Hi,

I'm trying to run a signal through 1 of the 4 analog switches that are in de CD4066B package. The signal is a sine wave of about 1V peak to peak. When I tie the control terminal to V+ (9V) the signal passes through with no problems. When I tie the control terminal to V- (-9V) I can still see a mangled version of the signal at about 50mV peak to peak on my scope. That's 5% of the original signal which seems alot. Why isn't it cut off? Could it get in through the power supply which is 2 9V batteries?

The switch measures ~100Mohm when off with nothing attached to the CD4066 except power and a control voltage. So if it creates a voltage divider with the 5k trimpot I would expect something like 1V * (1 - 100Mohm/(100Mohm + 5kohm)) which would be much less than 50mV.

All unused terminals are connected to V-. Attached is a circuit diagram.

 

Try terminating the output pin of the 4066 with a resistor (1K to 10K) to ground.

 

I assume you are measuring the output with a 20MOhm DVM, yes?

Then what you really have is a divider consisting of 100M and 20M resistors, and you are measuring the voltage across the 20M.

 

Try terminating the output pin of the 4066 with a resistor (1K to 10K) to ground.

I'll give it a try tomorrow. But that would probably add alot of distortion to the signal wouldn't it? I was thinking about making it switch audio later on (and see if I can live with the 0.5% distortion that the 4066 would add).

I assume you are measuring the output with a 20MOhm DVM, yes?
Then what you really have is a divider consisting of 100M and 20M resistors, and you are measuring the voltage across the 20M.

I looked at the signal with a Tektronix 475A and I'm totally clueless about what impedance etc that scope has. If I remember correctly I got a similar reading with my UNI-T UT61E but again I have no idea if that is a 20M meter or not.

 

But that would probably add a lot of distortion to the signal wouldn't it?

No.
Go with Lestraveled on this one. "Absolutely perfect" results with no load at all are not normal with this kind of switch. Add a termination resistor as recommended in the datasheet.

One more thing...I have heard that the TL07x series is a less noisy chip than the TL08x series. Can't prove it with the datasheets I can find, but if you haven't bought the chips yet...just sayin'. I started with the TL082 and never had a problem. It's rated the same as the TL072. Probably no significant difference, but there was a rumor among the nerds.

 

The "on" resistance of a 4066 is below 100 ohms. Adding a 10K termination resistor will have little impact on the level (or distortion) but will have a huge impact on the "on" verses "off" ratio.

Another way to view the switches in the 4066 are variable resistors. When they are on they will read about 85 ohms. When they are off they will read many meg ohms. So, simple put, the external resistors, and scope input impedance, form voltage dividers with the switch resistance.

 

..............One more thing...I have heard that the TL07x series is a less noisy chip than the TL08x series. ............... but there was a rumor among the nerds.

It is not a rumor
The TLO71 generates 18nV of noise and the TLO81 generates 25nV of noise (per sq rt of bandwidth (Hz)). Those number means the TLO71 is quieter than the TLO81.

 

It is not a rumor
The TLO71 generates 18nV of noise and the TLO81 generates 25nV of noise (per sq rt of bandwidth (Hz)). Those number means the TLO71 is quieter than the TLO81.

I know it's true, but the datasheets I have don't show that. Bad datasheets!

 

With the resistor to ground it works perfectly now!
Regarding the TL07x I'll keep it in mind for my next supply run.

Thanks everyone

 

Hi,

I'm trying to run a signal through 1 of the 4 analog switches that are in de CD4066B package. The signal is a sine wave of about 1V peak to peak. When I tie the control terminal to V+ (9V) the signal passes through with no problems. When I tie the control terminal to V- (-9V) I can still see a mangled version of the signal at about 50mV peak to peak on my scope. That's 5% of the original signal which seems alot. Why isn't it cut off? Could it get in through the power supply which is 2 9V batteries?

The switch measures ~100Mohm when off with nothing attached to the CD4066 except power and a control voltage. So if it creates a voltage divider with the 5k trimpot I would expect something like 1V * (1 - 100Mohm/(100Mohm + 5kohm)) which would be much less than 50mV.

All unused terminals are connected to V-. Attached is a circuit diagram.

If you need good isolation, its common to use a pair of transmission gates. The control is fed directly to one and through an inverter to the other - one gate passes the signal, the second gate is shunt connected and switched on to short the signal when the series pass gate is off.

If the load has other signals, such as using the 4066 for multiplexing - it may be necessary to have a 3rd transmission gate to isolate the load from the shunt connected gate.