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Old 01-05-2006, 04:11 AM
The Skeptic The Skeptic is offline
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Having read the basics on what a multiplexer does, I'd like to test a CD4053BE, which is a triple 2-channel. From the truth table, it seems that controls A, B and C decide between ax/ay, bx/by and cx/cy inputs, respectively, and I suppose the outputs are a, b and c. Apparently, the INH is somewhat of a general control (am I right so far?). But what are the in/out and out/in thingies on the connection diagram? And what are the VEE, VDD and VSS terminals?
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Old 01-06-2006, 06:15 PM
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beenthere beenthere is offline
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Hi,

The 4053 is three analog switches. As long as it is not inhibited, the a, b and c pins are tied to ax, bx and cx. If A, B or C is high, the a, b, or c pins are switched to the ay, by, or cy pins. Analog switches go either way, so the pins are either in or outputs. That way, you can go from one input to one of two outputs, or from one of two inputs to a single output.

Vdd is the logic level pin. It goes to the logic voltage bus. Vss is the logic common, and so goes to the logic common bus - usually 0 volts. Vee is the mux substrate. It is the negative level of the signal passed. If you're swtiching a waveform that goes from +5 to -5, then Vee will have to be at least -5 volts. If you're switching logic, then tie Vss to Vee. The maximum potential difference between Vdd and Vss/Vee is 15 volts.
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Old 01-06-2006, 08:01 PM
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Quote:
Originally posted by beenthere@Jan 6 2006, 05:15 PM
Hi,

The 4053 is three analog switches. As long as it is not inhibited, the a, b and c pins are tied to ax, bx and cx. If A, B or C is high, the a, b, or c pins are switched to the ay, by, or cy pins. Analog switches go either way, so the pins are either in or outputs. That way, you can go from one input to one of two outputs, or from one of two inputs to a single output.

Vdd is the logic level pin. It goes to the logic voltage bus. Vss is the logic common, and so goes to the logic common bus - usually 0 volts. Vee is the mux substrate. It is the negative level of the signal passed. If you're swtiching a waveform that goes from +5 to -5, then Vee will have to be at least -5 volts. If you're switching logic, then tie Vss to Vee. The maximum potential difference between Vdd and Vss/Vee is 15 volts.
Ahn... what is a voltage bus? According to http://www.st.com/stonline/products/...re/ds/8215.pdf , Vdd is the positive supply, Vss is the negative supply (those are things I understand) and Vee is simply the "supply voltage".

Beenthere, unfortuately I lack the knowledge to understand your answer on the voltages. I'm using DC and all I want is to be able to switch between two channels. Is that "switching logic"? Thanks a lot already.
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Old 01-09-2006, 12:13 AM
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beenthere beenthere is offline
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Hi,

When you have multiple devices in a circuit sharing a common voltage supply, the line is called the bus - everyone rides on it, so to speak.

For your purposes, Vdd is the positive logic voltage. Vss is the logic common, almost always 0 volts (ground). Vee is a substrate voltage for the analog switch. If you are passing voltages that go negative, then Vee has to be as negative, or the CMOS switch will fail.
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Old 01-09-2006, 09:59 PM
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Quote:
Originally posted by beenthere@Jan 8 2006, 11:13 PM
Hi,

When you have multiple devices in a circuit sharing a common voltage supply, the line is called the bus - everyone rides on it, so to speak.

For your purposes, Vdd is the positive logic voltage. Vss is the logic common, almost always 0 volts (ground). Vee is a substrate voltage for the analog switch. If you are passing voltages that go negative, then Vee has to be as negative, or the CMOS switch will fail.
Simple. self-explanatory and extremely useful. Thanks a lot, beenthere!
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Old 01-12-2006, 07:56 AM
Gadget Gadget is offline
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As with most 4000 series CMOS, supply voltage is 5 to 15 volts.
The 4051,52, and 53 (I think thats right from memory) as with the 4016 and 4066 (Quad bilateral switches) are normally used to switch analogue voltages... and commonly used as Audio switches.
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