high output power colpitts oscillator

many thanks WimRFP ..
since i have to submit my project within two weeks, so I'll study your circuit later on, and then I'll ask for illustration if any.
for this two weeks I'll use a 555 timer circuit or an ordinary signal generator to drive the coils, surely this two sources have a very poor power output, but I'll let the load to be a LED, and the distance will be a few centimeters instead of 2 meteres.
after the submission I'll be free to construct your class C oscillator and transmitting power for two meters.
@WimRFP many thanks again for your support and continuous cooperation.

Hello,

To increase the power output of a circuit or generator, just add a PNP/NPN complementary pair output stage with an output capacitor for blocking the DC component. If using a AC signal source, just add a capacitor between output of generator and input of this booster. You may add a small voltage divider at the input so that with no input the output (at the emitters) is half the supply voltage.

Connect the bases together and make sure the supply voltage is somewhat above the pp output (this avoids deep saturation of the BJTs). Also add a ultra fast recovery (or schottky) diode across CE of each transistor. Use transistors like BD135 (NPN) BD136 (PNP) or equivalent (don't use low power devices in TO92 or SOT23). It is important that the ft of the transistors is >30 MHz (assuming that you use about 500 kHz).

You may add a series resistance at the output to limit the power so that you don't blow your transistors. Such a PNP/NPN booster has no voltage gain, but when supplied from 12V and a drive of about 5Vp (so 10Vpp) will give you > 2W, sufficiently to power a LED over some distance (> some centimeters) via inductive coupling. Don't forget to add some heatsink

hi
think that the darlington pair will give a better current amplification, am i right ?
and please tell me, why the two diodes ? i put them in the circuit without knowing the reason ..
i tried it in the simulator, the current is greater after this circuit but the voltage is less than that of the output of colpitts oscillator..
please check the connections in the attached circuit, if there is any wrong..

Hello,

I am sorry to say but it is not going to work this way! It is highly recommended to find somebody locally with some knowledge on electronics

Please replace the 2N3055A by a (for example) a BD135, 2N2222A, BC547, etc. That should work better. Other people also suggested to drop the 2N3055A.

Your opamp isn't fast enough (slew rate limitation), remove the opamp and use a simple emitter follower (with BD135, run at about 30mA collector current). Use single supply. Give base bias of 7V (assuming 12V supply). put decoupling capacitor between oscillator output and emitter follower.

connect collector of BD136 to ground. put 10k resistor to ground from base of Q2 (leave bases connected), put 10k resistor from base of Q2 to 12V supply. these resistors put base of Q2, Q3 to half supply voltage.

D1 with kathode to Vcc, anode to emitter, D2 with anode to ground, kathode to emitter. These diodes are for avoiding high reverse stress on the transistors in case of inductive load.

You have still no decoupling between the collector and the base of the oscillator, so output voltage will be very low. .

To save time, build the final stage only (so the two transistors with base bias) and use a function generator. Dont forget decoupling capacitors between generator and amplifier and output and load. when you don't drive the amplifier, emitter voltage must my half supply voltage.

the output current as well as the voltage are still smaller than those of the input ..

here are the values measured ..

Hello,

your input signal is to high, with 12V supply you can have an output swing of about 11Vpp maximum.

Regarding low output current, that is correct for your circuit. You have no load as far as I can see, so there will also be no output current.

For better readability, I would suggest to change your drawing so that Q1 is under Q2.

Your amplifier with field generating coil (TX coil, L1) will look like the image in the attachment.

hello
thanx again for make it easier, i got the components of the circuit and will construct it soon, then will reply.
now i have another question,not related to the oscillator circuit nor amplification. it is about the impedance matching and how can i use it to obtain better results ?
i read about it, but still couldn't figure out how can i use it in my project -transferring power wirelessly.

The said purpose of your design is driving a coil to transmit power over some distance. In my opinion, it's advisable to analyse
the problem in terms of basic electromagnetic properties, so to understand the field strength respectively driving currents required
for the transmitter coil.

It's also basically a good idea to think about impedance matching.

The impedance of the coil can be expected mostly inductive with a (hopefully small) loss resistance and an (unfortunately much smaller)
transformed resistance of the distant load. Impedance matching of an inductive load simply means creating a resonant circuit.
So - in constrast to the statement in your above circuit.doc -you should have intentionally introduced circuit capacitances to put
the coil in resonance and reduce the reactant load of the generator. Furthermore, after eliminating the reactant load part,
impedance matching involves a transformation of the real load part to fit the generator output.

Hello,

I do NOT recommend you to build the circuit first. First you should do the difficult part. It is not useful to build the easy part if you didn't solve the difficult one (the inductor and the matching).

So first characterize your transmitting coil (that is inductance and Q factor at operating frequency). The Q factor you can determine by using resonance (with the actual capacitor you will use) and determine the impedance curve.

After that you have to do the matching to about 6…12 Ohms. Then put everything in the simulator to verify it. As FvM said you should do the EM issues also (I gave you info on that earlier). As you are now satisfied with cm distance, you may skip this as with a 30cm loop and this booster stage you can be sure to get sufficient power transfer for a LED (don't forget the rectifier circuit).

@FvM: as i mentioned am using the parasitic capacitance of the coil to hit the resonance, you are right i didn't use a seperate capacitance in the simulation since i am now concerned only with building a suitable oscillator.

@WimRFP: i didn't get the matching you're trying to explain ! since the oscillator circuit is inductively coupled with the transmitting coil, and the recieving coil is also inductively coupled with the load, so i couldn't get what matching is required and between what !
and why do i need the rectifier circuit ? in the experiment i did - 13 cm - i didn't use any rectifier !

