4066 bilateral switch unused inputs and maximum voltage through switches? - Page 1
Forum section: electronics / beginners

author: dentaku
author_rank: Frequent Contributor
timestamp: February 24, 2014, 02:27:46 am
content: I was looking at the 4066b - quad bilateral switch datasheetsthe On Semi one is clearer as to what pins do what than the TI one I've found.http://pdf.datasheetcatalog.com/datasheets/320/206765_DS.pdfI know about tying unused inputs to ground but I'm still wondering what exactly is considered an input in the case of something like the 4066?I'm assuming it means the four CONTROL inputs but what about the switch INs themselves.Are they inputs that you would connect to ground if they are unused?ALSO: If I power it with 9V and use square waves for CONTROL I'm assuming I can't run a signal that is adjustable to be anywhere between -9V to +9V through the switch?because the datasheet says..."The maximum analog signal level is determined by VDD and VSS. The analog voltage must not swing higher than VDD or lower than VSS."Basically, I want to use a counter IC of some sort to send pulses to the CONTROL inputs which in turn switches four varying voltages (four voltage dividers) to a VCO's control voltage input. The problem is that to get the maximum range of frequencies from the VCO I need to give it -9V to +9v and I don't think the 4066 can handle this.

author: GK
author_rank: Super Contributor
timestamp: February 24, 2014, 03:11:40 am
content: 4066 will do +/-7.5 V max on bipolar supply rails. Look at something from the DG or ADG range to switch higher amplitude analogue signals - eg DG308. These devices also more conveniently reference the logic control inputs to ground when operating on bipolar rails.

author: amyk
author_rank: Super Contributor
timestamp: February 24, 2014, 12:11:20 pm
content: The principle behind what to do with unused inputs is to avoid having a floating gate. The switch inputs don't connect to gates.

author: sync
author_rank: Frequent Contributor
timestamp: February 24, 2014, 01:02:39 pm
content: Look at the 4051,4052, 4053 analog multiplexer. These will work up to +/-9V or +/-10V (depending on manufacture).

author: dentaku
author_rank: Frequent Contributor
timestamp: February 24, 2014, 02:02:56 pm
content: Aha... That's what I was expecting so it's nice to see someone here explain it in a simple way like that.Quote from: amyk on February 24, 2014, 12:11:20 pm

author: dentaku
author_rank: Frequent Contributor
timestamp: February 24, 2014, 02:15:52 pm
content: I DO have a 4051BE and a 4051BP plus a 4052BP and a 4053BE so I guess I have some reading to doI got all of these from the collection of an old man who died last year so I'm never exactly sure what I have in my bins.Basically I'm seeing if I can make a little step sequencer with counters, analogue switches and voltage dividers connected to a VCO control voltage input.VCOs, VCFs, VCAs are easy enough (LM13700s are nice) but I'm still learning about these old 4000 series ICs.Quote from: sync on February 24, 2014, 01:02:39 pm

author: edavid
author_rank: Super Contributor
timestamp: February 24, 2014, 04:51:56 pm
content: You should probably put an op amp stage in front of the VCO control input, so it can accept a more convenient voltage range.

author: dentaku
author_rank: Frequent Contributor
timestamp: February 24, 2014, 05:04:09 pm
content: Yeah. I just might do that.Quote from: edavid on February 24, 2014, 04:51:56 pm

author: edavid
author_rank: Super Contributor
timestamp: February 24, 2014, 05:43:56 pm
content: Where did people get the idea that it's OK to top post? Is it a mobile thing?

author: GK
author_rank: Super Contributor
timestamp: February 24, 2014, 10:58:42 pm
content: Maximum "nominal" supply voltage for the 4000 series is 15V. You can push it by running them at the absolute maximum ratings, but that generally isn't a very good design practice.Unless you are using rail-rail in/out op-amps, they will need at least a volt or two of headroom, so you won't be able to power them from the same rail potentials as the analogue switches, if you intend to run the 4000-series analogue switches to their limits. The DGXXX series on the other hand can run on rails up to +/-22V, so they can be powered directly from the op-amp supply rails. That's generally a lot more convenient.

author: edavid
author_rank: Super Contributor
timestamp: February 25, 2014, 01:25:53 am
content: Quote from: GK on February 24, 2014, 10:58:42 pmCD4000B series is rated for 18V supply voltage.

