Im building a circuit to switch the phase of a signal. Im using a circuit very similar to this:
If the gif isnt working, Its a non inverting buffer with 10 input and feedback resistors with a switch to ground from the positive input. This is set up to switch the phase of the signal when the switch is shorted.
Ive replaced the switch with a 4066 between the positive input of the opamp and ground.
Im getting confused over the power requirements of the chip.
I thought that a plus 5v minus 5v supply should have been suitable. My signals are 10v pkpk (its for a synthesizer, but this would equally apply if it was in a mixer).
At plus minus 5 the circuit switches but the top volt of the signal gets distorted. Why its only the top volt and not also the bottom volt i'm not sure. I would think this is some clamping diodes dont this.
I tried putting +12v and -5v on the chip. This fixes everything. It even switches from 5v logic. I feel this shouldnt work and is a bad solution.
Normally, a lot of these analogue cmos chips are run -12 to plus 5 so you get the lower on resistance and still get the 5v switching, like the CD4053 for instancce.
But the 4053 does have a dedicated ground.
Some circuits also suggest that at +/-5v the 0 logic level is still at 0v, but other circuits (including my -12/+5v test) suggest its more related to the Vss pin.
I think currently my best way forward is to use some transistor level shifters to move logic 1 to 12v and power the chip from -5v to +12v.
What am I missing?
If the gif isnt working, Its a non inverting buffer with 10 input and feedback resistors with a switch to ground from the positive input. This is set up to switch the phase of the signal when the switch is shorted.
Ive replaced the switch with a 4066 between the positive input of the opamp and ground.
Im getting confused over the power requirements of the chip.
I thought that a plus 5v minus 5v supply should have been suitable. My signals are 10v pkpk (its for a synthesizer, but this would equally apply if it was in a mixer).
At plus minus 5 the circuit switches but the top volt of the signal gets distorted. Why its only the top volt and not also the bottom volt i'm not sure. I would think this is some clamping diodes dont this.
I tried putting +12v and -5v on the chip. This fixes everything. It even switches from 5v logic. I feel this shouldnt work and is a bad solution.
Normally, a lot of these analogue cmos chips are run -12 to plus 5 so you get the lower on resistance and still get the 5v switching, like the CD4053 for instancce.
But the 4053 does have a dedicated ground.
Some circuits also suggest that at +/-5v the 0 logic level is still at 0v, but other circuits (including my -12/+5v test) suggest its more related to the Vss pin.
I think currently my best way forward is to use some transistor level shifters to move logic 1 to 12v and power the chip from -5v to +12v.
What am I missing?
You can't have the signal go outside the PS rails on CMOS, especially the RCA 4000 kind. It blows them eventually. .5 v over Vdd or -.5 v below Vss are absolute maximum limits.
The control voltage to the 4066 has to be referenced to the Vss pin. The resistances shown on TI's table are for the control pin equal to the Vss pin off and Vdd pin on. So your transistor level switcher on control may work being 1 v higher than Vss transistor on but may not achieve the resistances shown in the table. A nmos level shifter like 1n7000 may get closer to the Vss but have its own gate voltage problems.
Another solution is to shift the op amp rails with a diode from the absolute power supply rails. Then run the level shifter bjt emitter to the power supply rail instead of op amp rail. So if the bjt Vce is 1 v and the op amp rail is .6 v higher than the PS rail, then the control voltage to 4066 is 0.4 v when it is supposed to be off.
Note clamping your input to the op amp rail with diodes is good practice in PA service. Somebody might plug an instrument amp output, 20 vac, to the input of your board. You protect the diodes and op amp PS with a leading 100 ma fuse, or much cheaper, a 1000 ohm tenth watt resistor .
The control voltage to the 4066 has to be referenced to the Vss pin. The resistances shown on TI's table are for the control pin equal to the Vss pin off and Vdd pin on. So your transistor level switcher on control may work being 1 v higher than Vss transistor on but may not achieve the resistances shown in the table. A nmos level shifter like 1n7000 may get closer to the Vss but have its own gate voltage problems.
Another solution is to shift the op amp rails with a diode from the absolute power supply rails. Then run the level shifter bjt emitter to the power supply rail instead of op amp rail. So if the bjt Vce is 1 v and the op amp rail is .6 v higher than the PS rail, then the control voltage to 4066 is 0.4 v when it is supposed to be off.
Note clamping your input to the op amp rail with diodes is good practice in PA service. Somebody might plug an instrument amp output, 20 vac, to the input of your board. You protect the diodes and op amp PS with a leading 100 ma fuse, or much cheaper, a 1000 ohm tenth watt resistor .
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Although your idea is good, there are some better analog switches like the DG series. They have their internal level converter having +/- 12V and 5V independent pins making them easy to use. DG212 and similar. Google for them at your needs. There are normal open, normal close, at high or low input signals.
Hi we are using a dg408 elsewhere in the design but they are significantly more expensive, too much so to swap out all the 4066.
Here we are a few only shorting to ground so the sound quality it's much of an issue. I was hoping to maybe use some muting transistors but that also seems to have opened a can of worms.
+12 / -5 is beyond 15V end to end supply capability but you can get, say, +9V from your +12V existing rail
As mentioned above, best is +/-7V5 rails but I guess you want to keep it very simple.
