In the test circuit I'm building there is a 1 kHz PWM signal passing through a DG467. As far as I know, the DG467 was placed there to interrupt the signal if needed, like an electronic on/off button controlled by a logic input. However, somehow this switch is deforming the signal (turning a perfectly square wave form in something that looks more like a curved sawtooth). A couple of (rather basic) questions of this component I'm not able to figure out for sure, so that I'm not missing something:

  • Is this component to be considered an 'ordinary' switch that should not impact the signal besides interrupting it? Or does it somehow do more to the signal that is going through that could explain the deformation I'm seeing?

  • The datasheet mentions the switch can operate with a dual power supply. Why would this be needed? I mean, shouldn't it be enough to use V+ or GND on the logic input to toggle it?

  • This specific component states "VL Logic Supply Not Required". But for example the MAX319 has a specific VL pin to support TTL 5 V level. Why would a third voltage (next to V- and V+) be desired? I mean, if this is an ordinary switch, why would it need this third voltage?


2 Answers 2


No, it's not an ordinary switch, quite far from it, so it is expected to deform waveforms for which is it is just unsuitable or it might be connected incorrectly.

An ordinary switch has almost zero resistance (in the order of milliohms) when connected and almost infinite impedance (above 10 megaohms) when disconnected, and can pass many hundred milliamps or amps even, depending on switch of course.

The switch you put there is basically an analog multiplexer which connects or disconnects two terminals with transistors.

The transistors are not perfect, they have multiple ohms of resistance, which is non-linear based on signal voltage, and can only pass about 10 milliamps of current. The chip also adds some capacitance, so it basically acts as an RC filter and limits bandwidth. The switch does have limited bandwidth, it's just rather high, but on the other hand, a square wave has bandwidth in theory up to infinity.

The transistors need positive and negative supply for the switch transistors that exceeds the signal voltage you are switching. This chip does not need a separate voltage for logic level input to control the switch, but some chips do have a separate digital voltage supply input, because you can also have different logic levels.

The chip does not need negative supply if you don't have a negative voltage in the switched signal. But the higher the voltages are beyond the signal, the FETs have less resistance and better switch the signal.

  • \$\begingroup\$ My heart took a small skip when I saw one end of the switch connected to a pin labeled NC. Had to drift my eyes down to the 2nd diagram where the same pin was labeled NO, before my heart returned to normal. ;) \$\endgroup\$
    – jonk
    Commented Aug 11, 2022 at 1:17
  • \$\begingroup\$ @Justme Thanks. So, if I understand correctly, the goal of this components is in fact to disconnect or connect the two pins, just like a mechanical switch. But the culprit is that it is using transistors and because of that there is more to it as the signal passes these transistors which might influence the signal (which of course won't happen with a normal switch). Is that corrrect (1) ? Also, given this, should this component specifically be able to handle a -12/+12v 1 kHz square wave when connected correctly (v- = -12, v+=12v) with no (or very minor) deformation of the signal (2) ? \$\endgroup\$
    – KSEEV
    Commented Aug 11, 2022 at 10:43
  • \$\begingroup\$ @KSEEV your signal source and load impedances, and thus current through the switch is unknown, and there is no headroom because supply voltage and signal voltages are same. \$\endgroup\$
    – Justme
    Commented Aug 11, 2022 at 10:51

CMOS switches are anything but simple. They are very easy to be made grossly non-functional as you've noted. They also take great care when switching precision analog signals.

The more ideal behavior you expect of the switch, the more careful you have to be in designing both the source that drives the switch input, as well as the load that the signal is input to.

  1. The "analog" power supply rails must completely encompass the signal range. Some switches also need a digital supply, some don't. But the analog supplies will always be there, and they determine the signal voltage range. Always refer to the specifications for the input voltage range vs. the supply voltage range. Some switches may require a margin between the signal voltage and supply voltage rails.

  2. There are two principal signals a switch can deal with, each with their own application requirements: voltage and current.

  3. In voltage switching applications, it is the value of the output voltage on the switch that matters. Then:

    1. For accuracy, the ideal source feeding the switch has low impedance, and the ideal load has high impedance.

      Accuracy and power transfer efficiency are two different things!

    2. "Zero" current should flow across the switch, due to its relatively large resistance compared to mechanical switches (many orders of magnitude higher!). Otherwise, there will be a voltage drop across the switch, degrading the accuracy.

    3. Switch resistance is non-linear and changes as the signal voltage changes. Thus, "zero" current operation is needed for highest voltage linearity.

