8
\$\begingroup\$

Figure 1.41, Art of Electronics

I’m reading through The Art of Electronics and am puzzled by the use of the term “current-sensing resistor” as it pertains to differentiator circuits. In the above figure, the author explains that a perfect differentiator could be made from a capacitor alone, although the current through the capacitor could not be used as an output. So, he says a “current-sensing resistor” must be added. My questions are:

1) What does the term “current-sensing” mean in this context?

2) Why is the resistor even necessary? I’m having trouble understanding intuitively why this is.

\$\endgroup\$
10
\$\begingroup\$

What does the term “current-sensing” mean in this context?

When a current flows through a resistor, a proportional voltage difference appears between its ends, according to Ohm's law. Thus, a resistor converts a current to a voltage. If one end of the resistor is connected to GND, as here, then the voltage with respect to GND is proportional to the current flowing to/from GND through the resistor.

There are caveats, of course:

  • If the current is not from an ideal current source, but from a circuit whose behavior depends on the voltage across it, then the behavior of the circuit changes compared to if the resistor were just a wire (0 Ω, 0 V voltage difference).

    In this particular case, the current is determined by the time derivative of the voltage across the capacitor. But by introducing the resistor, we've split the voltage \$V_{in}\$ between the capacitor and the resistor, so we've changed the current (see Chu's answer for the exact effect).

  • The resistor dissipates energy and gets hot. To minimize this effect (if the current is unchanged), you use a lower value resistor, but that means that the voltage is correspondingly smaller and more subject to noise. In particular, the resistor must be large compared to the resistance of the power supply wiring.

  • If you're measuring the voltage then that implies you're connecting it to some other circuit element, through which some amount of current also flows, disturbing the characteristic ("loading the circuit"). Hence, you want to connect \$V_{out}\$ to a high-impedance input like an operational amplifier or other buffer circuit.

Most of the above applies to any current-sensing resistor, not just ones that are part of a differentiator. Common uses of current-sensing resistors are in devices that measure current, like multimeters and energy usage meters. When they are large and low value, to handle large currents, they are often called "current shunts" instead.

Why is the resistor even necessary?

As I mentioned above, decreasing the resistance decreases the voltage output proportionally. Imagine decreasing it all the way to zero; then the output is zero volts — the upper \$V_{out}\$ terminal is connected directly to ground, and you've turned circuit B into circuit A. So you've eliminated the resistor but also eliminated your output voltage.

\$\endgroup\$
4
\$\begingroup\$

For a capacitor, \$\small i=C\large \frac{dv}{dt}\$, but we would like a voltage signal as the derivative because it's more convenient ... so pass \$\small i\$ through a resistor and measure the voltage across the resistor.

Unfortunately, adding the resistor to the circuit means that the current is no longer an exact derivative of voltage, so we must tolerate an error; but this is minimised if the resistor is a small value.

The transfer function of the circuit, \$\frac{V_{out}}{V_{in}}= \frac {RCs}{1+RCs}\$, approximates to \$\small \frac{V_{out}}{V_{in}}= RCs\$ if \$\small RCs<<1\$.

It's not all bad news, however. Differentiation is inherently a noise-amplifying process, and the \$\small (1+RCs)\$ denominator represents a low-pass filter with corner frequency, \$\small \omega_c=\frac{1}{RC}\$ rad/sec.

\$\endgroup\$
  • \$\begingroup\$ Good but I think you meant... \$RC <<\frac {1}{s}\$, \$\endgroup\$ – Sunnyskyguy EE75 Jun 24 at 2:10
  • \$\begingroup\$ @SunnyskyguyEE75 Yep, corrected. \$\endgroup\$ – Chu Jun 24 at 7:37
3
\$\begingroup\$

The circuit in A differentiates the input voltage signal but the output is a current that can't be easily used for following stages.

To convert the current into to voltage that can be easily used you can use a resistor shown in B - the voltage across the resistor is proportional to the current flowing through the resistor. This is a "current sensing resistor" its job is to convert a current into a voltage.

However the presence of the resistor destroys the perfection of the differentiator so it is now only approximate. If the voltage across the resistor becomes a significant fraction of the voltage (Vin) there will be an error in the differentiation.

There are circuit arrangements to minimize this error that are explained later in the book. Usually they involve an active circuit known as an operational amplifier.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.