All transistor radio circuits (not crystal) on the internet seem to use an antenna and parallel LC circuit. Why does no one use series LC circuits to select stations?

To my knowledge, series LC circuits at resonance let maximum current flow at resonant frequency. BJTs allow a current through their base control current from collector to emitter. Therefore, a higher base current could lead to greater amplification.

Experimentally, I have built the following schematic on a breadboard. I get much more audio output using antenna and series LC circuit over antenna and parallel LC.

Are other factors being considered for the choice to use parallel LC circuits in transistor radios?

My revised parallel LC circuit:

enter image description here

My initial series LC circuit:

enter image description here

  • \$\begingroup\$ You show an amplified crystal radio. The series LC is used when there is plenty of current but the antenna current is extremely low. You need a parallel LC that produces its highest voltage at resonance. Then your "radio" that has only one LC will pickup maybe 6 AM stations at the same time. Real AM radios have at least 6 parallel LC circuits for good selectivity. EDIT: The resistances of your base bias resistors are extremely low and are a short to a parallel LC. \$\endgroup\$
    – Audioguru
    Dec 19, 2021 at 2:13
  • \$\begingroup\$ Very interesting about series vs parallel but I suppose that makes sense. Forgive my lack of knowledge (I am just a hobbyist) but what is going on with my base resistors? Could you explain a bit more about how that works? \$\endgroup\$ Dec 19, 2021 at 2:29
  • \$\begingroup\$ Oh wait would R2 form an RLC circuit with a parallel tank that stops oscillations when its only 1kohms? \$\endgroup\$ Dec 19, 2021 at 2:51
  • \$\begingroup\$ You must learn about a simple transistor circuit to be able to calculate resistor values. Your very low value base bias resistors assume a transistor hFE of 1 instead of the hFE of 200 shown on the datasheet. R1 should be at least 20k and R2 should be at least 10k. \$\endgroup\$
    – Audioguru
    Dec 19, 2021 at 3:25
  • \$\begingroup\$ What do the blue numbers and the (unreadable) words next to them mean? \$\endgroup\$ Dec 19, 2021 at 4:01

5 Answers 5


Why does no one use series LC circuits to select stations?

The selectivity of a series tuned circuit is inversely proportional to the series resistance, ie. the lower the resistance the higher the selectivity. This resistance includes the source (antenna) and load (amplifier input) resistances. That means to get good selectivity from a series tuned circuit the amplifier input impedance needs to be very low.

NOTE: the following refers to the schemetic you originally posted with series tuned input (1 mH inductor in series with 10-100 pF variable capacitor):-

Your inductor and capacitor values correspond to a resonant frequency of ~500 KHz (when the tuning capacitor is set to 100 pF). If the source impedance was 0 Ω and the load was 75 Ω then the -3 dB bandwidth would be ~14 kHz, which is reasonable selectivity for AM broadcast. But your amplifier input impedance is not 75 Ω, it is ~725 Ω, which greatly reduces the Q of the tuned circuit and increases its bandwidth to ~120 kHz. That's not good.

But it gets worse. A 1/4 wave 'whip' antenna with perfectly conducting ground plane has an impedance of ~37 Ω at resonance, but at AM broadcast frequencies the impedance of your 3 foot length of 'random wire' is much higher and almost totally capacitive, with a small value of ~6 pF. This causes the resonant frequency to increase to ~2 MHz. The tuning capacitor will have little effect because antenna capacitance dominates, making your series tuned 'station selector' practically useless.

The schematic you now present with parallel tuned circuit and higher Base bias resistor values is better, but still not great. The problem now is that the Q of a parallel tuned circuit is directly proprtional to parallel resistance, ie. the higher the resistance the higher the Q. With the higher value Base bias resistors your amplifier input impedance is ~7.2 kΩ, much better then 725 Ω but still quite low for this parallel tuned circuit. The bandwidth is still very wide at ~57 kHz. Amplifier input capacitance detunes it to a lower frequency of ~410 kHz, which may be OK on the low end of the band but could prevent you from tuning into stations at the higher end.

So what can you do? The short antenna works 'OK' coupled directly to the top of a parallel tuned circuit, but the amplifier input impedance is too low for good selectivity. To fix this you can either:-

  1. Increase amplifier input impedance using a FET, or a BJT (Bipolar Junction Transistor) in Emitter Follower configuration. The ZN414 AM radio IC is an example of using BJTs this way.

  2. Tap into the inductor winding or add another winding with fewer turns to make a transformer. With the correct turns ratio this can match the lower impedance of the amplifier to maximize power transfer without compromising the Q of the tuned circuit. If you use a longer antenna then this should also be coupled in the same way.

Here's an example of a 'crystal set' radio receiver using adjustable taps for both the antenna and detector:-

enter image description here

  • \$\begingroup\$ Awesome explanation, can you provide some insight as to how to calculate input impedance of an amplifier? Is there a formula or some logic about how input impedance affects bandwidth? I really want to know more about that. \$\endgroup\$ Dec 19, 2021 at 14:58

I get much more audio output using Antenna and series LC circuit over Antenna and parallel LC.

That's because your low-impedance series-resonant circuit is a good match for the low-impedance input of the transistor amplifier, whereas your high-impedance parallel-resonant circuit is not.

Performance with the parallel-resonant circuit may be improved by feeding the amplifier from a low-impedance tap on the coil.

enter image description here


Assuming a "short" monopole antenna (i.e. basically capacitive in nature)...