As mentioned above, best is +/-7V5 rails but I guess you want to keep it very simple.
- CD4066 Pinout, Examples, Datasheet, Applications and Features
- Wide voltage supply range of 3V to 15V and maximum switching voltage is 20V
- Wide range of digital (15V) and Peak to Peak analog switching (±7.5 V)
- A high Output current of -0.4mA
- Control pin Voltage at low level is 0.9V (max) and at high level is 11V (min @ 15V Vcc)
- Noise immunity 0.45 VDD (typ.)
- High Output-Voltage Ratio of 80 dB typically at fis = 10 kHz, RL = 1 kΩ
- Very Low Internal Resistance
- Crosstalk between switches is –50 dB at fis = 8 MHz and RL = 1 kΩ which is very low
I am not much for opamps, they are useful for somethings but I prefer discrete devices. So how about a mosfet, used like one on Nelson Pass's designs, with constant current sources around it. Input on the gate, take the output from the drain or source for in-phases or out-of-phase signal, switch on either output. Just an idea. Enjoy.
Hi just 1 microrelay and it works OK. Opamps need input caps and RC filters.
CD4066 and TL072 are long in the tooth in 2021.
CD4066 and TL072 are long in the tooth in 2021.
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True.
But cheap and plentiful.
1 x 4066 replaces four microrelays, at far lower cost and is definitely easier to activate.
And in any case we are talking shorting to ground and Synthesizers, easy peasy non critical job.
Your signal levels are too high for that.
We are also using tl064 rather than 7series as they use about 1/10th the power. The voice card has around 90 opamps on and the polysynth will have 8 cards.
So if I run this Cd4066 minus 7.5 to plus 7.5, to make my level shifter, I would connect the transistor between those rails but still switch the transistor with 0-5v?
My transistor based level shifters work a treat (albeit inverted, but that's not an issue) to switch 12v logic elsewhere on the board.
So if I run this Cd4066 minus 7.5 to plus 7.5, to make my level shifter, I would connect the transistor between those rails but still switch the transistor with 0-5v?
My transistor based level shifters work a treat (albeit inverted, but that's not an issue) to switch 12v logic elsewhere on the board.
As the switches have one point grounded, perhaps a simple JFET whose gate is between 0 and say, -12V can do the job.
I tried simulating with transistors and fets but didn't have much luck. The other problem there is I would have to translate the 0 to 5v coming out the shift registers to 0 to - 12v.
It's starting to look as though the - 7,5 to +7.5 will be easiest with a transistor level shifter.
From what someone said, (I think this is where some of my confusion has come from), I can use a higher than rail voltage for the logic. So if I ran - 7.5 to 7.5v and I put a - 7.5 to12v logic line in that would be under 20v.
That way I can put the level shifter up by my shift registers and I would only have to run the minus 7.5 up there.
That may or may not be what I end up doing.
It's starting to look as though the - 7,5 to +7.5 will be easiest with a transistor level shifter.
From what someone said, (I think this is where some of my confusion has come from), I can use a higher than rail voltage for the logic. So if I ran - 7.5 to 7.5v and I put a - 7.5 to12v logic line in that would be under 20v.
That way I can put the level shifter up by my shift registers and I would only have to run the minus 7.5 up there.
That may or may not be what I end up doing.
If I only have 2 of these devices that need +/-7.5v, could one use a potential divider off the 12v rails? I imagine I could use relatively high values as I don't think this chip uses much power. Or is that a hideous suggestion.
Just follow the datasheets 🙂
They show how to connect them and, most important, needed control signals.
Here´s one example, using a single supply.
As in all cases , control signal must swing from most positive to most negative pins voltage.
CD4016 - A CMOS Chip With Four Analog Switches
They show how to connect them and, most important, needed control signals.
Here´s one example, using a single supply.
As in all cases , control signal must swing from most positive to most negative pins voltage.
CD4016 - A CMOS Chip With Four Analog Switches
I use zener diodes for low current loads. Like 6 4588 op amps, a couple of 1n5344 to turn 18 vdc from a hedge clipper wall supply to +-8. 22 ohm series resistors to hold the current down to wattage level. You could use 1 watt zeners probably. Couple of electrolytic filter caps out, plus maybe a .1 uf disk to the middle.
Don't know why you are reducing to +-7.5, 1n4066b will take 20 v Vdd-Vss or +-10. There are $2 analog switch ICs that will take 30 volts or higher. https://www.newark.com/renesas/dg411djz/analog-switch-quad-spst-dip-16/dp/79K6182
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I don't know why I would reduce to 7.5v, someone said I had to,but just before reading this, I've had a look at the data sheet and vss to vdd can be up to 20v although 18v is recommended.
I'll do some experimenting but if I can get away with - 5 +12 that would be good. Just need to make sure I don't end up clipping the negative side. I'm using this chip for several different purposes within the design so not all have this problem. The sync circuit is 0 to 12v with a transistor level shifter, for instance.
But I do have switching round the filters. This may be a problem there.
I'll do some experimenting but if I can get away with - 5 +12 that would be good. Just need to make sure I don't end up clipping the negative side. I'm using this chip for several different purposes within the design so not all have this problem. The sync circuit is 0 to 12v with a transistor level shifter, for instance.
But I do have switching round the filters. This may be a problem there.