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Audioguru's FM Tx Troubleshooting

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hi audio guru,
From your original circuit, what does the capacitor C2 does? i can recognise the common emitter configuration but can't seem to figure out the capacitor C2 which shorts base to emitter.
The first transistor is the preamp for the mic. C2 stops it amplifying the 100MHz radio frequency thet is very stong and if the 100MHz is not stopped then the first transistor would probably be overloaded by it.

what does the whole part do? pre emphasis?how does it work? if yes can i replace it with a simple high pass filter such as LC combination? thanx in advance!
The first transistor uses C4 in parallel with R5 for pre-emphasis. You don't want a huge inductor plus a capacitor to make pre-emphasis.

i am not able to get 5-35 pf cap
Use the 2pF to 22pf one and not be able to tune the lower half of the FM band.
 
hi mvs sarma,
Thanx for offering the eagle files, it would be great if you could do so, but i have this tendency to understand things before doing so...can't seem to grasp the preemphasis part, as in how does the common emitter like configuration gives pre emphasis? isn't it suppose to boost high frequency to reduce traingular noise? then why the trouble of using so many components and instead use a simple LC high pass filter? we can calculate the pre emphasis time to be 50uS or 75uS(US) using the LC via the formula Fout=1/2*pi*R*C
 
The voltage gain of a common-emitter transistor is roughly Rc/Re. But C4 bypasses the emitter resistor R5 at high audio frequencies which increases the voltage gain of the transistor which boosts high frequencies making pre-emphasis.
It is an RC circuit which makes the required 6dB per octave boost. If you use an LC circuit then the high frequencies are boosted at 12dB per octave which is wrong.
The formula for an LC filter is completely different from the formula for an RC filter.
 
hello there audio guru,
First of all thank you so much for clarifying my ambiguities and mistakes...the formula for LC is Fout=1/[2*pi*sqrt(LC)] whereas for RC filter is Fout=1/(2*pi*R*C)...back to the RC filter which gives 6db roll off per octave, here's the scenario:

let's set the pre emphasis to 50uS and the following can be calculated:
time constant,t=RC (simple capacitor + resistor time constant formula)
F(cut off)=1/(2*pi*RC)

since time constant,t=RC hence the formula can be converted to:
F(cut off)=1/(2*pi*t)

substituting the value of t=50uS:
F(cut off)=1/(2*pi*50x10^-6)
=3183.098862Hz

My question is, am i calculating this correctly cause from your circuit of preemphasis, the parallel RC filter which values are R=470ohm and C=100nF, would give:

time constant,t=470*100x10^-9
=47uS

OR

f(cut off)=1/(2*pi*47*10^-6)
=3386.275385Hz

There's a slight deviation from the ideal case, will it affect the receiver side or anything critical for the FM transmission?
I am aware that there's tolerance for the values of resistor and capacitance and perhaps that tolerance will help get somewhere close to the 50uS time constant range.btw, why do we need a 6dB roll off?

another question is that you mentioned that the capacitor C2 is used to filter off 100MHz radio frequency? Is this frequency from other radio station or in another words ambient noise? and what does transistor overloading means, does it mean that the slew rate(rate of change of output based on input) of the transistor cannot cope with the 100MHz signal? on a side note, is this how you calculate the cut off frequency of the RC filter of R3 and C2:
R3=30k ohm,C2=100pF

f(cut off)=1/(2*pi*30k*100p)
=53.051kHz

does this mean that R3-C2 forms a low pass filter with the calculated cut off frequency?

lastly, sorry for asking so many elementary questions, trying to understand it from the electronics point of view...and thank you very much for advising!Appreciate it!
 
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My question is, am i calculating this correctly cause from your circuit of preemphasis, the parallel RC filter which values are R=470ohm and C=100nF, would give:

time constant,t=470*100x10^-9
=47uS
Correct. 47us is 6% too low but you wll not notice.

why do we need a 6dB roll off?
FM radio stations boost high frequencies with pre-emphasis that has a 6dB/octave slope and FM radios us a 6dB/octave de-emphasis roll-off so that the audio is returned to normal but hiss is reduced by 6 dB/octave.

you mentioned that the capacitor C2 is used to filter off 100MHz radio frequency? Is this frequency from other radio station or in another words ambient noise?
The FM transmitter that you are talking about has an extremely strong signal at its radio frequency (about 100MHz) due to the very short distance from the antenna to the mic preamp.

what does transistor overloading means, does it mean that the slew rate(rate of change of output based on input) of the transistor cannot cope with the 100MHz signal?
Overloading means overloading. The input signal level is much too high.

is this how you calculate the cut off frequency of the RC filter of R3 and C2:
R3=30k ohm,C2=100pF
At 100MHz, the 100pF of C2 has a reactance of only 16 ohms which is almost a dead short so it reduces the RF signal to almost nothing.

does this mean that R3-C2 forms a low pass filter with the calculated cut off frequency?
R3 has nothing to do with it because the impedance of the mic is much lower and is in parallel with R3.
 
hi audio guru,
Thank you so much for enlighthening me with the details of your incredible design! Appreciate it!:) i browsed through the net a saw a post (in 2006) by you recommending that these are good FM transmitter for mp3:
BH1417 Stereo FM Transmitter
BH1417 Stereo PLL FM Transmitter

my main goal to is try and transmit audio (mono) not stereo but all of the transmitter is a stereo transmitter which is good for music/song transmission. is there any way i can fool the system and transmit mono audio? (isn't mono better than stereo for audio transmission?)

btw, in your circuit after the pre emphasis, i noticed that there are 2 transistors, Q2 is the high frequency colpitts oscillator but what is the purpose of the part Q3? Am i mistaken or something?
 
in lighter vein, the whole article is being prepared step by step from the mind of Audioguru. Not much of effort on the part of the poster, and incidentally i feel the O P's post appears hijacked.
 
sorry, didn't mean to hijack the post.....@audioguru, i found another post by you explaning the function of each part of your design...but not the BH1417 transmitter...thank you for sharing your knowledge:)
 
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Rohm makes a modern entire line of FM srereo transmitter ICs (BH141x) to replace their awful old dashioned BA1404 one.
These ICs are used in most MP3 to car radio adapters that are sold.

My main goal to is try and transmit audio (mono) not stereo but all of these transmitters are stereo transmitters which is good for music/song transmission. Is there any way i can fool the system and transmit mono audio? (isn't mono better than stereo for audio transmission?)
The BH141x FM transmitters have very good stereo. Older stereo transmitters were distorted and had poor frequency response. A cheap FM stereo radio also has poor stereo quality but high quality home stereos or car radios are excellent in stereo.

btw, in your circuit after the pre emphasis, i noticed that there are 2 transistors, Q2 is the high frequency colpitts oscillator but what is the purpose of the part Q3?
In my write-up about the problems with the original transmitter that didn't work, I mentioned that its frequency changed when something moved toward or moved away from its antenna. It was caused by the "something" that was moving changed the capacitance at the antenna and the antenna was connected directly to the tuned circuit of the oscillator so of course the varying capacitance at the antenna changed the tuned frequency. So I added Q3 to isolate the antenna from the oscillator and at the same time provide more RF output power.
 
Fault on your board:-
 

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