I am trying to measure the input resistance of the following multiplexer:


The reason I am measuring the input resistance of this multiplexer is that I have a signal coming out of my source and upon the signal reaching the input of my multiplexer the signal is distorted. My guess is that the output resistance of the source is high compared to the input resistance of the multiplexer. I want to be able to measure the input resistance of my multiplexer but I am not sure how. I have the following ideas but not sure which is correct:

I am using a push-pull circuit to model my input of my multiplexer:

Here VDD is 3.3V and VSS is 0V. To measure the input impedance of the entire multiplexer I tried the following:

enter image description here

Where the signal from my function generator was 1.8 Vpp, square wave, and I tried at frequencies from 1KHz to 8 MHz. This resulted in an decrease in impedance as the frequency is increased due to MOS gate capacitance. I measured V2 with respect to ground and V1 with respect to ground (V2 and V1 were measured peak to peak). Then Iin = (v2-V1)/R (in this case R = 4.7K). Then Rin = V1/Iin. Now i am not sure if this is the input impedance of my multiplexer or not? The reason I am not sure of this is I also wanted to add in a pull-down resistor to pull down my signal to ground. So I wanted to calculate the input resistance between the ground and input. Therefore I could add in a pull-down resistor to the input of the circuit.

That way this could help bring down my signal to ground (two resistors in parallel of about the same size would decrease the resistance in half) so i wanted to measure the resistance between the input and ground and then wanted to see what the resistance would be between the input and Vss(+) if I were to use a pull-up resistor as well (for another test). For input resistance between the input and VDD(+) I used a similar set-up as mentioned above expect instead of connecting my reference to ground I connected it to VDD(+) (3.3V). I am not sure if this is correct? Then for the resistance between the input and ground I realized that this would be the exact same as what I did for the input resistance of the entire multiplexer (first scenario). Therefore, I am confused as to how to measure the total input resistance of my multiplexer, the input resistance seen between my input and ground, and my input resistance seen between my input and VDD.

  • 1
    \$\begingroup\$ What does the data sheet tell you and, don't forget to consider dc bias currents of up to 1 nA. \$\endgroup\$
    – Andy aka
    Dec 7, 2015 at 23:08
  • \$\begingroup\$ I would like to know how to measure it since I could not find anything on the datasheet. thank you. \$\endgroup\$ Dec 7, 2015 at 23:13
  • 1
    \$\begingroup\$ I think the only resistance you need to be concerned with from the mux is the "on resistance" which is in the data sheet under the DC electrical specs, there's also a curve. \$\endgroup\$ Dec 8, 2015 at 1:11

2 Answers 2


An issue probably causing the signal distortion you describe is that the logic inputs must always be positive relative to ground but not more than the logic supply voltage, nominally +5V. 74LS TTL and 74HCT CMOS logic families should not have their outputs driven negative relative to ground nor more positive than the logic supply voltage. In fact, voltages above supply voltage or below ground may damage the input pins of the logic devices if the driving source impedance is low impedance.

Your signal generator output is probably bipolar, both positive and negative relative to ground. Most signal generators provide 50 Ohm output impedance which is possibly low enough impedance to damage the inputs of either LS TTL or HCT CMOS logic devices. If your signal generator provides output voltage offset capability, setting offset to +1.6V and amplitude around 3.3V peak to peak will make the output signal compatible with LS TTL or HCTL CMOS inputs and likely solve the signal distortion problem.

74HCT CMOS logic family is CMOS logic compatible as drop-in replacement for LS TTL logic. Input is capacitive around a few picofarads. The H means high speed comparable to or faster than LS TTL, The T in HCT CMOS logic means that the input switching thresholds and output drives are compatible with TTL logic. HCT CMOS logic device inputs include protection diodes from ground to the input pin and from the input pin to the logic supply voltage. This clamps the input signal to a voltage range of about -0.6V to logic supply voltage plus 0.6v. These diodes conduct to protect the device inputs in case of static discharge and are actually more important when the devices are not installed in circuit assemblies. Modern 74HCTL logic can operate with 3.3V supply voltage and inputs and outputs are still compatible with 74LS TTL running from 5V supply voltage. CMOS outputs drive both high and low levels very close to the supply rails.

CMOS logic which is not TTL compatible switches at half the supply voltage. These inputs might not switch or might respond erratically when driven from LSTTL logic device outputs.

A common ad hoc scheme for driving inputs from an external signal is to use capacitive coupling to the input with 220 Ohm pull up and 330 Ohm pull down resistors on the input pin. This biases the input to a 3.0V, a logic high level and provides impedance around 130 Ohms, which is close to the characteristic impedance of twisted pair cables or printed circuit traces. If you want a more symmetrical response to input signals or more sensitivity to small signals at the expense of greater noise sensitivity you may reverse the configuration of the pull up and pull down to bias the input pin to about 2.0V. Either way, this imposes a maximum dwell time at a low logic level input from the RC time constant of the blocking capacitor and the biasing resistors

The correct way to condition an input signal to TTL levels is a high speed comparator with output pulled up to the logic supply voltage.

The pull up resistor for the comparator or biasing pull up plus pull down resistors should be placed near the logic device input pins to prevent signal reflections, which can also distort the input signal. Comparator inputs should likewise use a termination resistor to absorb reflections at the signal inputs coming from the signal generator. These resistors should match the characteristic impedance of the cable connecting the signal generator to the comparators. The reflection occurs at signal level transitions and can make the transition as much as twice the transition voltage. The Schottky diodes added to the inputs of LS TTL, low power Schottky logic devices, are put there to damp the over voltage and under voltage spikes at the device inputs caused by the impedance mis match from the characteristic impedance of the circuit wires or circuit board traces to the input impedance of the driven logic devices.


I'm not going to tell you how to calculate it because the data sheet gives better ideas. There's usually something in the data sheet that gives a clue such as this graph: -

enter image description here

This graph tells you how much feed thru there is on a switch that is "open". The measured/predicted value is across a 50 ohm terminator on the "far side" of the input. As expected, as frequency rises feed-thru rises.

At 10 MHz there is about -55dB feed thru or in voltage terms, if the input were a 1Vp-p sinewave, the output is 1.78 mV. This implies a 50 ohm current of 35.6 uA RMS and an input impedance of 28.1 kohm.

So, input impedance can be calculated but it depends on several factors such as what is connected to the output and input capacitance. Given also that there are input leakage currents to consider (circa 1 uA), it's a complex picture.

If you take the 1 MHz scenario, attenuation is ~ 75 dB and this is equivalent to 178 uV across the 50 ohms and a current of 3.56 uA RMS. If you take this figure and the previous figure you can see that in one decade the feed-thru has dropped to one tenth. This tells me that the dominant force behind the transfer of leaking signals is capacitive.

So, you have an impedance of 28.1 kohm at 10 MHz and this translates to a capacitance of 0.57 pF.

So now you have a fairly reasonable model of the input pin - there's about 0.57pF to the output common pin and there's typically 5pF down to ground. Leakage currents are also present.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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