author: GK
author_rank: Super Contributor
timestamp: February 25, 2014, 01:39:34 am
content: 18V is the "absolute maximum" rating – device specifications are characterised to 15V max. You generally don’t run logic IC’s at their absolute maximum supply ratings as that gives little margin for tolerating things like switching glitches on the rails or any other possible glitch/variation.

author: edavid
author_rank: Super Contributor
timestamp: February 25, 2014, 03:07:16 am
content: Quote from: GK on February 25, 2014, 01:39:34 amNo, that is not true - 18V is allowed - see attached snippet from the CD4001B datasheet.

author: GK
author_rank: Super Contributor
timestamp: February 25, 2014, 03:29:50 am
content: I didn't say it wasn't allowed - I said 18V is the absolute maximum.Pay particular attention to the note about "nominal" voltages in that snippet you just attached.

author: edavid
author_rank: Super Contributor
timestamp: February 25, 2014, 05:51:31 am
content: Quote from: GK on February 25, 2014, 03:29:50 amThat is incorrect. The datasheet states clearly that 18V is a recommended operating condition, and 20V is the absolute maximum.

author: GK
author_rank: Super Contributor
timestamp: February 25, 2014, 06:17:01 am
content: And which manufacturer are you snipping bits of datasheet from?The absolute maximum given by ON Semi (just one example) for the 4000 series is 18V.If you are designing a circuit to be compatible with any 4000 series part of a specific part number, you are stuck to the lower max rated voltage.

author: amyk
author_rank: Super Contributor
timestamp: February 25, 2014, 10:27:46 am
content: Fairchild: 18V absolute max, 15V recommendedIntersil: 20V absolute max, no recommendation (although 18V and 15V are specified)NXP: 18V absolute max, 15V recommendedROHM: 20V absolute max, 18V recommendedST:22Vabsolute max,20VrecommendedTI: 20V absolute max, 18V recommendedToshiba: 20V absolute max, 18V recommendedUTC: 18V absolute max, 15V recommendedSo it would seem that not all 4000-series CMOS is created equal.

author: vk6zgo
author_rank: Super Contributor
timestamp: February 25, 2014, 10:42:10 am
content: Many years ago,I used 4066's in a project,& initially left the unused switch control pins open.The 4066's got very hot!I returned the unused switch control pins to zero volts.The bloody things still got hot!Finally,I organised a minus 6.8 volt supply,& returned the switch control pins to that.No more overheating!!Edit:-Sorry about that--hope it is less confusing now!

author: GK
author_rank: Super Contributor
timestamp: February 25, 2014, 11:42:36 am
content: Quote from: amyk on February 25, 2014, 10:27:46 amFor compatibility reasons a nominal supply voltage of 15V max. has endured as pretty much an unofficial standard when designing with 4000 CMOS.

author: SeanB
author_rank: Super Contributor
timestamp: February 25, 2014, 06:26:21 pm
content: As well with analogue switches connect the unused analogue switch pins to the common ground, as these are actually inputs to the internal gates and not driven outputs. They can float and cause latch up with stray voltage fields.

author: dentaku
author_rank: Frequent Contributor
timestamp: February 25, 2014, 10:10:04 pm
content: Quote from: SeanB on February 25, 2014, 06:26:21 pmInteresting, I didn't think that was the case.edit... oops I forgot the worddidn'tin my original post

author: edavid
author_rank: Super Contributor
timestamp: February 25, 2014, 10:46:13 pm
content: Quote from: SeanB on February 25, 2014, 06:26:21 pmWhat do you mean by this?Look at the schematic on the TI CD4066B datasheet - the in and out terminals are not tied to any transistor gates.http://www.ti.com/lit/ds/symlink/cd4066b.pdf

author: Neganur
author_rank: Unknown
timestamp: February 26, 2014, 05:26:39 pm
content: you should relax a little and focus on the content of discussionsQuote from: edavid on February 24, 2014, 05:43:56 pm

author: edavid
author_rank: Super Contributor
timestamp: February 26, 2014, 06:22:30 pm
content: Quote from: Neganur on February 26, 2014, 05:26:39 pmYou should stop being so relaxed and look at the format

author: vk6zgo
author_rank: Super Contributor
timestamp: February 27, 2014, 02:26:03 am
content: Quote from: edavid on February 25, 2014, 10:46:13 pmNo,but no matter what the mechanism,if you leave thecontrol pinsopen,the switch can draw current---& it's not always good enough to return them to common ground,see my posting above.

