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I am trying to understand more about electronics. To this end I started studying the Ben Eater videos. Brilliant. But there is something I don't understand about this circuit. Specifically it's to do with the reason why that capacitor will shorten the clock pulse.

schematic

simulate this circuit – Schematic created using CircuitLab

Link to a video that shows the capacitor. You can fast-forward to 3:19 to see it.

Another video where he's got the clock pulse up on the oscilloscope. This may be seen at around 9:21.

Why does the voltage after the capacitor slope down like that, and then stay low? It's as though the capacitor's resistance increases to infinity as it charges, but that can't be right. Otherwise I'd expect the inverse when CLK goes low and the cap starts to discharge and release its pent-up energy somehow.

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    \$\begingroup\$ The input circuit he's using is a differentiator, hence you see (approximately) the derivative of the clock input. There will also be a negative going pulse when the clock turns off unless there's protection circuitry inside the logic gates that prevents signals from going below 0V. \$\endgroup\$ – Captainj2001 Nov 25 '19 at 14:41
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    \$\begingroup\$ "capacitor's resistance increases to infinity as it charges" mostly correct. Capacitors will increasingly impede current flow due to a constant voltage applied to them hence they are used for DC block and AC pass. A mechanical analogy is pressure in an air tank pushing back as you try and fill it. \$\endgroup\$ – DKNguyen Nov 25 '19 at 14:41
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    \$\begingroup\$ The cap has a low value so he can charge very quickly before the pulse ends. \$\endgroup\$ – G36 Nov 25 '19 at 14:41
  • \$\begingroup\$ electronics-tutorials.ws/rc/rc-differentiator.html read this. \$\endgroup\$ – Captainj2001 Nov 25 '19 at 14:42
  • \$\begingroup\$ Caps. may be charged or discharged with a resistor and voltage with an estimate of T= RC (seconds) . They block DC since they are insulators unlike resistors which are conductors. But the behavior is exponential with T=RC as the estimate for time. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 25 '19 at 14:45
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To understand intuitively the operation of this CR differentiating circuit, think of the capacitor as of a kind of "rechargeable battery". So two voltage sources are connected in series... and this "battery" of two voltage sources is loaded by the resistor R.

1. Initial zero input voltage. Assume that initially there was no input voltage long enough and the capacitor is fully discharged. So there is no voltage applied across the resistor... and there is no current flowing through it...

2. Positive input voltage. The input voltage rises sharply but the voltage across the capacitor cannot change immediately... and is zero at the first moment. The input voltage source faces no obstacle; the whole input voltage is applied through the capacitor to the resistor and appears at the output (forming the positive "tip of the "needle").

CR differentiating circuit_1

The capacitor quickly charges... its voltage increases thus increasingly hampering the input source... and the voltage across the resistor (the output voltage) decreases. It is an interesting situation - the input source supplies the capacitor "source"... but the latter opposes it... and finally completely neutralizes its voltage (as in the proverb, "Feed a dog to bark at you":) The "battery" consists of two equivalent but opposing voltage sources. There is no current flowing... there is no output voltage.

3. Zero input voltage. The input voltage drops sharply to zero (short connection, not open circuit!). The capacitor "anti-voltage" cannot change immediately... and is equal to the previous input voltage. There is only one acting voltage source - the capacitor... its voltage is applied through the input source (short connection) to the resistor and appears at the output (forming the negative "tip of the "needle"). The current flows in an opposite direction... and the output voltage is negative...

CR differentiating circuit_2

The capacitor quickly discharges... its voltage decreases up to zero. There is no current flowing... there is no output voltage.

