Mạch phát FM 1.5W -- Tầm phát từ 1-2Km

[hungthinhqni]

Tôi xin giơi thiệu thiêu cho các bạn mạnh này tham khảo nha :





Thông số kĩ thuật :


R1 220K
R2 4.7K
R3,R4 10K
R5 100ohm
C1,C2 4.7uF electrolytic
C3,C4 1nF
C5 2-15pF
C6 3.3pF
Q1 BC547C
Q2 2N2219A
P1 25K
MIC ecm series

Description:

This is a well designed basic FM transmitter that you can easily recive the signals transmitted from this transmitter in a 1-2km range with using a normal FM reciever. Another property of this circuit is that the bobin is placed on the printed circuit board. The input sound's amplitude can be adjusted by using the P1 potentiometer. You can easily prepare the PCB by using any PCB editor and apply it to the board by using the ironing method.

[queduong]

Tôi xin giơi thiệu thiêu cho các bạn mạnh này tham khảo nha :





Thông số kĩ thuật :


R1 220K
R2 4.7K
R3,R4 10K
R5 100ohm
C1,C2 4.7uF electrolytic
C3,C4 1nF
C5 2-15pF
C6 3.3pF
Q1 BC547C
Q2 2N2219A
P1 25K
MIC ecm series

Description:

This is a well designed basic FM transmitter that you can easily recive the signals transmitted from this transmitter in a 1-2km range with using a normal FM reciever. Another property of this circuit is that the bobin is placed on the printed circuit board. The input sound's amplitude can be adjusted by using the P1 potentiometer. You can easily prepare the PCB by using any PCB editor and apply it to the board by using the ironing method.

Con 2N2219A lấy đâu ra 1,5W vậy pa ??? hơn nữa lại là 1 tầng dao động.

Xem mạch phải suy luận chứ .

[queduong]

quên , cái mạch FM này ở việt nam chắc giỏi lắm phát đến 50 mét là hết đất

[leduytnt]

Tôi xin giơi thiệu thiêu cho các bạn mạnh này tham khảo nha :





Thông số kĩ thuật :


R1 220K
R2 4.7K
R3,R4 10K
R5 100ohm
C1,C2 4.7uF electrolytic
C3,C4 1nF
C5 2-15pF
C6 3.3pF
Q1 BC547C
Q2 2N2219A
P1 25K
MIC ecm series

Description:

This is a well designed basic FM transmitter that you can easily recive the signals transmitted from this transmitter in a 1-2km range with using a normal FM reciever. Another property of this circuit is that the bobin is placed on the printed circuit board. The input sound's amplitude can be adjusted by using the P1 potentiometer. You can easily prepare the PCB by using any PCB editor and apply it to the board by using the ironing method.

bac co rap mach nay chua hoat dong nhu the nao tam xa toi da bi nhieu?

[duyphi]

cái này nhiều lắm được 20 mét!

[dam_me]

a hungthinh cho e hoi ti nha? no giong nhu cai micro ko day fai ko a? vay moi lan a thu tin hieu la fai lay radio ra ma` do` ha? vay e mun dung de hat karaoke thi sao,sao a ko post lun bo thu len lun cho ze sai???

[httung]

Có bài viết tính cự ly của máy phát gửi các bạn tham khảo, khi tính toán ta cần biết độ nhạy của máy thu, cường độ trường tại điểm thu, công suất của máy phát => cự ly .
Đây là những phép tính đơn giản, mong cácc bạn thông cãm vì bản English.
Estimating Transmitter Distance:
Here is a much simplified equation for analysing low power radio transmitters, for line of sight. It does not take into account probagation conditions or other limiting factors, but does include a variable for the losses in the antenna and tank circuit of a transmitter. It may be applied to low power transmitter circuits such as the circuits on this site. In deriving this equation, I have had to estimate two unknowns, the loss and inefficiency of a telescopic whip antenna, and the small signal high frequency collector current of the transistor in this circuit. Here is an example circuit:



The general equation for estimating transmitter field strength is calculated from the equation below:





Where d is distance in meters, E is the field strength in V/m and Pt is the total power from the transmitter. By finding your radio receivers field strength,(usually in the manual) then the equation can be transposed to solvedistance:





The next step is to work out the power from the transmitter. The 2 stage circuit above works from a 9 volt battery, its output frequency was measured to be 107.2MHz. The final common emitter stage of this circuit, develops power in the tank circuit, which is transferred to the antenna, in this case a 30cm telescopic whip. Most of the power is developed in the coil, there are three ways to calculate this:





