Let's say I have mentioned circuit and I want to be able enable/disable one of the 1uF capacitors to change the frequency of the LC circuit.

I would like to do this electronically with a micro-controller.

  • I know a relay is ideal for this but is bulky and mechanical.
  • Another option could be a an "analog switch ic" but they have a relative high internal resistance, and a low max current. The low max current is a problem for a parallel LC circuit because the current can go quite high.

Are there other solutions to do this?

Thank you in advance

Example circuit


  • bigger picture: This is just a picture from the interwebs, but what I would like to make is very close to this, the inductor will be a large solenoid for vehicle detection. If the inductance changes (because of the presence of a car) the resonant frequency will change. The comporator will convert this attenuated oscillation in a block wave, the microcontroller can than determine the frequency of the block wave and so detect any changes (traffic).
  • \$\begingroup\$ Hi @dim I added a bigger picture at the bottom of my post, if you know a way how I could measure the inductance-"change" in the digital domain I'm interested. \$\endgroup\$ – Bruce Nov 27 '18 at 9:18
  • \$\begingroup\$ What "Q" do you wish? Large MOSFET drivers may have 1 to 3 ohms resistance. 1uH and 1uF resonate at 160KHz, with Z of 1 ohm. If a 1ohm driver, your Q would be one. 100uH and 0.01uF also resonate at 160KHz, and 1 ohm provides Q=100. \$\endgroup\$ – analogsystemsrf Nov 27 '18 at 9:20
  • \$\begingroup\$ @analogsystemsrf As far as I understand you can't use a mosfet to connect/disconnect another capacitor to GND. I might be wrong. \$\endgroup\$ – Bruce Nov 27 '18 at 9:37
  • \$\begingroup\$ @dim right I'm trying to measure (a change in) inductance and would like to be able to do this at different frequencies. Any other suggestion on how to approach this with a microcontroller? Measuring the L/R time constant might be an option but with 20µH-2000µH this will be a very short time. \$\endgroup\$ – Bruce Nov 27 '18 at 9:40
  • \$\begingroup\$ At about the frequencies you are using, you might find that the oscillation frequency doesn't change that much (rather the losses extracted from the magnetic field will possibly dominate and these might be a better "signal"). \$\endgroup\$ – Andy aka Nov 27 '18 at 12:10

I think you will have to evaluate the trade-offs of the mechanical relay type switching versus the electronic analogue switching component. The relay as you say is ideal from the circuit switching standpoint so the real analysis has to be made regarding some type of electronic switching.

It is not generally true that all analogue switches have relatively high on resistance. There are devices available that have surprisingly low on resistance so certainly do a search for them so you can include them in your evaluation.

The same comments can be made regarding current rating of analogue switches. Do not think that all analogue switches fall in the same realm as the CD4066.

There are some important considerations to be made regarding the semiconductor switching however. Here are a couple of them:

  1. Voltage Level - Semiconductor switching devices will have specific specified voltage levels that you will have to consider for the safe operation of the silicon.
  2. Voltage Range - IC type switching components in general will have a range limit over which they can support the switching of the signal. For switches with just one power supply this range will typically be between VDD and GND. For switches with an additional negative power supply this range will be between VDD and VEE. You must consider the cases when the switching is element is both ON and when it is OFF.
  • \$\begingroup\$ This is why I used the word "relative" most of them have a internal resistance of more than 10Ohm but indeed there are plenty in the milliOhm range. But I couldn't find them higher than 500mA peak current. \$\endgroup\$ – Bruce Nov 27 '18 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.