author: edavid
author_rank: Super Contributor
timestamp: February 27, 2014, 04:01:20 am
content: Quote from: vk6zgo on February 27, 2014, 02:26:03 amI was asking about SeanB's statement, but actually I didn't understand yours either. Surely the unused control pins should be tied to either VSSor VDD, like any other unused CMOS inputs?

author: SeanB
author_rank: Super Contributor
timestamp: February 27, 2014, 04:21:26 am
content: The control pins are CMOS gate inputs, but the analogue inputs and outputs are internally going to structures with parasitic diodes, and current through these can destructively latch up the chip, or cause the other switches in the package to misoperate. Thus these need to be connected to a defined logic state as well, and the best is to use the Vss pin.

author: edavid
author_rank: Super Contributor
timestamp: February 27, 2014, 06:11:10 am
content: Quote from: SeanB on February 27, 2014, 04:21:26 amIf the input and output pins are not connected to anything, how can current flow through them?Why wouldn't the same problem occur with e.g. a disabled three state output?

author: dentaku
author_rank: Frequent Contributor
timestamp: February 28, 2014, 05:03:27 pm
content: Quote from: sync on February 24, 2014, 01:02:39 pmI've looked into the 4051B, which I have two of, and it's exactly what I've been looking for but I'm still not sure if it's OK to send the output of 8 voltage dividers (8 pots) into the eight Y0 to Y7 independent inputs/outputs pins and out the Z common input/output pin then into the pitch control of my LM13700 based VCO.From what I can tell by looking at various manufacturer's datasheets I might only be able to put 10mA through the switches of the 4051?I'll probably need to buffer or amplify the signals before they go into the VCO with an opamp but I'm more concerned about knowing if I'm putting too much current through the switches.Obviously I'm making a simple 8 step sequencer here that sends anything from -9V to +9V into the VCO

author: edavid
author_rank: Super Contributor
timestamp: February 28, 2014, 07:29:50 pm
content: Quote from: dentaku on February 28, 2014, 05:03:27 pmIf you look at the resistance vs voltage curves, you'll see that even 10mA will give really weird results.QuoteWhy does your VCO draw more than 10mA at the control input? That sounds more like a CCO.

author: dentaku
author_rank: Frequent Contributor
timestamp: March 01, 2014, 12:13:21 am
content: Quote from: edavid on February 28, 2014, 07:29:50 pmThat's true, this VCO/CCO doesn't draw much current into the pitch control input (and through a 27K resistor) but I want to be able to use this sequencer for other things.If the Z common output pin of the 4051B is connected to an opamp buffer or non-inverting amplifier would thatensure that very little current could ever flow through the switch?I never though of calling it a CCO but I guess that's appropriate . It's pretty much exactly fig. 19 from this .pdf.http://www.nutsvolts.com/uploads/magazine_downloads/11/May%202003%20Ray%20Marston%20-%20Understanding%20And%20Using%20OTA%20OP-Amps.pdf

author: edavid
author_rank: Super Contributor
timestamp: March 01, 2014, 12:45:34 am
content: Quote from: dentaku on March 01, 2014, 12:13:21 amSure, no more than the op-amp bias current, which is negligible.

author: dentaku
author_rank: Frequent Contributor
timestamp: March 01, 2014, 01:27:19 am
content: Quote from: edavid on March 01, 2014, 12:45:34 amAha. I was driving back home and I thought about using an opamp which of course has a very high input impedance .

author: SeanB
author_rank: Super Contributor
timestamp: March 01, 2014, 03:55:51 am
content: Quote from: edavid on February 27, 2014, 06:11:10 amSame again, but generally a trisate output will have something like a terminator or other logic defining state, and these often have issues with disconnected cards and ESD events blowing them, or hot plugging killing them without protection added. Unused switches float, and if there is an electric field they can float to above latch up level just from the pin or stub traces on the board. Driven outputs are safe as they have a defined voltage and a relatively low impedance to keep the state defined and close to a rail.