    4. Incremental switch resistance grows with switch current. It will go up by several orders of magnitude as some maximum current is reached - this current may be quite small, say on the order of 10mA. When you load the switch heavily enough, it turns into a soft current source. The voltage accuracy goes straight out the window :(

  4. On the other hand, in current switching, it is the value of the output current that matters, or it may be the total output charge. Then:

    1. For accuracy, the ideal source feeding the switch has high impedance, and the ideal load has a low impedance - opposite of what voltage switching called for.

    2. The operating voltage is nearly constant, and selected to minimize charge injection. It is determined by the load side of the switch, since the source side only controls current.

    3. The switch resistance becomes a higher-order error source vs. in a voltage switching application.

    4. The current limit of the switch, as well as its input voltage range, must be respected by the current source. Keep the maximum current limit and voltage compliance range inside of what the switch can tolerate.

    5. If the switch drives an op-amp integrator input, the switch's charge injection is an error source. The relative contribution of charge injection to charge transfer error is highest when the charge is close to zero, i.e. when the current is around zero.

    6. If the switch drives a current-to-voltage converter, then it should be an op-amp summing amplifier input, since it maintains fixed switch voltage, just like an integrator input would.

      A resistor-to-ground is, to the first order, just as good, but unfortunately if you need precision, then higher order effects will matter, and variable switch voltage will degrade the performance. The charge injection affects the settling time, and with sufficient settling time (switch ON time), it will have a negligible effect.

    7. If the switch is considered the sole source of error, then - as a rule of thumb - current-signal switching is more accurate than voltage-signal switching. In general purpose switches, this difference can be easily an order of magnitude.

      But the switch is not the sole source of error. There are practical limits to the accuracy of the signal source and the switch load. The switch-induced errors may be more- or less-important than they look. Design and evaluate switched signal paths in target application, not as independent building blocks!

  5. Switch resistance of a particular switch model grows as the supply voltage drops. Most switches will operate at lowest resistance at the upper end of the recommended supply voltage range. Charge injection may be compromised though.

  6. Between switch models, the lower the maximum supply voltage, typically the lower the minimum switch resistance.

  7. Driving any switch channel outside of the power supply range may not only strongly leak from that channel to one of the supply rails, but may also cause multiple channels to become shorted, turned on, turned off, nonlinear, leaky, etc. A fault condition on one channel can propagate to other channels on the same chip! Some switches are protected to an extent from this: then it is an important feature and will be mentioned explicitly in the "features" section of the datasheet.

  8. Switch action necessarily injects some charge into the switch channel. Some switches are actively compensated for charge injection, and the charge injection is almost zero around some "ideal" switch voltage operating point - see relevant specs in the datasheet. Even switches without active charge injection compensation can be found with a variety of charge injection specs, spanning an order of magnitude or two at least.

    Charge injection is of particular concern in sample&hold circuits, integrators, slow load-side op-amps. It can be a source of nasty surprises in voltage switching when the load-side op-amp is subject to phase reversal.

  9. Multiple switches acting in unison may require pre- or post-delays relative to the controlling event, i.e. they may have to switch before or after the ideal switch command. They may also require dead-times when no switches conduct, to prevent cross-conduction, leakage/crosstalk between channels, etc.

    This is a complex enough subject to warrant its own Q&A. It's not critical in every application, but needs to be considered and determined to be non-critical. Usually if it's ignored, the circuit won't perform well - often at the extremes of operating voltage/temperature that are hard to test in.

  10. Always have a plan for what the load will do when the switch is open. Voltage-switching applications may benefit from an added small hold capacitance to at least absorb the charge injection - but if insufficient settling time is allowed, this will increase the crosstalk between channels in multiplexed applications!

  11. A/D input multiplexing switches on modern converters are nasty loads all by themselves. The transients they produce are often orders of magnitude faster than the bandwidth of the op-amp feeding the source, and if the source doesn't settle quickly enough, there will be apparent noise and/or nonlinearity and/or crosstalk when channels are switched.

    You may have an ADC that samples 8 channels at 10kHz/channel, for a total sampling rate of 80kHz, but the op-amps driving those inputs may require GBW of tens of MHz, slew rates of tens of V/us, and, preferably, specified (and quick enough!) settling time to the accuracy you expect of the ADC. The speed of the op-amp driving the input is independent of the sample rate: you need just as fast of an op-amp to drive that same ADC sampling at 1Hz as you would at 100kHz. That's because the transients are stereotypical, and independent of sample rate: they just repeat when each sample is taken.


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