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  • \$\begingroup\$ There is no "anti-voltage"; it's just voltage. \$\endgroup\$ – Hearth Nov 25 '19 at 21:46
  • \$\begingroup\$ This is only a figurative name that helps understanding. "Anti-voltage" means opposing, contrary voltage. We can use it also when explaining op-amp circuits with negative feedback (e.g., the "antivoltage" at the inverting input of the non-inverting amplifier)... bootstrapped circuits, etc. \$\endgroup\$ – Circuit fantasist Nov 25 '19 at 22:01
  • \$\begingroup\$ It seems to me more likely to hinder understanding than help it. \$\endgroup\$ – Hearth Nov 25 '19 at 23:49
  • \$\begingroup\$ Thinking in this general way, you can "see the forest behind the trees"... i.e., the general idea behind different (at first look) specific implementations. Once seen the common idea, we can understand other unfamiliar circuits ... even invent new ones. This is how inventors think. Of course, both approaches - general and specific, are useful and needed. \$\endgroup\$ – Circuit fantasist Nov 26 '19 at 7:17
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As others have mentioned in the comments, the circuit you are looking at is called a differentiator.

The differentiator makes use of the properties of a capacitor to mimic mathematic differentiation.

Capacitors are (more or less) "transparent" to changes in voltage, but "opaque" to constant voltages.

That is, a changing voltage will cause current to flow through the capacitor.

In a square wave, the voltage only changes on the rising or falling edge. In between, the voltage is either constant high or constant low. On the edges, current flows through the capacitor because the voltage changes. In between, no current flows because the voltage is constant.

It is as you suspected: the capacitor sometimes acts like an open circuit, and sometimes like a low value resistor.

There is in fact a mathematical expression for the resistance (or, rather, impedance) of a capacitor for a given frequency:

$$Z = \frac{1}{2 \times \pi \times f \times C}$$

Where:

  • \$f\$ is the frequency in hertz
  • \$C\$ is the capacitance in farads.
  • \$Z\$ is the impedance in ohms.

A differentiator is also a high pass filter.

This circuit from the Wikipedia "High pass filter" page is the same as your differentiator:

enter image description here

Since any signal can be viewed as a sum of frequencies of many frequencies, you can view a differentiator as a high pass filter that removes the lower frequencies (and DC) from your signal. It passes only the high frequencies, and the sum of the high frequencies is a spike.

The wikipedia page on square waves has a nifty animation of building a square wave from many sine waves:

enter image description here

If you imagine leaving out the low frequency parts (the wider sine wave at the beginning) then you can see how a high pass filter can result in the spikes that a differentiator makes out of a square edged clock.


Note that the differentiator can only make spikes from a square wave.

If you need short spikes from a signal that looks like a sine wave, then you must first make a square wave out of it and then run it through a differentiator.

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An ideal capacitor does not have resistance, it has capacitance. A real capacitor does have some leakage resistance, but it is usually very high, far too high to make a significant difference in your circuit.

The fundamental equation of an ideal capacitor is.

$$I=C\frac{dv}{dt}$$

Where \$\frac{dv}{dt}\$ is the rate of change of voltage with respect to time. Note that the steady-state voltage does not appear in this equation at all. Under conditions of un-changing voltage no current flows in an ideal capacitor. Equally the voltage across an (ideal) capacitor cannot change instantaniously, since such a change would require infinite current.

So lets say the circuit is in steady state, then the input goes instantaniously from 0 to \$V\$. The voltage across the capacitor cannot change instantaniously, so initially all the voltage is dropped across the resistor. That means inititially the current is \$\frac{V}{R}\$

But that means the voltage across the capacitor starts to rise, initially at a rate of \$\frac{V}{RC}\$. More voltage across the capacitor means less voltage across the resistor, which means less current. The result of this is that the voltage on the output exponentially decays towards zero.

When the incoming clock returns to zero a similar thing happens. The voltage across the capacitor cannot instantaniously change, so the output voltage becomes negative, before again decaying towards zero.

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    \$\begingroup\$ I think they use "resistance" figuratively... in the sense of some opposition, impediment, obstacle... Indeed, it is not ohmic resistance since it does not dissipate power. Strictly speaking, both resistor and capacitor "steal" energy... but while the resistor immediately "spends" it, the capacitor "saves it for a rainy day"... \$\endgroup\$ – Circuit fantasist Nov 25 '19 at 22:21

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