a very simple FM Tx

[walid1]

the following FM circuit

I have tree questions:
1- what the function of the 1 n cap
2-why tapping the coil, why not connecting the antenna directly to the collector
3-what the function of the 1.8p cap, and why it is so small

thanx

View attachment 40066

 

[audioguru2]

the following FM circuit

I have tree questions:
1- what the function of the 1 n cap
2-why tapping the coil, why not connecting the antenna directly to the collector
3-what the function of the 1.8p cap, and why it is so small

Hi Walid,
That is an extremely simple FM transmitter circuit. Did you make it? How is its performance?

1) The 1nF cap makes the transistor a common-base amplifier at radio frequencies where the emitter is its input and the collector is its output.

2) The coil and capacitor in parallel with it is the tuned circuit that determines the RF frequency. The antenna is connected to it so anything that gets near or touches the antenna changes the capacitance and changes the tuned frequency. When the antenna is connected to a tap on the coil then the change of tuned frequency isn't as much, but then the range of the transmitter isn't as far.

3) The 1.8pF cap provides positive feedback so the transistor oscillates. It is in parallel with the transistor's capacitance and the wiring capacitance so the total capacitance might be about 10pF. 10pF is not small at 100MHz where it has a reactance of 160 ohms.

 

[walid1]

Thank you guru u are a good man

1) The 1nF cap makes the transistor a common-base amplifier at radio frequencies where the emitter is its input and the collector is its output.

why it need to be a common-base, what would be happen if we remove this 1n cap.
I know that

 

[audioguru2]

why it need to be a common-base, what would be happen if we remove this 1n cap?

This circuit uses a common-base amplifier. There are many other kinds of oscillator circuits. The capacitor makes the base at 0V for RF frequencies. If it is removed then the circuit probably won't oscillate.

I know that

 

[walid1]

Hi guru
i read carefull your last reply but it was not sufficient to make me understand so, i now reading this doc.
http://www.maxim-ic.com/appnotes.cfm/appnote_number/2032
may it help me
then i'll back to u to finish this hard subject
thank u guru

 

[walid1]

i read it but i not understand any thing
it is not as i expect
can u guru point me to another article more close to what i need
can u help me in understanding carfully how this vco operate
thank u

 

[audioguru2]

Hi Walid,
Maxim's VCO IC uses a varactor diode (it changes its capacitance when its voltage is changed) to frequency-modulate and to tune its carrier.

 

[walid1]

Hi guru

Maxim's VCO IC uses a varactor diode (it changes its capacitance when its voltage is changed) to frequency-modulate and to tune its carrier.

ok, i agree
my circuit did not using a varactor diode, there is no varactor diode
i think that a p-n junction do this, is this true and how?

 

[audioguru2]

my circuit did not using a varactor diode, there is no varactor diode
i think that a p-n junction do this, is this true and how?

The datasheet for the transistor that is used in your project and for most other small transistors shows the transistor's change of its capacitance with voltage change, like a varactor diode.
The input audio causes the transistor to conduct more and less which changes the voltage across it and changes its capacitance.View attachment 40067

 

[walid1]

Hi guru

This circuit uses a common-base amplifier. There are many other kinds of oscillator circuits. The capacitor makes the base at 0V for RF frequencies. If it is removed then the circuit probably won't oscillate

why the circuit probably won't oscillate, can you please, explain this point more.

The antenna impedance match to the circuit's output impedance is poor, but it is simple and it works

What would u do to improve the matching between o/p and antenna?

The base input impedance of the transistor at low audio frequencies is its beta of about 200, times the emitter resistance of 220 plus its internal emitter resistance = 44.5k ohms. It is in parallel with the two 47k resistors also in parallel, so the load on the mic is 15.4k ohms.

beta = 200, are u guess it or depend on some source or you consider any general purpose transistor's beta as approx 200.
also you calculate the internal emitter resistance (re) as beta X re =500 ohm, that is re = 500/200 =2.5 ohm, frome this:
25/Ic=2.5 ==> Ic = 10 mA, is these values from datasheets, I search for BC547 datacheets and find many copmanies, please provide me with your datasheet or tell me from any comany.

It is complicated to calculate how much is the output voltage and current because tapping down the coil reduces the output voltage into the low impedance antenna but provides a better impedance match so the collector voltage swing is much higher

But with tap, the coil acts as a step down transformer?

The input audio causes the transistor to conduct more and less which changes the voltage across it and changes its capacitance.

Are you mean the capacitance between B and E?

thank you alot

 

[audioguru2]

why the circuit probably won't oscillate, can you please, explain this point more.

The transistor needs a change of base to emitter current to be an amplifier and to oscillate. If the base voltage is not secured with a capacitor to ground then it will also have its voltage changed when the positive feedback changes the emitter voltage, then there is no change in base to emitter current.

What would u do to improve the matching between o/p and antenna?

I would add an RF amplifier that would have a lower output impedance than the oscillator. It would also isolate the antenna from the oscillator so the frequency wouldn't change when anything gets near the antenna. Then I would add a series or pi LC impedance-matching network at the output which would also reduce harmonics.

also you calculate the internal emitter resistance (re) as beta X re =500 ohm, that is re = 500/200 =2.5 ohm, frome this:
25/Ic=2.5 ==> Ic = 10 mA, is these values from datasheets

The Re of a silicon transistor is approx. 26/Ie in mA, which is about 2.6 ohms. The beta ranges from 110 to 800 so I picked 200. You can check its accuracy by looking at the input impedance of a 2N3904 transistor on its datasheet since it is about the same as a BC547. At 10mA the input impedance of a typical 2N3904 is 500 ohms.