Your series tuned antenna

  • Produces no voltage gain (due to extremely low Q-factor) and
  • No channel selectivity and
  • Barely any tuning capabilities from the variable capacitor and
  • The centre frequency of tuning is wholly redefined by the length of the antenna

Your parallel tuned antenna

  • Produces a significant voltage gain (due to much higher Q-factor possibilities) and
  • Reasonable channel selectivity if the antenna is quite short (below a length of about 0.1λ a short antenna produces a constant capacitance
  • Reasonable tuning span from the variable capacitor but,
  • The centre frequency of tuning is affected by the presence of the antenna

What I'd recommend is this

enter image description here

The above provides decent tuning at 1 MHz and, the presence of the 10 pF capacitor in series with the antenna (shown as voltage source V1) largely prevents significant retuning effects when the antenna length changes considerably i.e. the reactance of C1 swamps the reactive elements of the antenna: -

enter image description here

If you didn't have C1 and you fitted a quarter wave monopole you'd have a wide flat tuning like this: -

enter image description here

The above would be excellent for a superheterodyne receiver but, for an amplified crystal radio (your circuit despite you thinking it isn't a crystal radio), it would be useless.

  • \$\begingroup\$ Fantastic information! I would like to know more about the "returning effects" of the antenna, can you break down what is going on there at a simple level? Or perhaps just provide the name of that phenomenon so I can learn more? \$\endgroup\$ Dec 19, 2021 at 14:52
  • \$\begingroup\$ @YousifAlniemi can you break down what is going on there at a simple level? - there is nothing at all going on at a simple level with antennas. Anyway, you've closed down this Q and A session now. \$\endgroup\$
    – Andy aka
    Dec 19, 2021 at 14:58

OP's antenna is an open end (whip), combined with a good ground. The RF signal near 1 MHz. is delivered to a transistor whose equivalent input resistance is near 1000 ohms (perhaps somewhat less).
This scenario is covered in AUDIO Handbook, 1976 National Semiconductor Corp, Technical editor Dennis Bohn

The short 3 -ft whip antenna's electrical property is equivalent to a 10 pf capacitor at 1 MHz.
One wishes to deliver RF power to the 1000 ohm load (RL), and also to select one station while rejecting others. Selectivity involves the resonator quality factor Q. The author has decided on a Q of 80 (that is loaded Q), which means that the resonator has a bandwidth of 12,500 Hz - plenty wide enough to accommodate both sidebands of one AM channel. Audio from the detector would be restricted to 6250 Hz.

At 1MHz, the author assumes that the variable capacitor is set to 90 pf.
The variable capacitor (90 pf) adds to the antenna capacitance to make 100 pf. The inductor must resonate:
\$ L=\frac{1}{4\pi^2 f^2 C}=253uH\$

Loading this inductor with 1k transistor base not only kills selectivity, but doesn't transfer power efficiently from antenna-to-base. An impedance transformer is required to achieve selectivity and efficient power transfer.
Let's first find the equivalent resistance in parallel with the inductor to achieve a Q of 80:
\$ R_P = Q\times X = 80\times 2\pi1x10^6\times253x10^{-6} = 127k\$

We're going to have to transform this to 1k, so the inductors' Q will have to double (254k for the inductor itself, and 254k for the transformed 1k load). So the design choice is to waste half of the received power in the resonator to provide selectivity, and deliver the other half power to the transistor's base.


simulate this circuit – Schematic created using CircuitLab

We'll use a transformer or perhaps a tapped inductor to achieve a 254-to-1 impedance ratio: \$\sqrt{254}\$ requiring a turns ratio of 16:1 (assuming tight coupling). This step-down transformer will efficiently couple resonator voltage to transistor base, while still providing selectivity.
As a sanity-check, how difficult is it to construct a 253uH coil having the required unloaded Q of 160? An air-wound coil would be quite large, and difficult to estimate coupling coefficient between main coil and step-down coil. Ferrite material can shrink size and # turns, but watch out for ferrite losses: Q of 160 may be difficult to achieve, especially for tiny epoxy-coated ferrite inductors.

Is 1k a reasonable estimate of the transistor's input impedance? Let's say collector and emitter current are both near 1mA, and the emitter is at AC GND.
A rough stab at room temperature uses the "26ohms/mA" rule for equivalent resistance at emitter - this is boosted by \$h_{fe}\$ at transistor base. So guess at transistor's current gain of 50...yield 1300 ohms. There will also be some capacitance, and a little loading from collector feeding back to base. So we're in the right ballpark.

The Audio Handbook goes on to estimate how much voltage the transistor would see if a RF signal presents 100uV/m to this short antenna. Rather involved, but the short answer is 26uV.


With the right high turns ratio ferrite rod, you can use both parallel to boost voltage and transform into low impedance series resonant front end.

This came from a kit you can buy.


  • \$\begingroup\$ The AM radio kit has only one LC for hearing a few stations at the same time. REAL AM radios have about 6 LC pairs. The radio has no AGC so each station will have a different loudness. EEK! The audio amplifier has no negative feedback so there will be lots of horrible distortion. \$\endgroup\$
    – Audioguru
    Dec 20, 2021 at 16:17
  • \$\begingroup\$ Yes lots of room for improvement with only 5 Q's but its // & series on the front end which is all I wanted to point out. @Audioguru my Xtal radio had 0 Q's and only 1 coil \$\endgroup\$ Dec 20, 2021 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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