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I think this is a question is some very elementary circuit design but... I want to use a varactor diode to tune the resonance frequency of a resonator I have built.

I understand that a varactor is a voltage dependent capacitor, but I can't get my head around the correct configuration of using them.

Usually if I want to tune my resonator I insert a capacitor between the hot end of the resonator and the ground this will obviously shift the resonance frequency according to $$\nu_0 = \frac{1}{2\pi \sqrt{L(C_p+C_{capacitor})}}$$

I want to do the same but with the Varactor diode, my current thinking feels very wrong as the DC voltage source is surely just going to go straight through the resonator and into ground.

I know this is basic but I am a little stuck!

schematic

simulate this circuit – Schematic created using CircuitLab

user27119
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  • You've got a voltage source. That will keep the voltage constant, in theory, no matter how low the DC resistance of your (L || Rp) is (and how much current thus flows) – Marcus Müller Nov 12 '16 at 21:04
  • @MarcusMüller I must be missing something, I want to use the voltage source to tune the varactor and hence the resonance frequency. I'm unsure how to integrate the voltage bias for the varactor and then how to place this in series with the inductor . – user27119 Nov 12 '16 at 21:06
  • Should I perhaps place a high value resistor after the cathode which then goes to ground? – user27119 Nov 12 '16 at 21:09
  • Yes, use a high value resistor or a bias choke to feed the DC control voltage to the varicap, other than that it looks ok (but you might want to put a DC block capacitor between the varicap and the tank as the tank inductor will short DC voltages), just remember that *any* voltage will affect the varicap diode, so the RF amplitude needs to be kept low to avoid the capacitance changing in sync with the RF – Sam Nov 12 '16 at 21:33
  • Rp usually goes between Varicaps and VCO. Or use FET's instead of Varicap with a twin L VCO – Tony Stewart EE75 Nov 13 '16 at 00:38

2 Answers2

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The circuit you're intending to use is not one of those commonly found in actual devices – exactly because the L has (nearly) zero DC resistance, so a lot of current will flow through it due to the DC bias.

It's important to notice that your (R||L||C) resonant circuit is not usually useful in this constellation in isolation. If you want to use it as the resonating part of an oscillator, you'd typically have some kind of amplifying element that feeds energy back into that oscillator; since we're talking AC here, the coupling can usually be done through a capacitor, and that means that you can isolate the biasing; you might want to think more in terms of:

schematic

simulate this circuit – Schematic created using CircuitLab

More common, however, is the configuration where the bias voltage is applied over two antiparallel diodes (from German wikipedia):

varicap based oscillator

Notice the choke in the bias voltage path – since the ideal voltage source has zero impedance, it would completely "swallow" any energy stuffed into your oscillator, and Lb avoids that.

Marcus Müller
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  • Marcus, your first circuit *still needs* a bias choke from voltage source "adj", as you've included (as Lb) in your second circuit. A large-value resistor is often substituted for Lb. – glen_geek Nov 12 '16 at 21:40
  • @glen_geek I hoped the second circuit would clarify that! – Marcus Müller Nov 12 '16 at 21:42
  • @MarcusMüller Thanks for your explanations, I should have more clearly defined my circuit. The RCL component of the circuit is in actual fact one piece or component - such that its components of C, L and R cannot be changed. – user27119 Nov 16 '16 at 16:22
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In the end this is what I needed...

schematic

simulate this circuit – Schematic created using CircuitLab

user27119
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