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I'm going to use a DG403 analog switch, supplied with +/-15V, to switch an audio signal below that range.

datasheet

The logic signal would be 0V/5V, so I need 5V on the VL pin of the switch. The logic signal is coming from another PCB, and there is no 5V source on the board with the switch.

I'm trying to understand if I can simply use a voltage divider from +15V (which is already stable and regulated) to provide 5V to the switch, or do I actually need to use a 5V regulator?

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A voltage divider would be fine, the part draws only a few microamps from V_L. V_L is used as a reference to decide the voltage at which the part decides if an input is High or Low. So it doesn't need to be exact.

As Spehro Pefhany notes, there's a possibility that the current drawn will be larger if your signals are slowly rising. Decouple your voltage divider with a capacitor (with a diode to V+ so you don't force SCR latchup when the power is turned off), to handle that.

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I would be a little careful with this one.

According to the datasheet, the current drawn from VL is maximum 5uA when Vin = 0V or 5V. If Vin is somewhere in between (and when switching) it might draw considerably more. This is one of those cases where the datasheet does not exactly lie, but you shouldn't read it with an optimistic mindset.

I suggest you either test it, especially if you expect inputs such as 3.3V/0V, or put in a little voltage regulator to prevent possible trouble.

Edit: See, for example, the 74HCT00 specification- power supply current with inputs stable at Vcc or GND is 20uA maximum, however with Vcc-2.1V inputs, the supply current can increase by 675uA per input (or 735uA per input over a wider temperature range), so you could have that chip drawing almost 6mA without any dynamic current at all.

enter image description here

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  • \$\begingroup\$ That's fair, looking at the equivalent circuit, it's certainly possible there could be a shoot-through. But some decoupling should easily handle that. \$\endgroup\$ – james Jan 10 at 21:40
  • \$\begingroup\$ @james Not just shoot-through, if the inputs are not close to the 0V/VL logic levels there could be (substantial) DC current flowing. \$\endgroup\$ – Spehro Pefhany Jan 10 at 21:59

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