Series resistor sets impedance?

neazoi

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Hi, I have a 50 ohm output HF (1-30MHz) signal generator.
What if I put a series 10k resistor on it's output (center lead of the coaxial)?
Will it increase the impedance seen by external devices connected to it, to about 10k?

For the time being please ignore the effects of signal attenuation, just focus on the impedance set.
 

neazoi said:
What if I put a series 10k resistor on it's output (center lead of the coaxial)?
Click to expand...
A resistor has two legs.
Your description only tells where you connected one leg .. but what did you do with the other leg?

And if you say "increase" ... this means there are two situations: one with the resistor, one without the resistor. You should give clear informations for both situations..
Where exactly are the "external devices" connected?

As so often... I tiny effort to take a pencil and a paper .. would be so helpful.

Klaus
 

    neazoi

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1st case: 50 ohm generator connected to a filter
2nd case: 50 ohm generator connected to a 10k series resistor and the other resistor lead to this filter.

Will the input of the filter "see" a 10k input impedance?
 

neazoi said:
Will the input of the filter "see" a 10k input impedance?
Click to expand...
Your previous question was clear enough, the answer has not changed: yes, the filter will "see" 10kOhm source impedance (10 kOhm + 50 Ohm in series)
 
This is completly true only in theory, because any connection, even short, between the 10k resistor and the filter will alter the impedance, since you system is not matched.
The worst case is the resistor directly connected at the output of the generator, then a coax to go to the filter.
The best case is the coax cable connected to the generator, then the 10k connected as close as possible to the filter.
Furtermore any parasitics capacitance to ground at filter side will play a role.
Of course low frequency behaves better.

What's about the purpose of this added 10k resistor ?
 

Which filter has 10k input impedance? Intended frequency range?
 

Hi,
volker@muehlhaus said:
Your previous question was clear enough, the answer has not changed: yes, the filter will "see" 10kOhm source impedance (10 kOhm + 50 Ohm in series)
Click to expand...
you surely have more HF experience than me ... so usually I rely on your informations.

I´m totaly fine with your answer .. in an idealistic environment ... or in a simulation.

But the OP most probably talks about a real circuit.
Thus - according my idea - the wiring plays a big role. Even which wires he uses and where he places the resistor.
In a coarse simulation you don´t simulate the wires.

Tell me if I´m wrong: in a real circuit the 10k impedance the OP talks about ... is rather easily overruled by the impedance of the wiring.
Thus to really get 10k impedance ... the wiring needs to be done carefully. (and the selection of the resistor itself)
(it´s rather easy to get the 10k impedance when generator and filter is on one small PCB, but still there the GND plane will have influence. I mean talking about 10k as source impedance is rather high for HF).
If I´m not mistaken a stray capacitance of 0.5pF causes a parallel impedance of 10k @ 30MHz (making a total impedance of 7k). And I think 0.5pF are more than realistic on a real circuit.

--> Thus I would vote for the OP to show a photo of the total circuit including generator, wiring, filter... for a more detailed discussion.

Klaus
 

If the cable is terminated with the same impedance (50 or 75) The cable input appears as Zo = sqrt(L/C) as it gets no reflection.

Thus if the cables is lossless, the load sees the same signal attenuated at source by the voltage ratio of impedances.
Ar = 50/(10k+50) over the bandwidth of the cable which I assume is far greater than your 30MHz.

Coax being around 100 pF/m compared to a ~3.5 pF 10:1 scope probe but some attenuation will occur so measure the low R source or load only.

Normally 10~90% step response, Tr yields BW=0.35/Tr but for Tau=RC using 0~63% risetime, it's a bit more BW.

Compute 0.5/RC=BW= 0.5/(1e4*3e.5-12)=1.4 MHz for 1 m or 14 MHz for 10m so be mindful of how you probe high impedance sources with RF. It needs a FET probe or don't try.
--- Updated ---

a thousand less words

 

D.A.(Tony)Stewart said:
If the cable is terminated with the same impedance (50 or 75) The cable input appears as Zo = sqrt(L/C) as it gets no reflection.
Click to expand...
This is true if both sides of the cable are matched to the cable impedance. In any case the OP asks for the impedance at filter side that is on the right side of the coax in your simulation, not on the left side where you placed the measurement probe. In the point you measure is correct to see 10k (as per your simulation) since the 50 ohm termination are transferred to the left side (reflection is present but negligible). The 10k impedance, instead, is not transferred to the right side.
 