I have an idea of building colpitts oscillator using OPA548 op-amp which have a very high output current - 5 A - but didn't work perfectly in the simulator yet. any informations on the OPA548 will be nice.[/quote]

using the parasitic capacitance of the coil to hit the resonance

Click to expand...

I can hardly imagine, that this approach results in a meaningful dimensioning. Most inductive power transmitters (several W power)
in the low MHz range, as used for electronic article surveillance (EAS) or RFID, have only one or very few turns, but they have to handle already
coil voltages of several hundred volts. (I have 50 V even in a 8 cm mW powered 13.56 MHz RFID reader coil). If you don't manage to put your
transmitter coil in a high Q resonance and match the generator to feed the coil losses, your project can't work.

You may want to review existing application notes for large distance 13.56 MHz RFID readers to understand how they do.

If you don't manage to put your
transmitter coil in a high Q resonance and match the generator to feed the coil losses, your project can't work.

Click to expand...

am using a colpitts oscillator inductively coupled with the transmitter coil, i can't understand the matching required, can you illustrate ? please ..

I suggest to do the following: Take your previously shown "circuit.doc" circuit, add realistic series resistances for the coils (you have to
consider skin effect for the resistance calculation) and determine also the coupling factor k according to your intended geometry. Than play
around with the circuit and see, how much power you can transfer to the remote load and try to maximize it. If the simulation uses realistic
assumptions, I have no doubt that you can verify it in a real setup.

P.S.: A few data can explain the need for impedance matching, I think:

A single turn 30 cm loop has an inductance of about 1 uH. At 3 MHz, you can achieve Q ≈ 100 with 1 mm copper wire (or Q ≈ 1000 with
10 mm copper tube or 16 mm band). For Q = 100 and input power of 100 W, you get a voltage of 425 Vrms across the single turn.
corresponding to an input impedance of about 1800 ohm.

Hello,

Please read the specs of OP 548 (slew rate), and you will see it is useless for your application.

Just as an example I did some simulation on a simple circuit. The TX coil is single turn about 30 cm diameter out of 80 mm wide 0.5mm FR4 with one side copper. The coupling coil is just a two winding somewhat small coil. I selected low Q factor for all coils (below the values stated by FvM), so sufficient power is available for the load.

There is 32Vp across the coil, that is about 10Ap through the 0.3uH coil, sufficient to power a led over some distance, even without tuning the secondary circuit.

Peak output current for the booster is about 1A, that is about 4 W DC input to the booster.

When you are not able to do the math, you have to experiment with the risk that you will not be able to get a useful result. The tuning and inductive coupling is far from easy if you don't understand resonant circuits and inductive coupling. Also selecting the capacitors for making the resonant capacitor (here about 38 nF) has to be done carefully. Use MKP (polypropylene) or better capacitors that can handle the current (so study the datasheet).

Instead of tuning the circuit, you may change the frequency. Start with a large distance between the coils. When changing frequency very slowly, you will notice an increase in supply current to the amplifier. Slowly reducing the distance (and changing the frequency to find the peak supply current) will transfer more power to the TX coil. For these transistors it is advised to keep the supply current of the booster below 0.5A. When you make the distance too small, you will draw too much current and destroy your booster. While the circuit shows a ground between the two coils, this is not required (it is better to have no ground).

The disadvantage of such circuits is that when you remove the inductive load, the voltage across the transmitting coil (TXcoil) will rise and may overload the amplifier. So you need to have good heatsink for the transistors and current monitoring.

I selected low Q factor for all coils, so sufficient power is available for the load.

Click to expand...

But increasing the coil Q means increasing the transmit efficiency, particularly for the intended large distance. At the receiver coil, you'll
get maximum output power for a given loss resistance with a real load equal to the coil losses (= power matching). At high Q, you can
get a problem of exact tuning, however.

The coupling for 30 cm coils with 2 m distance is about 1e-4, by the way.

Hello

FvM:
Fully agree with higher Q is higher efficiency. By using low values for the Q in my simulation, you can be sure that the actual result will be better (that means more current through the TX coil).

As I didn't want to do the full calculation for the complete TX to RX link (he has to do that himself), I chose a low Q. In fact the energy extracted by the RX circuit I put into the Q of the TX coil. We also don't know what type of capacitors he will use.

For your interest, my experience with this matter is from CW Tesla coils, 8 MHz EAS and 134 kHz, 13.56 MHz RFID antenna design.

thanx WimRFP.
but why the input is square wave ? and in which frequency should the oscillation be ? because the maximum amplitude got in 500 KHz -not the rsonant frequency of any LC circuit- ?

i asked you earlier about the rectifier circuit and why do i need it ?

Hello,

When you use a square wave, this emitter follower booster (with the Schottky rectifiers) has very good efficiency (over >85%). It does work with a sinewave with somewhat less output.

You could derive it from the time scale above the simulated waveforms, but I used 1.5 MHz input frequency. The optimum frequency is almost equal to the resonant frequency of the TX coil. Best is to determine this by experimentation as your capacitors (and coil) will have some tolerance also.

Regarding the rectifier for the LED. The LED is made for giving light, not for being a 1.5 MHz rectifier with high efficiency. Depending on the type of receive circuit, you may put too much reverse voltage to the LED. A center-tapped secondary winding with two schottky rectifiers will do the job. It saves you one diode drop with respect to a full bridge rectifier.

am trying to understand the impedance matching technique and having some problems.
for the circuit attached, what is the load impedance that achieve maximum power to it ?