# one transistor lc oscillator

Discussion in 'Wireless & RF Design' started by hkntrt, Jun 28, 2012.

1. ### hkntrt Thread Starter New Member

Jun 20, 2012
13
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There are many simple fm transmitter project using that oscillator type on the web.I am trying to understand how it works and how can i calculate their values ?

take a look to attachment for my design

Thanks.

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2. ### Wendy Moderator

Mar 24, 2008
20,735
2,499
This is a common base configuration, the base is considered grounded. Feedback is between the emitter (which is the input) and the collector (the output). The LC circuit selects the frequency. You could tie a capacitor from the base of the transistor to ground, and it would work the same.

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3. ### hkntrt Thread Starter New Member

Jun 20, 2012
13
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according to my calculations 888 Ω (Rb2) must be replaced 16.9 pF. is that true ? and C2 what must be ? by the way hfe=200 and transistor is bc547b

4. ### FlashFire New Member

Nov 18, 2009
5
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You should be able to consider DC biasing separately from what happens during oscillation. At DC the inductor will appear as a short circuit and the capacitors open.
The oscillation frequency should be determined by the values of the inductors and capacitors. The oscillation frequency will be determined by the resonant frequency of the LC circuit.

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5. ### hkntrt Thread Starter New Member

Jun 20, 2012
13
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Please can you tell me what must i do ""exactly"" and can you write your calculations ?

6. ### FlashFire New Member

Nov 18, 2009
5
2
Im not an expert on LC oscillators so my input comes from my limited experience. What I normally do is make sure I design using an inductor of a decent size, a 225nH inductor looks to be fine, i've had trouble getting circuits using anything less than or equal to roughly 100nH to oscillate.
Normally I let C1 = C2 and calculate the value of C1 and C2 using the resonance formula (w0 = 1/[sqrt(L*Ct)]) where Ct is the total capacitance given by the series combination of both capacitors. The ratio of C2 to C1 (it could be the other way around though, check a textbook) however is often set at around 4 instead of 1.

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7. ### hkntrt Thread Starter New Member

Jun 20, 2012
13
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thanks a lot @FlashFire , Who wants to give me more hint

8. ### upand_at_them Active Member

May 15, 2010
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L1 and C1 comprise what is known as a "tank" circuit. You can find more about them in the tutorials on this website (see the tabs at the top of this page).

They oscillate at some frequency, determined by their values. This is modulated by the input at the base of the transistor, which should have a capacitor to decouple the DC. The base resistors RB1 and RB2 are a voltage divider that determine the operating point of the transistor.

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9. ### vk6zgo Active Member

Jul 21, 2012
677
85
It certainly looks like a Common Base circuit,except that there is no bypassing of the biasing resistors,so the base isn't grounded,but is connected to ground by (for AC),RB1//Rb2,or 674 Ohms.
I think if the base is bypassed,it might work.

May 28, 2009
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11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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While this is most likely intended to be a common base oscillator [hence the query re missing base bypass cap] I doubt it really qualifies as a Colpitts. The Colpitts is characterized by a capacitive divider across the tank inductance with the feedback originating at the junction of the two capacitors - which is not the case in the OP's original schematic. The resonant frequency for the Colpitts is related to the series equivalent capacitance of the voltage divider capacitive pair - which again, is not the case with the OP's schematic.

Your second link does indeed point to a Colpitts based schematic which differs from the OP's posted circuit topology.

The values of 1nF & 225nH give 10.61Mhz as the resonant frequency so 1nF is OK.

Last edited: Aug 15, 2012