# Analog Switch question

Discussion in 'Electronic Basics' started by MRW, Jun 27, 2007.

1. ### MRWGuest

Hello

I have an analog switch question. For this example, I would like to
use Texas Instrument's TS5A3159A SPDT Analog Switch.

For the digital input (triggers the switch ON or OFF), it has the
following electrical characteristics:

Input Logic High - 2.4 to 5.5 V
Input Logic Low - 0 to 0.8 V

I have the following questions:

1. What happens to the switch at 0.8 < Voltage at digital input < 2.4?
Does the switch either turns ON or OFF?

2. What happens to the switch if I leave the digital input open?

3. What other ways can I use to control the digital input
electronically? As of now, I can only think of using a comparator.

4. Would it be okay to leave the NC or NO terminal floating until a
signal source is connected to it?

Thanks!

2. ### John PopelishGuest

Possibly. Or possibly leaks a bit of signal through a high
resistance, and sucks power supply current. It is an ill
defined operating point, so generally you try to get through
it pretty quickly.
The input leakage current spec is a couple nanoamperes at
room temperature, and in either direction, which implies a
very high input impedance. If left floating, the voltage on
the input is indeterminate and may float to anywhere between
the supply rails, and if any static electricity is nearby,
it may get driven a bit beyond that. This is not a good
idea. If it is possible that the input may be disconnected
from its logic signal, even briefly, with the power on, I
would connect a pull up or pull down resistor that is a low
enough resistance to make sure the input voltage stays on
one side or the other of the .8 t o2.4 volts no man's land.
For instance, if you want it to default to the low voltage
state (less than .8 volts with the absolute maximum leakage
current of 100 nA trying to pull it positive) I would
connect a .8V/100nA=8 meg resistor from input to ground.
Actually, I would probably use a 1 meg resistor, since that
Logic gates and inverters can also be used.
Yes.

3. ### MRWGuest

Awesome, thanks again for your help, John!

4. ### MRWGuest

I actually have another question regarding analog switches. I have
this image:

http://bayimg.com/OAcOnaABk

I'm trying to convert the voltage reference of a source in another
reference for the analog switch. If the source is referenced to
ground, then I would like to be able to reference it to Vcc/2 instead.

In my diagram, does it accomplish the task?

Also, for the capacitor values, I am assuming that I have to pick the
value based on:

- minimum frequency to pass thru the low pass filter
- frequency relationship is also dependent on the parallel combination
of the voltage divider resistors
- capacitor voltage rating is based on the voltage difference between
VCC/2 and reference voltage of the source

Are these assumptions correct?

Also, for audio bypass capacitors, what other characteristics should I
know?

Thanks!

5. ### John PopelishGuest

It is one general way to do it.
Yes, C1 and the parallel resistance of the two resistors
connected between it and fixed voltages (assuming the analog
switch sends signal to a high impedance load) have an RC
time constant. At frequency 1/(2*pi*R*C) the response of
this low pass filter will be 3 db down. More for lower
frequencies.
That would be the minimum. More doesn't hurt anything.
Yes.

But there are other considerations.

How much supply current will these dividers consume? How
regulated is that Vcc supply?

This divider scheme will add some of whatever that noise is
to the incoming signals.

load on the source impedances feeding it?

How high can these resistor values go, before leakage
current through the input capacitors and the switch shift
the DC operating point enough to cause some trouble?
Otherwise, higher resistance has only good effects.

In many cases, it is better to use an opamp to make a low
pass filtered Vcc/2 supply, and tie all the nodes that need
this reference voltage to that output, with single resistors.
They don't catch fire? It is a fairly undemanding
application. Perhaps you might define what *you* mean by
the term, "audio bypass" so we are all talking about the
same thing.

6. ### MRWGuest

This would just be a simple Ohm's law of the voltage divider right?
i = Vcc / (R_div1 + R_div2) ?

So there is a balance between picking a higher resistance and limiting
the voltage divider supply current draw, right?

I am assuming this would be the parallel combination of the voltage
divider resistors in series with the source impedance. In order to
resistance (i.e. source impedance and parallel combination in voltage
divider configuration). Is this a correct assumption?

I'm not to familiar with leakage current, yet. But is this the same
concept of adding a high resistance resistor to the input nodes of an
opamp? The input bias current of the opamp would "drop" a higher
voltage if the resistor value is high... (? .. I think).

Something like this one? I figured if I put a capacitor at the output
to ground, then that would filter out the higher frequency components.
I picked the capacitor value based on the parallel combination of the
three output resistors. Did I make the right assumptions for this
setup?

I was just referring to the capacitors at the input terminals (Mic in
and CD in).

Thank you very much, John!

8. ### John PopelishGuest

Yes, times the total number of such dividers in the system.
Both the current draw and the ability of supply noise and
ripple to get into the signals through the divider get
better as the divider resistors go up in resistance.
That's right. The parallel impedance of your divider forms
a second signal divider with the source impedance as the
input side. For instance, if your two sources each have a
1k ohm impedance, then your two 1k resistors in parallel
form a divider with that source impedance that divided the
signal voltage down to .33 of its unloaded amplitude.
You want the bias divider to look like a high impedance,
compared to the signal source impedance. Having the bias
divider have a parallel impedance at least 10 times the
source impedance wastes less than 1/10th of your signal
amplitude.
Right. if you want to provide an accurate bias voltage to
an opamp input, you need the bias network to have a low
impedance compared to the opamp bias current, so that the
bias system dominates the voltage.

In this case, the leakages are the input capacitor, anything
that leaks from the supply into the switch channel (the data
sheet should spec that) and any bias current injected into
this node, by opamps or whatever that are downstream of the
switch, when it is switched on.

You may be able to use 100k or 1 meg resistors for your bias
network and still control the average bias voltage of the
signal passing through. You have to total up all the leakage
sources and also come up with an acceptable DC error voltage
you can tolerate, to calculate the maximum resistor values
you can use.
Yes, something like that. Though I think I would use 100k
or 10 k resistors for R1 and 2, and add a capacitor across
R2 to form a low pass filter for power supply noise.

9. ### John PopelishGuest

Oops. Get rid of the cap from the opamp output to ground.
It doesn't filter much, since the opamp output impedance
approaches zero, but it does tend to destabilize the opamp.

If they are large enough to hold essentially a constant
voltage during the audio signal swing, then linearity is not
important. And since there is a definite DC voltage across
them, electrolytics (aluminum or tantalum) could be used.
But if the divider impedance is high enough, film caps are
probably the best way to have low leakage through them.

10. ### MRWGuest

Thank you again, John!

11. ### neon

1,325
0
Oct 21, 2006
this is a logic switch takes vcc to turn it off and 1 to .8v to on. if the input is 2.4 or whatever it will just pass it thruogh or open output