But with tap, the coil acts as a step down transformer?

Yes, it is an auto-transformer (single tapped coil).

Are you mean the capacitance between B and E?

The transistor's collector is connected to the tuned circuit. The collector to emitter capacitance changes which changes the tuned frequency at the collector.

 

[walid1]

Hi guru

The transistor needs a change of base to emitter current to be an amplifier and to oscillate. If the base voltage is not secured with a capacitor to ground then it will also have its voltage changed when the positive feedback changes the emitter voltage, then there is no change in base to emitter current.

I feel that i not understand what you want to say.
Lets analyze that statement:
I assume that VE = 3v
VB = 3.7v
positive feedback changes the emitter voltage to say 4 v
so VB = 4.7v
now what that cap do to maintain VB to 3.7 v
thank u guru

 

[audioguru2]

Hi Walid,
A filter capacitor needs time to charge or discharge in order for its voltage to change. At 100MHz its value makes a good filter so its voltage doesn't change.
At 7.5kHz and lower audio frequencies then the value of the capacitor makes a very poor filter so its voltage can change to allow audio to modulate the base of the transistor.

In my FM transmitter I used a 470pF filter capacitor to allow modulation up to 15kHz.

 

[walid1]

Hi guru

The base input impedance of the transistor at low audio frequencies
is its beta of about 200, times the emitter resistance of 220 plus
its internal emitter resistance = 44.5k ohms. It is in parallel with
the two 47k resistors also in parallel, so the load on the mic is
15.4k ohms.

This is good, but I want to introduce the effect of the 1nF cap on the lower 47Kohm.
The reactance of that cap at freqs from 300Hz t0 15Khz is 10.6 Kohm to 53 Kohm
47K // 10.6K  = 8.65 K, and
47K // 53K = 24.9K
so VB is not constant at VB = 9 * 47/(47+47) = 4.5v as i was think
it is really changes from 1.4V to 3.12V with the audio freq

Now Zin (total) = ZB//RB1//RB2
since RB2 equivalant is change from 8.65 K to 24.9K
so Zin change
please comment.

It is complicated to calculate how much is the output voltage and current because tapping down
the coil reduces the output voltage into the low impedance antenna but provides a better
impedance match so the collector voltage swing is much higher.

I agree that it is complicated to calculate how much is the output voltage and current, lets
assume them and investigate the tapping effect on them:
If the output voltage = 5mV rms at freq 100MHz as shown in the Fig. below, how much voltage reach the antenna?
and what the value of the impedance the antenna see, if the coil is 330nH and its XL = 207 ohm at 100MHz?

thank u guruView attachment 40173

 

[audioguru2]

Hi Walid,
I started to calculate the effect of the 1nF capacitor on the frequency response of the electret mic at 15kHz but I made a simulation instead. It makes no difference!

View attachment 40174

 

[walid1]

Hi guru
First thank u for the figures from which i can see that the 1 nF has no resonable effect at AF.

but u may not see this question

It is complicated to calculate how much is the output voltage and current because tapping down
the coil reduces the output voltage into the low impedance antenna but provides a better
impedance match so the collector voltage swing is much higher.

I agree that it is complicated to calculate how much is the output voltage and current, lets
assume them and investigate the tapping effect on them:
If the output voltage = 5mV rms at freq 100MHz as shown in the Fig. below, how much voltage reach the antenna?
and what the value of the impedance the antenna see, if the coil is 330nH and its XL = 207 ohm at 100MHz?

thank u guru

these flowers are for youView attachment 40175

 

[Kevin Weddle]

How do you know you have crossed the finish line with an oscillator? I seem to think that general test lab equipment cannot tell the whole story. I cannot believe these mass produced behemoth bluetooths with Intel Pentium 4's and Windows spyware can measure accurately enough for me.

If you know the oscillator frequency, it's probably better to make your own measuring circuits. An embedded application.

 

[audioguru2]

Hi Walid,
Thanks for the white roses. They are fresh and smell nice. ;D

The tuned circuit at the collector of the transistor is resonant and tries to be a high impedance. The coil is an auto-transformer and the 50 ohm antenna is multiplied by the square of the turns ratio so appears as 5 x 5 x 50= 1250 ohms at the collector.
I think the power output is higher than if the antenna is connected directly to the collector.

 

[walid1]

Hi guru

lets do it again

at 106MHz, L =330nH  and C=6.8pf, we have

XL = 2 pi f L = 220 ohm
XC= 1/(2 pi f C) = 220 ohm  (resonance)

Ztot = (XL*XC)/(XL+XC) = 110 ohm

I think it is wrong calc, Ztot must be zero because XL = j w L
and XC = -1/j w c

I don't know what to do............

IF the LC impedance = 110 ohm as before, and the antenna impedance = 50 ohm

110/50 = 2.2

the turns ratio = sqr(2.2) = about 1.5:1  which equivalent to 3:2  = 6:4

if the coil was 6 turns then the tap must be at the fourth turn and not the fivth turn to have better impedance matching.

please guru correct me