I disagree.
Zin to coax is not affected by left side or impedance of the source. Thus matching source is not required for my statement to be true.

The Zout will also be 10k + 50 Ohms but with a delayed attenuated source Vs and the return loss at both source and load will be low (meaning high reflections.

To match all impedances and attenuate may be desirable.
The proper method is to delete the source 10k to match the source and make an output pad with desired Av=attenuation, Zin = 50 and Zout as desired to match the filter unless otherwise specified as low Ohm source.

If using a series LC BPF. you want a low Rs for high Q and tap out between LC for BW. e.g. 50.1: 0.1 attenuator
If using a Parallel LC BPF, you want a current source CE amp with some load or series hi-R for desired Q
If it is a ceramic resonator , follow specs.
 
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From the equation of the transmission lines:

Zin = Zo*[(ZL+jZo*tg(beta*length)/(Zo+jZL*tg(beta*length)]

as you can see Zin becomes independent from the length of the cable only if ZL = Zo. Looking into the cable from the termination resistor we have to set ZL = 10k, so the impedance will vary with the length of the cable than at least will not be a constant 10k
 

But he does not want to make a cable filter as you suggest. I said the load should be matched to the cable.
Your previous objection about the source side affecting cable s11 is not valid.
 

D.A.(Tony)Stewart said:
But he does not want to make a cable filter as you suggest. I said the load should be matched to the cable.
Your previous objection about the source side affecting cable s11 is not valid.
Click to expand...
As far as I've understood the question is what's the impedance seen by the filter; so we are interested to S22 (right side) and not to S11 (left side). If the the cable is terminated onto Zo, then S11 will be Zo. But we are interested in S22 where the termination on the other side of the cable is 10k. This means the load doesn't match the impedance of the cable, this means the impedance seen by the filter will not be 10k.
Possibly I was not been clear enough in my previous posts.
 

All I'm saying is if the cable is classic transmission theory. If the cable length is more than 10% wavelength one should match both ends and then transform the impedance on the output after this is correct if it is 10k.

 

albbg said:
Looking into the cable from the termination resistor we have to set ZL = 10k, so the impedance will vary with the length of the cable than at least will not be a constant 10k
Click to expand...
Without 50 ohm termination near the 10 k resistor, you still see constant impedance into the cable as long as it has correct source termination. There are many reasons to implement both side termination for high performance RF and pulse transmission. Source side termination with high impedance load (Zload >> Z0) is however a valid scheme and has the advantage of receiving full generator voltage at load side.

Your considerations are not right if the cable is terminated with Z0 at the source side. You see 10k + 50 ohm, independent of cable length and frequency, as already stated in post #2 and #5.
 

Hi,

we can guess and guess, endlessly ...
--> only a photo will tell the real arrangement

Maybe it´s done rather optimal and the filter will see close to 10k...
or the arrangement is non optimal and the filter will see a much lower impedance.
The truth will be somewhere between 50 Ohms and 10050 Ohms


Klaus
 

D.A.(Tony)Stewart said:
All I'm saying is if the cable is classic transmission theory. If the cable length is more than 10% wavelength one should match both ends and then transform the impedance on the output after this is correct if it is 10k.
Click to expand...
This is correct and corresponds to the connection "best case" of my post #6. The post #9, instead, refers to the opposite situation: "worst case". The problem is the transmission line between 10k and load.

Please find attached an image with some consideration.
 

There's mismatch at A and B, but only a lumped 10 k resistor, thus no complex impedance involved. There's zero reflection at C, thus the impedance at B seen into the cable is Z0, independent of cable length. The impedance seen into the cable at port C is however complex.
--- Updated ---

The original question is unclear because doesn't consider a transmission line and it's position in the setup. I was reading the question so that there's no cable length between 10k resistor and filter, as you did in post #6
albbg said:
The best case is the coax cable connected to the generator, then the 10k connected as close as possible to the filter.
Click to expand...
--- Updated ---

I'm presuming that we have 50 Ohm generator impedance but different and usually frequency dependent filter impedance.
 
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