Hi ! I am making an RF Transmitter and i have a problem with power leakage for LO port to IF port and RF port. I don’t know what to do and i am little confused about how can i fix this.
Source(-10 or -20dBm) 10KHz --------->MIxer conv.loss=8db--------->RF port
and
L.O=70MHz and 7dbm
I put an oscilloscope to mixers input IF and i don’t see a sin wave from my Source.it is a sine wave with a lot of phase noice and amplitude variations…when i decrease the L.O power the sine becomes better.
I have heard about attenuators but i cannot understand how they will solve the problem. They attenuate maybe the standing waves that maybe create ?Also and my signal right?
Resistive attenuators can help maintain to impedance matching in/out of the mixer. This then help reduce reflections on the transmission lines between circuits.
Also, many RF circuits have a specified impedance at a narrow range of frequencies and a Mixer produces many other frequencies: L+R, L-R, 2L+R, 2L-R, etc.
You may also be seeing all of these other frequencies which need to be filtered out before being sent to the next amplifier stage. So typically the mixer is followed by a resistive pad (attenuator) to ensure impedance matching then a Band-pass filter before the next amp stage.
Also, an O’scope is helpful when examining these signals but a better instruments is a Spectrum Analyzer.
Do you have any software to simulate these RF circuits?
Even SPICE can be used although one must work to properly setup the circuit and to do the analysis.
According to the specs, you should expect around 5dB (5.2) of insertion (conversion) loss from the RF to IF. And about 40dB LO-to-IF isolation. So with a +7dBm LO, the leakage to the IF should be around -33dBm; still well below the minimum input you had of -20dBm at the RF. What does all this means? If you’re really using the levels you state, the LO leakeage should have little effect on your oscilloscope measurement (even granting that the oscilloscope is really not the right tool for this). Now you could be having issues with reflections, or you could be overdriving things without knowing. Ideally you like to have 3dB pads at each mixer port to assure good matching in broadband conditions (because mixers are non-linear, and can have unexpected responses in a wide band). 6dBs would be better yet (3dB pad provides 6dB return loss worst case, 6dB pad provides 12dB RL worst case - really nice), but many times this is not a luxury that can be afforded. But for you initial experiment I would definetly go with the 3dB pads at each port. Now that will mean that you must had compensate the LO signal for this loss. Another point is that the IF port should not be directly driving the oscilloscope, it should drive a 50-ohm load, and you should use the oscilloscope as a high impedance load (1M) which is “sampling” the voltage across the 50-ohm load.
Now the last thing (and perhaps most important in the setup you have) is that given your LO (70MHz) and RF (10kHz), you will have an IF at exactly the same level at 69.990MHz and also at 70.010MHz. That is to say, you will have two main products out of your mixer conversion. And the oscilloscope has no means to discriminate against either one. That will definitely show as a “fuzzy” signal in the oscilloscope. To discriminate, you need a filter, but good luck finding a filter that can discriminate 20kHz in 70MHz center frequency (this would be a filter with a selectivity, Q, on the order of 3500). If such a low IF is required, a double conversion topology is usually looked at (but you best ask yourself is such a low IF really required).
@waltr: I have a crystal filter after mixer’s RF port so i cut the unwanted. i have a spectrum analyzer too,in my lab. So i must use a pad to make a good matching ?
@languer: -33dBm the leakage and -20 dBm the signa’s power so i am ok with this…ok…
about the matching now…i don’t have so much money to my lab right now to buy lot of attenuators so you must say to me where is the most important places to put pad. i have two mixers to my transmitter and two mixer to the receiver.
The greater the attenuation from a resistive pad the better the RL (return loss) or matching.
Why not just make the attenuator pads you need? Two SMA connector, a piece of copper cad PCB and three resistor is all that’s needed at 10kHz. For 70MHz solder a box together with some more copper clad PCB board.
(1) Between the 1st mixer (30MHz RF port) and the 30MHz XTAL filter input. Why? XTAL filters usually present a good match within the passband, but are highly reactive outside of the passband. These usually causes nasty reflections in the mixer and creates mixer spurious with much higher levels than expected.
(2) Between the 30MHz XTAL filter output and the 2nd mixer (30MHz IF port). Same reason as above.
