analogmosfetswitches
I have this circuit taken out from a book.
The book says that :"The circuit as shown will work for positive signals up to about 10V; for larger signals the gate drive is insufficient to hold the FET in conduction (Ron begins to rise)"
What does this mean? That I can't switch a higher voltage than my gate voltage? And why would Ron rise if the input signal passes 10V ?
Thanks in advance.
As the comment says, you can get a much better answer to this question if you give us some more information. One thing that's not clear is, are the signals your talking about actually meant to convey power to their receiver, or just control power that's already available there.
I'll assume these are just control signals.
I can think of two simple ways to accomplish what I think you want.
Simply sum the two signals. Because the digital signal has such a high magnitude, you'll have a signal between 0 and 10 V when the digital value is low, and between 24 and 34 V when the digital value is high. At the receiver, use a comparator with a 17 V threshold to extract the digital value, then subtract it off to recover the analog value. You will likely have some glitches in the recovered analog value whenever the digital value switches -- how you deal with those will depend on details of your requirements that you haven't shared.
Modulate one or the other of your two signals onto a high-frequency carrier. For example, use the digital signal to switch a (for example) 100 kHz sine wave onto your signal line. (I'm assuming your analog signal is only changing slowly. If your analog signal has high frequency content, then the details of how to do this will depend on what are the frequencies in the analog signal). At the receiver end, use a low-pass filter to extract the analog signal; and a high-pass filter and a rectifier to extract the digital signal.
Edit
Your edit helps quite a bit. It sounds like you can just use a switch like the ADG202A you mentioned, with no other op-amps or anything needed. Use the digital signal to control the switch. Route the analog signal through the switch to the machine under control. Use a pull-down resistor to pull the control voltage to 0 when the switch is opened. If you are worried that the analog voltage might be accidentally programmed below 0 and this could damage the machine, you could add a diode between the control pin and ground to prevent the control voltage dropping (much) below ground.
Some things you still need to explain to know if this will work: How much current can your programmable analog source generate? How much current is required at the control input of your machine?
** Edit 2 **
Here's a diagram of what it sounds like you want:
This will set the machine power to 0 when the digital signal is 0. If you want it the other way, you can use the ADG201A NC switch instead.
The ADG202A has a fairly high 60-Ohm on resistance. If the machine draws significant current at the control input you may need to locate a switch with lower on-resistance, or buffer the signal with an op-amp between the switch and the machine.
All MOSFETs inherently have a body diode between source and drain. This diode is reversed biased under normal usage, but if you put the drain at a lower potential than the source on an N-channel FET, you will, as you have discovered, forward-bias this diode.
simulate this circuit – Schematic created using CircuitLab
More on the body diode: MOSFET: Why the drain and source are different?
You will have to find a way to place the MOSFET in your circuit such that the drain is at a higher potential than the source. Or, you can use a P-channel MOSFET, where the drain must be at a lower potential than the source.
Either way, you will likely need some sort of level-shifting circuit to translate your 0V-3V logic input to whatever levels are required by the MOSFET arrangement you select. This can be as simple as another transistor (usually a cheap BJT) and a resistor, or a complicated arrangement designed to switch the MOSFET at very high speeds which might include dozens of components and maybe even additional voltage supplies. It all depends on your application.
Best Answer
You can't switch voltages higher than the gate voltage, no. You must always have the gate-to-source voltage (Vgs) greater than the transistor's threshold voltage if you want it to conduct well. If you want "pass a 10V signal" then you want the source and drain to be at 10V, which means that the voltage on the gate must be (at least) 10V plus Vth. Lower values of Vgs will result in higher drain-to-source resistance, so you really want the gate voltage to be much higher than the voltage you want to switch.