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I have a series resonant circuit where I will have a lot of capacitors. I want to select a capacitor value for resonance for a particular frequency and want to turn on that capacitor using a MOSFET.

To do that I am using two N-channel power MOSFETs in series. I don't want to control the current flow through the circuit, I just want to switch the circuit on and off. As I understand it, that means I have to switch the MOSFETs in saturation and cut-off mode to achieve that.

I know for cutoff Vgs < Vth (which for my MOSFET is 3 V). This I can do. But in saturation mode it should be Vds > Vgs-Vth. I am not sure how to achieve this. My supply voltage is AC with around +3 V to -3 V, so there will be times when Vds won't be greater than Vgs-Vth and the MOSFET will go into linear mode. How do I overcome this?

the basic circuit should be like this

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    \$\begingroup\$ Things get much simpler when you apply a DC Bias voltage such that the Vds of the MOSFETs is always positive, then you don't need two NMOS in anti-series, you only need one NMOS. Apply a 3 V (or more) DC bias to the drain of the top NMOS through a resistor. Make the resistor high value so that it doesn't interfere with the LC tank. \$\endgroup\$ Dec 10, 2019 at 14:15
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    \$\begingroup\$ You shouldn't be worrying about saturation mode or linear mode. You want to use the NMOS as a switch and that means applying as much Vgs as possible which will put the NMOS in linear mode. \$\endgroup\$ Dec 10, 2019 at 14:20

3 Answers 3

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Firstly I believe you don't mean "saturation" but the linear part of a MOSFET's characteristic: -

enter image description here

Secondly, you don't need two series connected MOSFETs; a single N channel MOSFET with a grounded source will do the job of switching in a capacitor and, to prevent the MOSFET's parasitic capacitance having too much of a detuning effect when "off", bias the drain to some voltage (like 12 volts to 24 volts) via a 100 kohms resistor.

Tried and tested idea.

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  • \$\begingroup\$ Thats also my question. to operate the MOSFET as just a switch so when i turn it on it should give the lowest resistance path for the current. For that shouldn't i operate it in saturation region? If i just use one MOSFET during the negative half cycle the ground potential would be higher so won't the body diode of the MOSFET allow the current to flow from ground to circuit... that will turn on all my capaciotrs \$\endgroup\$
    – zak3877
    Dec 10, 2019 at 14:46
  • \$\begingroup\$ @zak3877 look at the graph - look where MOSFET saturation is. Don't confuse MOSFET saturation with BJT saturation. BJT saturation is equivalent to MOSFET linear region. Don't ask this question again! The bias resistor holds the drain voltage highly positive and always reverse biases the body diode even when moderate AC voltages are present. \$\endgroup\$
    – Andy aka
    Dec 10, 2019 at 14:59
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This is a traditional application from CRT deflection circuitry: S-correction.

I have no idea if design resources exist anymore (probably handbook scans can be found somewhere?). See for example this patent:
US 595584 Horizontal deflection S-shaped correction signal circuit with variable capacitance
Or the explanation here: Electronics Now, March 1999, page 8, Service Clinic - Monitor Deflection Circuits.

For a practical case, consider this excerpt from a Sony Trinitron monitor, N3 chassis:

Sony N3 Horizontal Deflection - Highlighted

Power enters top-left (+200V or so), and passes through the pincushion output Q507; T502 is a choke supplying power to the output transistor Q508; deflection current follows the top and right lines, through T503, to the deflection coil, far right. Deflection current returns through L503 and L504 (center) (depending on the relay state; this gives the clicking sound heard when switching between high and low resolution screen modes), and probably a parallel path through T504 and the H-center circuit (bottom right). The bulk of the current returns to ground through C537 and C538, I think, while a parallel path sinks a harmonic current through a variable capacitance -- S-correction (bottom left).

The remainder of this discussion concerns the bottom-left block.

Note that each individual channel is simply a common-source switch, driven from a logic-level signal (well, gate level, certainly 5 to 15V; the 2SK2098 appears to be logic-level capable so it may well be 5V), with a 47k pull-down resistor on the drain, and the series capacitor to the common node.

When each transistor switches on or off, a big gulp of charge is injected into the deflection circuit. For a monitor, this doesn't matter; it'll take a second or so for the display to settle on switching modes anyway.

Note that bidirectional switches aren't required, as long as the drain self-bias and charge injection at switch-on, is acceptable. Bias is achieved by the internal body diode (note the arrow inside the MOSFET symbol), and the resistor simply prevents the voltage from floating away (excessive voltages could damage the transistor).

The biasing values are all rather inconsequential: 1k series gate resistors (R586, etc.), 100k gate pull-downs (R588, etc.), 47k drain pull-downs (R534, etc.; note that R534 is in parallel with two MOSFETs, whereas R532 for example does only one; yet they're the same values).

Note also the approximate binary sequence of capacitor values: C524-C528 include 56, 100, 220, 390 and 820 nF. Two MOSFETs are used in parallel for the largest values, to help share current. This makes a sort of "power DAC", where for a 5-bit binary code input, 32 unique output values can be selected.

I don't know how many of these lessons can be applied to the present question, but this historical application should at least provide some food for thought.

As for the case when a bidirectional switch is required (that is, to more or less answer the question proper): use an isolated gate drive, referenced to the (now floating) pair of source pins. This will pull down to GND when the switch is on, but floats at or below GND when off.

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An alternate method is to use PIN diodes instead of MOSFETs. PIN diodes are used extensively in switching elements of RF circuits. In effect, typically a forward biased PIN diodes behaves like a small value resistor(1-2 ohms), while a reversed biased PIN behaves like a very small cap - of the order of a few pF at most. Check it out !

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