(3) Between the LO inputs and the mixer. This can be debated. The mixers do like to be overdriven, but depending on the driver (i.e. LO generator), high reflections can really mess up the drive and cause higher spurious levels, higher noise, and in some cases parametric oscillations on the driver amp. So if a pad can be afforded (and this usually has nothing to do with the money, but more with the power budget of the LO path), it will help.
I would place pads at all ports first, and then take them out as you can deem then not necessary.
The image frequency, even though the filter is good (Q ~ 3000) is still going to show up (I think) on the oscilloscope more than the LO leakage. Since you have a SA (spectrum analyzer) you can check the levels.
Ok i understand that at stopband of a filter the impedance is not 50 ohm.something else…
but, these reflected for stopband frequencies finally where are they going? why i will have benefit if i match the stopband with mixers output?
They will get reflected back into the Mixer. Then add/subtract to the LO and get readmitted out the IF port. This becomes a total mess and all kinds of frequencies and is called 'intermodulation'. Here is a calculator see some of the frequencies that can be generated:
Since you have a SA you can see these for your self. And what you’ll see on the SA is spurious spurs that are not the primary (1st order) mixer products. You then increase the attenuation at the mixer ports until the spurious spurs go away.
(2) The leakage of concern (which you mentioned) was on the 1st mixer from the LO to the RF. Not sure what else to say about that. The image frequency I mentioned is not leakage, it is the other fundamental (or main) mixer product which in your case is undesired (@ 30.02MHz). It is undesired because it contains the same information as the desired product(@ 30MHz) and your receiving system usually does not care about it. It will also cause its own set of frequency (intermodulation) products (after it passes through another mixer - like the 2nd mixer, or a non-linear device) which will interfere with your desired signal at one point or another.
Ok i understand that at stopband of a filter the impedance is not 50 ohm.something else…
but, these reflected for stopband frequencies finally where are they going? why i will have benefit if i match the stopband with mixers output?
They will get reflected back into the Mixer. Then add/subtract to the LO and get readmitted out the IF port. This becomes a total mess and all kinds of frequencies and is called 'intermodulation'. Here is a calculator see some of the frequencies that can be generated:
Since you have a SA you can see these for your self. And what you’ll see on the SA is spurious spurs that are not the primary (1st order) mixer products. You then increase the attenuation at the mixer ports until the spurious spurs go away.
Ok…so i match the mixer output with a pad input and pad output to filter input to have 50ohm to transfer the max power in the passband.
At the stopband the reflected will cut for the attenuation of the pad.3dB power down is good maybe…
:ugeek: …Stopbands signals are very low power… I will give you some number from my lab
mixer’s input 10khz -10dBm power. LO=30.010KHz and +7dBm.
ouput mixer: 1st :=LO=30.010=-60dBm
2nd:LO+IF=30.020 =-18Dbm and LO-IF=30.000 (desired)=-18dBm
3rd,4th…etc…
Crustal filter fc=30MHz BW = +/- 10Khz
Ouput of filter: 1st:-55 dBm ( a little higher i don’t know why)
2nd:LO+IF=30.020=-52dBm and LO-IF=30.000=-18.50dBm
My desired is 30MHz at -18.50dBm… the 30.020 and 30.010 signal that are in the stopband are much low after the filter.
Dear languer…i am 24 years old beginner at RF and i have to complete a 2x2 RF-MIMO system on my own…the help i have from the laboratory stuff is poor…they are very rude people…i have find the forum unfortunately the only solution.sorry about the stupid questions…
You do need to do a lot of experimentation on your own in the lab. Use the SA and other equipment in the lab to test and measure the issues we have discussed and run through the inter-mod’ math to compare with the SA measurements.
So what have you found from measuring the mixer output in the lab with the SA.
Some corrections to your above post:
mixer’s input 10khz -10dBm power. LO=30.010KHz and +7dBm.
ouput mixer: 1st :=LO=30.010=-60dBm
You also have the difference as well as the sum output from the mixer’s IF port. So:
mixer output : 1st : LO = 30.010 at -60dBm AND29.090MHz at -60dBm
This other mixer product that can cause problems in the later stages and must be included in the frequency calculations.
waltr:
You do need to do a lot of experimentation on your own in the lab. Use the SA and other equipment in the lab to test and measure the issues we have discussed and run through the inter-mod’ math to compare with the SA measurements.