author: SeanB
author_rank: Super Contributor
timestamp: March 01, 2014, 04:06:52 am
content: Quote from: dentaku on March 01, 2014, 01:27:19 amBuffer opamp is a very good thing to have at the output if you want best linearity out of the switch. Actually at each input as well if the source is a high impedance as well, but really only if you are switching at a faster rate and need fast sampling. Look at the datasheet where you will see on resistance versus applied voltage, where it is shown as a box with somewhat slanted sides. The on resistance of the switch is going to vary as the input voltage changes from -10 to +10V, and will give rise to a varying voltage drop across the switch if you are trying to draw any current through it. This resistance is not constant at all, and even depends on the current flowing as well. Fine for switching digital signals or for high impedance inputs changing slowly but a big trap if you are driving an unbuffered ADC input, and gives rise to an error that depends on the last input's voltage difference and the current switch input unless you allow a long settling time after selecting to starting the conversion/hold of the ADC from the charge balancing. This also gives a charge dump back out of the switch to the selected input which can be very confusing.

author: dentaku
author_rank: Frequent Contributor
timestamp: March 01, 2014, 04:04:15 pm
content: Quote from: SeanB on March 01, 2014, 04:06:52 amWow, that's alot to think about for a beginner but I think I get it looking at the graphs. I'm only going to connect a string of voltage dividers to it with a pot for each step which then get switched in sequence by the multiplexer into the VCO. Noting high speed at all OR accurate.You mention -10 to +10. Does that mean that I CAN switch the full -9V to +9V my oscillator uses for pitch control?Does this mean that I can have VDD=+9V, VSS=0V and VEE =-9V? (would it have to be -9V?)I know this thread is going on forever but the only 4000 series ICs I've ever used are the 4029 counter I'm using in his circuit and a 4511 BCD to 7 segment driver.

author: fcb
author_rank: Super Contributor
timestamp: March 01, 2014, 07:11:52 pm
content: It would help if you published your proposed circuit.Depending on your exact VCO config with the LM13700, I would expect you are feeding a control current via the control bias inputs (pins 1 & 16). If this is the case, then you shouldn't hammer 10mA through there!What it sounds like you are building (8+ step sequenced synth) is usually done with a V>C converter between the sequencer and the VCO.So a typical implementation would be:8x potentiometers wired across 0v & 5v, pot wipers feeding the inputs of a 4051, output of 4051 feeding (possibly by buffer opamp) a V>C converter (this can be as simple as a resistor!) - I use current mirrors though as this makes a nice transition between the 0-5V world of the controls and the +/-Vcc world of the 13700.

author: dentaku
author_rank: Frequent Contributor
timestamp: March 02, 2014, 02:20:15 am
content: Quote from: fcb on March 01, 2014, 07:11:52 pmYou explained exactly what I'm trying to do. I'm guessing you've built synth modules before.It's an 8 step up/down sequencer that puts out + and - voltages built out of just parts I happen to have.The oscillator I'm using at the moment isn't the most important thing. I hope to be able to use the sequencer for modulating the pitch of different oscillators or the cutoff of any VCF or the volume of any VCA etc... A sequencer that's built to only control one VCO wouldn't be terribly useful.Anyway.- The 74c14 controls the clock speed (tempo of the sequencer) and I've already got that working.- The 4029B counts up and down just fine because at the moment I already have it set up and connected to a 4511 BCD to 7 segment driver and it all works- The LM13700 oscillator seems to work just fine and gives me nice square and triangle waves over a huge range of frequencies running off of +/- 9V- The 4051 is the only part I wasn't sure about.Here's what I came up with for images of this thing.

author: sync
author_rank: Frequent Contributor
timestamp: March 02, 2014, 03:11:49 am
content: Some remarks.Don't let the 4051's INH input floating. Connect it to ground.The TL072 is not a rail-to-rail op amp. You can't use the full +/-9V range with them.The output voltage varies with your power supply voltages. If you use batteries this may become a problem.Your VCO has a linear relationship voltage to frequency, eg. 1000Hz/V. This is unfavorable for music applications because the semitones don't have same voltage difference to each other.Normally analog synths uses a logarithmic relationship. Standard is 1V/octave. Then a semitone interval is always 83.3mV. This also eliminates your problem that you need such a wide range for the control voltage.You forget LEDs for indicating with pot is currently active.

