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question about rf transmitter circuit

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tanzil_dhk

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hi,
i have some question about rf transmitter circuit for the rfid reader attached here.(this circuit is designed by some students of coronell university):

1. what is the significance of r8 here....
2. the circuit is giving very low current in the output ..isn't that a problem for signal transmition
3. probably the two diodes are used for recoverying distortion
but in pspice simulation not getting any significant improvement...

(if anyone want i can give the .sch format)
 

Well, if its from Cornell, it has to be sweet!

The two diodes are part of the bias network. They keep the two output transistors at roughly the same quiescent current over temperature variations.

If you need more current, you can make the two 100 ohm resistors smaller.

I might filter the square wave input a little more, depending on what the eoutput spectrum looked like.
 

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    tanzil_dhk

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well,
thanx for ur sugession,one thing need to b clear....
the two resistance r7 and r8 are not 100 ohm,they are 10 ohm
would u plz help me a little more.....
in the pspice transient simulation i m getting very low vlotage(1.633mv) in output at the 125khz frequency (i have attached simulation wave)...what does it mean???.....
would u plz explain the transient output from the attached graph...
 

10 ohms is more like it!

What are the voltages on the bases of Q1 and Q3 like (including the DC component)? It is possible that the transistors are not at the correct bias points and are either shut off or turned on fully.
 

sorry for not attaching the wave simulation.....
no...probably thats not the problem....
the voltages at the bases of transistor Q1 and Q3 are respectively 6.875v and 5.420v
 

The circuit you show us is often used as a wide band audio amplifier :)

R7 and 8 are to improve distortion i.e. have better linearity. Diodes are used to have temperature compensation and supply proper bias. I.e. resistors are not needed if you are not afraid of distortion.

You cannot have DC output, as this will short circuit your output stage, you must have a capacitor to remove the DC voltage.

Make sure that the filter on the input is resonant on input frequency or you will attenuate your wanted signal.

As your circuit is DC coupled you can remove the input circuits for a moment and try adjusting the bias network on the input transistor and make sure you do have varying voltage on the output!
 

well,could u plz explain the transient analysis i posted here. isn't it a must to get better voltage at 125kz range(as my motto is to transmit a 125khz signal)..in the simulation u see the voltage is in mv range....my question is should i increase the voltage at 125khz level.
thanx.
 

That is a good point. He should put a 1 uF series capacitor at the output port and simulate again. His simulation program might be be putting a 50 dc coupled resistor to ground at the output.

Added after 12 minutes:

BTW, what happens if you change the 820 ohm series resistor to 0 ohm?
 

well.....
thats not giving an improvement....rather i have connected a capacitor (1nf) in series with R13. this increased the value at 125khz in a satisfactory level but increased distortion in the output wave.
 

Your circuit is giving 2.5mW on 50Ω. Gain is -25dB at 125kHz.
 

thanx a lot...
but could u plz explain the ac analysis of the circuit..why the voltage is too low at 125khz?? ...isn't that aproblem??
 

You made some mistake at simullation. Frequency response of the amplifier depends only on LC series circuit at the input.
 

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