So what have you found from measuring the mixer output in the lab with the SA.
**example** my mixer RF port has VSWR=3:1 (Preflect=30%) and next component is a filter with VSWR=1.5 (Preflect=4%).
I read that the total VSWR is 4.5:1 so ~Pref=40%. It is seems to me not right. For example. I have 1 watt signal at RF port. the reflected is 0.25watt. transmitted is 0.75 watt. then at filter the relfected=0,75*0,04=0,03watts. so total reflect signal to 1 watt is 0,28 watt .that is not VSWR 4.5
So a pad can remove the unwanted reflected waves that can cause problems to my device. that’s what i understand :geek:
This is the power loss between stages not the power loss through a stage.
You can not combine the VSWR of the next output stage (filter) but must consider each interface between stages separately. The power loss (in dB) of each interface and stage can be added to determine to total power loss (or through put) of the system. This is typically done by drawing a block diagram of the system (like what you have posted links to) and the Gain (loss) of each stage and interface written on the diagram to determine to system Gain.
This is different than trying to determine inter-modulation due to mis-matched interfaces.
1) First: I don't understand the loss=25% but the reflect power is 20%...
2)from mini circuits "When attempting to determine the total VSWR cr eated by the connection of two components,
it is generally safe to assume that two VSW Rs will tend to multiply rather than add. For example, when connecting a co
mponent with a VSWR of 3.0:1 to a second component with aVSWR of 1.50:1, the resulting maximum VSWR will be 4.50:1, with a corresponding return loss of 3.93 dB. Because of this, the effects of mismatch can be minimized by selecting an attenuator with the lowest possible VSWR."
I don’t understand the combination of two stages and how a attenuator can minimize VSWR… i understand all these about reflections we said but not this one
Fovakis:
I don’t understand the combination of two stages and how a attenuator can minimize VSWR… i understand all these about reflections we said but not this one
The higher the attenuation value, the closer the attenuator becomes to a 50 Ohm load. Find an online calculator and check out the resistance values for a 40 dB attenuator. Also, the attenuator works in both directions, so the reflected power is reduced as well.
Here’s a worst case example: A 50 Ohm transmitter feeding a 3 dB attenuator, with nothing connected to the attenuator output.
If you put 1 W into the attenuator, you would think you would have 100% reflected and therefore 1 W reflected at the transmitter. However, the signal is reduced in each direction.
1 W into the attenuator, 1/2 watt out of the attenuator gets reflected back.
1/2 W coming backwards through the attenuator, 1/4 W reaches the transmitter. So a 3 dB attenuator made a 100% reflection seem like only 25%.
Using an attenuator between stages reduces your power, but improves the impedance for each device.
Look, it is somewhat evident from the questions that you’re not too confortable with RF in general. And that is just fine. If you want to really understand it, then you really have to read it and start doing the math. You can see it empirically (ie. hook up an SA with a directional coupler,measure transmitted and reflected power, and calculate VSWR), but until you do the math it really won’t click.
However, the big picture is this. If you don’t care much about reflections then you don’t really care much about reflections. You want to have as good a match as you can afford, but not much better really. If you are using pads to improve matching you are throwing power away. So you want to make sure you’re doing this only when you need to. Many times when we are measuring devices we use pads to isolate and improve matching between generators, devices, etc. But there are sometimes where we cannot tolerate the loss. In these instances we need other means (wheter an isolator, circuilator, or nothing). At the circuit level you try to use lossless matching when you can, you use pads when you can, and you use nothing when you can.
Back to your test case:
Ouput mixer:
LO => -60dBm
Desired => -18dBm
Undesired => -18dBm
Output of filter:
LO => -55dBm (~ 37dB below desired)
Desired => -18.50dBm
Undesired => -52dBm (~ 34dB below desired)
So the question you have to answer is: “are undesired signals 35dB below my desired signal matter to my receiver?”. We don’t know the answer to this because we do not know what your requirements are. The other question you may need to answer is: “will undesired signals 35dB below my desired signal meet my transmitter requirements?”. Before answering those questions you may want to take a look at these same signals out of your 2nd conversion to see how they fare? You then may want to analyze your frequency plan, figure out which frequencies may give you problems, and concentrate on those. You can use something as simple as Excel to analyze your frequency plan and which frequencies you need to concentrate on.
