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Electronics Forum / Basics / June 2007Analog Switch question
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! > Hello > [quoted text clipped - 11 lines] > 1. What happens to the switch at 0.8 < Voltage at digital input < 2.4? > Does the switch either turns ON or OFF? 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. > 2. What happens to the switch if I leave the digital input open? 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 would add an insignificant load to the logic driver. > 3. What other ways can I use to control the digital input > electronically? As of now, I can only think of using a comparator. Logic gates and inverters can also be used. > 4. Would it be okay to leave the NC or NO terminal floating until a > signal source is connected to it? Yes. > > Hello > [quoted text clipped - 46 lines] > > Yes. Awesome, thanks again for your help, John! :-) 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! > I actually have another question regarding analog switches. I have > this image: [quoted text clipped - 6 lines] > > In my diagram, does it accomplish the task? It is one general way to do it. > 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 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. > - frequency relationship is also dependent on the parallel combination > of the voltage divider resistors Yes. > - capacitor voltage rating is based on the voltage difference between > VCC/2 and reference voltage of the source That would be the minimum. More doesn't hurt anything. > Are these assumptions correct? 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. What is the loading effect (attenuation) of this divider 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. > Also, for audio bypass capacitors, what other characteristics should I > know? 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. > But there are other considerations. > > How much supply current will these dividers consume? How > regulated is that Vcc supply? This would just be a simple Ohm's law of the voltage divider right? i = Vcc / (R_div1 + R_div2) ? > This divider scheme will add some of whatever that noise is > to the incoming signals. So there is a balance between picking a higher resistance and limiting the voltage divider supply current draw, right? > What is the loading effect (attenuation) of this divider > load on the source impedances feeding it? I am assuming this would be the parallel combination of the voltage divider resistors in series with the source impedance. In order to minimize the loading effect, the parallel combination must be a higher resistance (i.e. source impedance and parallel combination in voltage divider configuration). Is this a correct assumption? > 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. 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). > 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. 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? > > Also, for audio bypass capacitors, what other characteristics should I > > know? [quoted text clipped - 3 lines] > the term, "audio bypass" so we are all talking about the > same thing. I was just referring to the capacitors at the input terminals (Mic in and CD in). Thank you very much, John! > > 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 [quoted text clipped - 5 lines] > three output resistors. Did I make the right assumptions for this > setup? Forgot the link: http://bayimg.com/nACfNAaBl >> But there are other considerations. >> [quoted text clipped - 3 lines] > This would just be a simple Ohm's law of the voltage divider right? > i = Vcc / (R_div1 + R_div2) ? Yes, times the total number of such dividers in the system. >> This divider scheme will add some of whatever that noise is >> to the incoming signals. > > So there is a balance between picking a higher resistance and limiting > the voltage divider supply current draw, right? 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. >> What is the loading effect (attenuation) of this divider >> load on the source impedances feeding it? > > I am assuming this would be the parallel combination of the voltage > divider resistors in series with the source impedance. 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. > In order to > minimize the loading effect, the parallel combination must be a higher > resistance (i.e. source impedance and parallel combination in voltage > divider configuration). Is this a correct assumption? 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. >> How high can these resistor values go, before leakage >> current through the input capacitors and the switch shift [quoted text clipped - 5 lines] > opamp? The input bias current of the opamp would "drop" a higher > voltage if the resistor value is high... (? .. I think). 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. >> 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 [quoted text clipped - 6 lines] > setup? > http://bayimg.com/nACfNAaBl 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. >>> Also, for audio bypass capacitors, what other characteristics should I >>> know? [quoted text clipped - 7 lines] > > Thank you very much, John! >> http://bayimg.com/nACfNAaBl > > 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. 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. >> I was just referring to the capacitors at the input terminals (Mic in >> and CD in). 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. > >>http://bayimg.com/nACfNAaBl > [quoted text clipped - 15 lines] > But if the divider impedance is high enough, film caps are > probably the best way to have low leakage through them. Thank you again, John! :-) |
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