represents a positive derivative. A fast falling edge represents a negative derivative.

You can choose which edge you want to trigger your next circuit by shifting the DC bias up or down.
A pull-down resistor will select the positive going pulse.
A pull-up resistor will select the negative going pulse.


1651860960561.png
 

AnalogKid

Joined Aug 1, 2013
10,054
those <internal> diodes usually don't want to see much over a few milliamps of current.
That is the reason for an external high-speed diode. If the application is to turn a constant signal into a loooong pulse, a 100 uF capacitor can pop an internal diode, while an external 1N914 can take the hits all day.

AND - not all devices have internal protection. A classic, two-transistor multivibrator circuit running on 12 V can place over -11 V across the base-emitter jucnctions of the transistors, but typical small signal transistors are rated for only -5 V Vbe. Not good. Again, an additional diode clips the negative Vbe to -0.6 V, but in doing so it changes the oscillator frequency calculations.

ak
 
Last edited:

RPLaJeunesse

Joined Jul 29, 2018
213
AnalogKId replied "That is the reason for an external high-speed diode". I don't disagree but how do you guarantee the external diode take the current if the internal diode is in parallel? I always use an isolation resistor. And if you are using 100uF you better be using it into a Schmitt trigger or the circuit will never perform correctly. That's why monostable multivibrators and the NE555 were created.
 
Top