4066 bilateral switch unused inputs and maximum voltage through switches?

[edavid]

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.

What 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

No,but no matter what the mechanism,if you leave the control pins open,the switch can draw current---& it's not always good enough to return them to common ground,see my posting above.

I was asking about SeanB's statement, but actually I didn't understand yours either.  Surely the unused control pins should be tied to either V_SS or V_DD, like any other unused CMOS inputs?

[SeanB]

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.

[edavid]

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.

If 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?

[dentaku]

Look at the 4051,4052, 4053 analog multiplexer. These will work up to +/-9V or +/-10V (depending on manufacture).

I'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

[edavid]

Look at the 4051,4052, 4053 analog multiplexer. These will work up to +/-9V or +/-10V (depending on manufacture).

I'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?

If you look at the resistance vs voltage curves, you'll see that even 10mA will give really weird results.

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.

Why does your VCO draw more than 10mA at the control input?  That sounds more like a CCO.

[dentaku]

Why does your VCO draw more than 10mA at the control input?  That sounds more like a CCO.

That'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 that ensure 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

[edavid]

Why does your VCO draw more than 10mA at the control input?  That sounds more like a CCO.

That'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 that ensure that very little current could ever flow through the switch?

Sure, no more than the op-amp bias current, which is negligible.

[dentaku]

Why does your VCO draw more than 10mA at the control input?  That sounds more like a CCO.

That'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 that ensure that very little current could ever flow through the switch?

Sure, no more than the op-amp bias current, which is negligible.

Aha. I was driving back home and I thought about using an opamp which of course has a very high input impedance .

[SeanB]

If 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?

Same 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.

[SeanB]

Why does your VCO draw more than 10mA at the control input?  That sounds more like a CCO.

That'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 that ensure that very little current could ever flow through the switch?

Sure, no more than the op-amp bias current, which is negligible.

Aha. I was driving back home and I thought about using an opamp which of course has a very high input impedance .

Buffer 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.

[dentaku]

Buffer 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.

Wow, 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.

[fcb]

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.

[dentaku]

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.

You 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.

[sync]

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.

[dentaku]

Some remarks.

• Don't let the 4051's INH input floating. Connect it to ground.

Yup, glad you caught that

• The TL072 is not a rail-to-rail op amp. You can't use the full +/-9V range with them.

I 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.

• The output voltage varies with your power supply voltages. If you use batteries this may become a problem.

Someday I will have a real power supply

• 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.

I 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.

• You forget LEDs for indicating with pot is currently active.

I'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.

[edavid]

I 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.

Maybe you don't - what supply do you want to use?  Try hooking the VCO up to that.

• You forget LEDs for indicating with pot is currently active.

I'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.

You could use another 4051 if you have one.

[fcb]

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.

[dentaku]

I 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.

Maybe you don't - what supply do you want to use?  Try hooking the VCO up to that.

I'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.

• You forget LEDs for indicating with pot is currently active.

I'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.

You could use another 4051 if you have one.

Considering the 4051 switches look like they can only handle 10mA wouldn't that be a little low for lighting LEDs?

[edavid]

I'd love to use a simple 9V 1.5A regulated adapter I have. Most decent oscillator designs use +/- voltages though.

You could think about just using a 4046 VCO for now.

Considering the 4051 switches look like they can only handle 10mA wouldn't that be a little low for lighting LEDs?

Should be plenty unless they are really old LEDs.

[dentaku]

I'd love to use a simple 9V 1.5A regulated adapter I have. Most decent oscillator designs use +/- voltages though.

You could think about just using a 4046 VCO for now.

Considering the 4051 switches look like they can only handle 10mA wouldn't that be a little low for lighting LEDs?

Should be plenty unless they are really old LEDs.

I 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

[dentaku]

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?

[edavid]

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?

Whatever you want...

V_SS of the 4051 connects to V_SS of the logic circuit driving the select lines (the 4029, in your case).

V_EE connects to the negative supply voltage of the analog circuit driving the mux inputs (the pots, in your case).

[dentaku]

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?

Whatever you want...

V_SS of the 4051 connects to V_SS of the logic circuit driving the select lines (the 4029, in your case).

V_EE connects to the negative supply voltage of the analog circuit driving the mux inputs (the pots, in your case).

OK, that's good to know. Now I have a better image in my head of how it works.