analog-switchschematicsswitchingtexas instruments
I want to switch between an output of 12V and 5V. I'm using a CD4066B (datasheet)
Here's the pinout:
This is my schematic:
When I build this circuit, I end up frying the IC. What am I doing wrong? What can I do to make this circuit work as desired?
Your circuit concept is starting in the right direction but it requires some important changes for it to work.
You will require series base resistors for all of the PNP transistors. As drawn you will cook them with no base current limitation.
Secondly the NPN transistor is drawn backwards. Its emitter needs to connect to the GND.
Third you need to do some serious work reorganizing the way the bases of the various transistor are biased. You need to separate the 12V and 5V PNP bases so that the 12V one can shut off when the base drive goes away to the 5V one. After this fix it is necessary to provide also a separate pullup of the NPN transistor.
Don't cringe. Resistors are incredibly cheap.
Lastly this schematic, to be bluntly honest with you, is horribly drawn being that it does a bad job of following conventional standards for how a good schematic is drawn.
The analog switch by itself will not introduce a DC offset into the signal. That is, it does not add electrons from its Vdd pin to the signal channel. Leakage currents are sub-nanoamp in Fig. 9. Only you can say whether or not this error is significant.
You don't say what the signal bandwidth or amplitude is, and the FET switch characteristics can affect both of those.
A single channel ON resistance varies between 40 and 125 ohms in the tables, but is worse in the chart for your conditions. This is in series with all of your termination options. For example, two of these channels are in series with your 50 ohm differential terminator, significantly affecting the effective termination impedance seen by the signal.
Datasheet fig. 7 shows how the resistance varies instantaneously with signal amplitude. If the signals are source-terminated (example: 50 ohm output resistance at the source), this creates a voltage divider that varies with signal amplitude. The series leg is the 50 ohms at the source, and the shunt leg varies between 210 and 260 ohms (50 ohms + (2 x switch resistance)).
Based on nothing but what you've said so far, I'd look for a better switch. ADI and others have parts with 5 ohm (ish) ON values.
Best Answer
You're burning it up, because with VSS connectted to 5V ground is outside the alloable range for an input or outputs. (they must all be between VSS and VDD
there's probably a better way to do whatever it is you are trying to do. but I have no way to guess what that might be.
simulate this circuit – Schematic created using CircuitLab
that has the same 120 ohm impedance as the CD4066B switches, but later in comments you say that you need to power a device with the output, in this case an adjustable regulator may be a better choice.
simulate this circuit