author: dentaku
author_rank: Frequent Contributor
timestamp: March 02, 2014, 04:01:24 am
content: Quote from: sync on March 02, 2014, 03:11:49 amYup, glad you caught thatQuote from: sync on March 02, 2014, 03:11:49 amI noticed that it was more like +/- 7.5V but I guess that's not a big deal at the moment, especially at the top end because the VCO goes way up into not terribly useful frequencies anyway. I would be nice to be able to go rail to rail but I don't think I have an opamp that can do that. I've noticed that the plain old 358, 324 can at least got all the way to the negative rail unlike the TL072/082.Quote from: sync on March 02, 2014, 03:11:49 amSomeday I will have a real power supplyQuote from: sync on March 02, 2014, 03:11:49 amI see people talking about exponential converters and such but I haven't gotten that far yet I guess.It would be great to not need +/- 9V to get the full rage of frequencies though.Quote from: sync on March 02, 2014, 03:11:49 amI'll have to add that to the schematic tomorrow. I wish I had a nice transistor array or something like that so I wouldn't need to use 8 transistors.

author: edavid
author_rank: Super Contributor
timestamp: March 02, 2014, 04:14:13 am
content: Quote from: dentaku on March 02, 2014, 04:01:24 amMaybe you don't - what supply do you want to use? Try hooking the VCO up to that.QuoteYou could use another 4051 if you have one.

author: fcb
author_rank: Super Contributor
timestamp: March 02, 2014, 03:56:17 pm
content: Build your sequencer around 0-5v or at most 0-10v, try and keep to a 1v/octave standard internally - you'll thank me later. Use a 7805.At least that way you won't have your control voltages being dependent on battery state-of-charge, and you won't waste as much power heating up your pots.

author: dentaku
author_rank: Frequent Contributor
timestamp: March 02, 2014, 07:14:58 pm
content: Quote from: edavid on March 02, 2014, 04:14:13 amI'd love to use a simple 9V 1.5A regulated adapter I have. Most decent oscillator designs use +/- voltages though.Most of the time people use +/- 12V or +/- 15V for synth modules but since this is just an experiment I can use whatever is available to me.Quote from: sync on March 02, 2014, 03:11:49 amQuote from: dentaku on March 02, 2014, 04:01:24 amQuote from: edavid on March 02, 2014, 04:14:13 amConsidering the 4051 switches look like they can only handle 10mA wouldn't that be a little low for lighting LEDs?

author: edavid
author_rank: Super Contributor
timestamp: March 02, 2014, 07:22:33 pm
content: Quote from: dentaku on March 02, 2014, 07:14:58 pmYou could think about just using a 4046 VCO for now.QuoteShould be plenty unless they are really old LEDs.

author: dentaku
author_rank: Frequent Contributor
timestamp: March 02, 2014, 07:35:28 pm
content: Quote from: edavid on March 02, 2014, 07:22:33 pmI DO have a little SAW wave VCO on a breadboard right now that runs on one 9V battery just fine but it's very susceptible to noise on the power rails.I guess all I can do now is built it and test it out on my oscilloscope and see if I get square steps at the voltages I'm expecting. Maybe start with using a 7805 and only sending 0V to +5V through the switches at first to prove it works.This thread has to end sometime

author: dentaku
author_rank: Frequent Contributor
timestamp: March 03, 2014, 05:21:06 pm
content: One last thing about the 4051B.In my simulations it all works fine but VDD, VSS and VEE are hidden pins so it gets confusing.In the Toshiba TC4051B datasheet it says"The digital signal to the control terminal turns "ON" the corresponding switch of each channel, with large amplitude (Vdd-Vee) can be switched by the control signal with small logical amplitude (Vdd-Vss). For example, in the case of Vdd=5V, Vss=0V and Vee=-5V, signals between -5V and +5V can be switched from the logical circuit with single power supply of 5V. As the ON resistance of each switch is low, these can be connected to the circuits with low input impedance."Does this mean that I should connect the Vss pin to 0V and the Vee pin to -9V? Or should Vss and Vee just both be 0V?

author: edavid
author_rank: Super Contributor
timestamp: March 06, 2014, 06:14:40 am
content: Quote from: dentaku on March 03, 2014, 05:21:06 pmWhatever you want...VSSof the 4051 connects to VSSof the logic circuit driving the select lines (the 4029, in your case).VEEconnects to the negative supply voltage of the analog circuit driving the mux inputs (the pots, in your case).

author: dentaku
author_rank: Frequent Contributor
timestamp: March 06, 2014, 12:08:17 pm
content: Quote from: edavid on March 06, 2014, 06:14:40 amOK, that's good to know. Now I have a better image in my head of how it works.