At resonance the voltage and current in the oscillator tank circuit will be in phase. Therefore all that is needed is to find the impedance of the tank circuit and either the voltage across it or the current flowing through it. The problem in measuring the ac voltage across the tank circuit is that most meters will not give accurate results at high frequencies. This is the same for high frequency currents. To estimate the ac collector current in the tank circuit, I have worked out the dc collector current. The two values will be slightly different, but as this is only an approximation, the error will not be significant. To find the dc collector current, measure the dc voltage across the emitter resistor and use ohm's law. In my circuit, this measured 2.99V across the 470 ohm emitter resistor, the dc collector current is therefore :

2.99 / 470 = 6.362mA

This value will be substituted for the ac collector current. The impedance of the tank circuit at resonance is given by the following equation:
=100K


The R is the dc resistance of the coil in the tank circuit. At VHF, this is small as coils have only a few turns. In this circuit the dc resistance was measured at 0.1 ohm.

Small signal Analysis:
The equivalent output circuit for the transmitter is now worked out and drawn as below. The impedance of the tank circuit (100K) is in parallel with the output impedance of the transistor. This value, around 40k can generally be ignored, but in this case it is in parallel with the output circuit and makes an appreciable difference. Also the 3.3pf capacitor is in series with the 470 ohm resistor. This is also considered at short circuit as the power supply is decoupled with a capacitor. The capacitive reactance of the 22nF capacitor is a short circuit. The effective load or impedance of the output will be as below:





The overall output circuit is the parallel combination of these components. The 3.3p capacitor has an impedance of around 450 ohms at 107.2 MHz. The combined impedance is therefore:-
40k // 100k // (450+470) = 891.3 ohms

Having now found the impedance, the approximate power in the tank circuit can be calculated:-





Having now found a value for total transmitter power,Pt and using a radio receiver with a known sensitivity of 20uV/meter the distance the signal would be received is worked out:





This equation assumes that all the power in the tank circuit, 36mW is transferred without loss to the antenna and that the antenna has a gain of unity. The result also assumes there are no losses incurred from transmitter to receiver due to probagation effects as well. Using a 30 cm length of telescopic antenna , I have modified the equation to compensate for losses in the antenna and coupling circuit . I have assisgned a variable called AL into the equation and estimated its value at 1%. The modified equation is then:





The new result calculates effective distance from transmitter to a radio receiver with 20uV/m sensitivity. This is clearly a vast reduction in distance from the first result. To test this result, i went to a large field. Holding the transmitter at roughly 1 meter high from the ground, i walked away carrying the receiver. The signal was clearly audible 300 meters from the transmitter giving a strong reading on the signal strength meter of the receiver. This was about the length of the field. I must stress again that the above calculations are ONLY approximate, but if anyone repeats this experiment, i would like to hear rom you.

[tuand9]

sao không nói rõ tần số phát bao nhiêu cho em để đo vậy

[lanhuong]

a hungthinh cho e hoi ti nha? no giong nhu cai micro ko day fai ko a? vay moi lan a thu tin hieu la fai lay radio ra ma` do` ha? vay e mun dung de hat karaoke thi sao,sao a ko post lun bo thu len lun cho ze sai??? -dam_me-

1/. Anh dam_me làm ơn viết chữ Việt có dấu nếu không muốn làm bạn với anh MHz .
2/. Cái mạch ở đây là máy phát FM trong dải tần thông dụng 88MHz - 108 MHz, chứ không phải là micro không dây, nên dùng radio mà thu là tất yếu. Với công suất 1,5W thì pin nào chịu cho nổi.

Micro không dây chỉ chừng 15 - 50 mili W, FM hay AM và cả PLL, do đó thường là không nằm trong các dải tần thông dụng.

Với những giải pháp đặc biệt, các mạch phát micro không dây có cự ly hiệu dụng rất cao đến hàng trăm mét chỉ với công suất bé tí đó, bộ thu dĩ nhiên là phải ráp riêng rồi.

Mạch thu đã có rải rác trong luồng, anh chịu khó tìm vậy.
3/. Riêng về cái "máy phát" này thì đúng là có nhiều "quấn đề", nói 1 - 2Km là ... lạc quan tếu. Những mạch tương tự thế này, Lan Hương đã ráp khá nhiều, không cái nào phát xa hơn ... 30 mét cả, chất lượng phát lại tồi.

Vì sao thì còn phải chờ anh em có thắc mắc không đã.

Lan Hương.