# Preparing Input and Selection voltages to a Multiplexer

I am trying to make a fast switching circuit that is able to switch a current (0-20mA) supplied by a current source (open-circuit voltage 100V) between two channels (i.e. switching between two pairs of electrodes). The load has a low resistance (100-250 ohhm). The current signal is composed of square pulses of (5-20ms) duration, one pulse every 300ms.

I have a CD4067BE 1x16 multiplexer operated at 15v/0v Vdd/Vss to minimize its on-state resistance. At 15v, channel selecting voltage should be at min. 11v for logic 1.

my questions:

How to limit the input voltage coming from the current source to MUX such it does not exceed 15v? can I use zener diode without any harm to the current source?

I am using MCU signals (0-5v) to control selection inputs of the MUX. However, at least 11v is needed to activate selection. What a driving circuit do you suggest?

You can put (say) a 12V zener diode across the output of the current source to limit the open-circuit voltage. It won't affect the 20mA current much at 5V. It will dissipate 1/4W if the output load was disconnected, so maybe use a 500mW zener. For example, for a BZT52C12, the maximum reverse current at 8V will be 100nA. Put it across the current source, not at the output of the mux.

20mA is really a bit high for the 4067 (absolute maximum is rated at 10mA) so I would suggest you pry the purse open a bit wider and use a proper analog switch with built-in level shifters, which will also solve your other problem. For example, DG408/9, but there are others.

• Thank you for the suggestion of DG409. However, I still need to limit the input voltage to the MUX using a zener diode, right? Do I need a series R before zener to limit max possible current through zener, or I should not worry as the current is already limited by 20mA?
– n.na
Aug 28, 2016 at 13:41
• The zener alone should work. Aug 28, 2016 at 13:54

... to switch a current (0-20mA) supplied by a current source (open-circuit voltage 100 V ... The load has a low resistance (100-250 ohm). The current signal is composed of square pulses of (5 - 20ms) duration, one pulse every 300ms.

20 mA into 250 Ω means that maximum load voltage will be $V = 20m \times 250 = 5~V$. This makes it unclear why you need a 100 V source voltage. I suspect that it is because that's the voltage you will need to strike or spark on the electrode.

In any case, at 100 V this chip is not the solution to your problem. You could use a Zener diode to limit the voltage but your electrodes may not work at the low voltage.

simulate this circuit – Schematic created using CircuitLab

Figure 1. An alternative control strategy.

You could consider digital switching of the source using a suitably high-voltage transistor driven by an opto-isolators which in turn are controlled by your micro.

• The 100v is the open-circuit voltage of the current source that I am using. This current source is capable of providing current for a wide range of load resistance, and thus it has this high open-circuit voltage. If I use a 250 ohm load, the voltage becomes 5V. It is an already designed current source circuit.
– n.na
Aug 27, 2016 at 18:48
• Yes, I've represented both by specifying a 100 V supply and a constant current source in the schematic. You haven't supplied enough detail in your question for me to go any further. Aug 27, 2016 at 18:56