100KHz low pass with constant 50R input impedance

A down-conversion mixer from up to 200MHz down to 50KHz low IF likes to see a solid 50R load, and not see signals reflected from a following low-IF filter. It should not be necessary to use a really fast op-amp (that can deal with image frequencies) in the LPF, if a passive LPF is used first. So that's what this filter is intended for.

The filter sketched has 100KHz corner frequency, and a very constant 50R input impedance. I got this working nicely in LTSpice, but can't find the ASC file now.

It seems it should be possible to do away with or reduce some of the resistors (at the expense of greater dependence on the 50R nature of the load) and in the process to reduce the loss... but I can't see how to calculate that.

Any thoughts on reducing the loss while staying close to 50R in and out of the passband?

Clifford Heath.

Reply to
Clifford Heath
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Jeroen here posted some designs for constant-input-impedance filters. I did some simplified versions, not as constant as his.

The idea is to make an LC filter, and hang a network across the input to absorb the stopband drive that the filter wants to reflect. The simplest such network is a series RC to ground.

Your filter looks pretty lossy to me. A pure LC filter has theoretically no losses in the passband.

Here are Jeroen's notes:

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Do you need wideband termination on both ends of the filter? Nothing outside of the passband makes it through the filter, so the output side may not need to be wideband matched.

My versions were lopsided too, only wideband 50 ohms on one end, and not so good a match as Jeroen's. I only needed to do a pretty good job of absorbing cable reflections.

--
John Larkin         Highland Technology, Inc 

lunatic fringe electronics
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Reply to
John Larkin

Do you need a constant impedance up to a few gigahertz or just some nice 50 ohm load around the 400 MHz (RF+LO image frequency) ?

Use a 100 kHz HPF to feed a 50 ohm dummy load (idler) to dissipate the unwanted power.

Reply to
upsidedown

Am 15.06.19 um 04:20 schrieb Clifford Heath:

Yes, LTspice has its own ideas what to present in the "recently used" history and what not.

Back in the time when shortwave receivers still were of some interest,they had crystal roofing filters with truly weird impedances. The ring mixers did not like that: collapse of IP3. The solution was a large JFET in CG between the mixer and the filter that could be set up for 50 Ohm impedance, wideband.

TI P8000 / P8002 or Teledyne Crystalonics CP643. (RIP) That also made up for the 6-8 dB loss of the ring mixer. The FET ran at 55 mA and was quite linear, unimpressed by the filter Zin.

regards, Gerhard

Reply to
Gerhard Hoffmann

Well, up to GHz really, since this is for instrumentation use I cannot know what antenna might be used, just that I don't want rectified spuria upsetting things. We plan to digitise I & Q from the output of the LPF, so can reduce the resolution BW below 200KHz in DSP. Can't get rid of something odd that aliases into that band though, so the LPF should be clean and flat-looking.

Clifford Heath.

Reply to
Clifford Heath

To stay resistive, it has to match the order of the-band.

It's necessary to "lose" the stop-band signal somewhere, so that has to be resistive. The question is how not to lose so much pass-band.

Filed away for reference thanks, but these are all band-pass filters, and >=3rd order, which I don't need.

I don't think the kind of LPF I will follow this with (100KHz) will show much impedance variation from 1MHz up, so no, I don't think I do. If I needed matching at the passive filter output I would use a buffer stage.

I just want to stop most of the RF from reaching the steep op-amp filter after the passive filter.

Clifford Heath.

Reply to
Clifford Heath

What's wrong with the traditional diplexer approach. LPF to IF path and HPF to resistor?

piglet

Reply to
piglet

Jeroen's filters are all lowpass and start at 3rd order.

I was thinking that an ideal LC lowpass can be paralleled at its input with an ideal LC highpass, both filters load-end terminated. The paralleled input end must look like a perfect wideband 50 ohms.

(derived by the maximum power transfer theorem and conservation of energy)

So the best front-end correction network for a lowpass might be a terminated highpass.

For the first-order case, the lowpass is just series L into the 50r load, and the absorber is C+50r to ground. If L/50 = C*50, the input looks like 50 ohms wideband.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

Yes, you can do that for Butterworth filters. For other filters, there will be some ripple around the cross-over frequency. The trick is to minimize that.

There are also constant-resistance bridged T or L sections. These have first-order frequency responses. Let me know if you're interested.

For examples and more, see N. Balabian, T.A. Bickart, Electrical Network Theory, Wiley 1969, ISBN 0471-04567-4 and G. Amsel, R. Bosshard, R. Rausch, C. Zajde, Time domain compensation of cable induced distortions using passive filters for the transmission of fast pulses, Rev. Sci. Instr. Vol. 42, No. 8, August 1971, pp. 1237-1246

Jeroen Belleman

Reply to
Jeroen Belleman

The low-stress "LTSpice-accelerated" design procedure could be to design a lattice filter lowpass with a constant input impedance (Za and Zb conjugate duals of each other.)

Then use the transformation equations to transform to the equivalent single-ended bridged-T network which will have the same properties. tweak that form of it in LTSpice using the available off-the-shelf inductance values as parameters.

Reply to
bitrex

I'd like to see that, if it's not too much trouble. We are doing some work with distributed amplifiers and it may relate.

Mini-Circuits has some fancy tiny constant-R filters.

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but only down to 150 MHz.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
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Reply to
John Larkin

Conversion to a bridged-T network including the 50 load to help define the wideband input impedance will help reduce the loss, consider:

How Zin is not dependent on the source impedance, and:

For how the much easier-to-analyze lattice structure can be mechanically converted to an unbalanced equivalent.

Reply to
bitrex

Ahh, right you are, I saw series LCs and only glanced at the curve, thinking it was S21. Still, 3rd order is unnecessary as I will follow it with a steep op-amp filter. Just want to keep RF out of the op-amp.

Yes, that's the basic idea. There's a few different ways to do it.

That might work. It must have been too obvious for me to see it :)

Clifford Heath

Reply to
Clifford Heath

That's almost exactly the filter I showed, but with the shunt R across the capacity instead of the output. It's pretty lossy. Maybe that's unavoidable for constant-R.

Clifford Heath.

Reply to
Clifford Heath

Thanks, that helped. I found my LTSpice file, and modified it to use the exact topology from that page and to calculate the resistors for a given

0 < k < 1. What I had before was a result of a fixed k=0.25, and I didn't know the formulae for varying k. The result is here for anyone else to use.

With k=0.98 you get close to the minimum 6dB loss. At that point, you can short the 1R Rseries, remove RShunt, and Zin still stays within an ohm of 50, which is good enough for the mixer.

Clifford Heath.

---- Cut Here for ConstZ_LPF.plt ---- [AC Analysis] { Npanes: 2 { traces: 1 {2,0,"-V(Input)/I(Rsource)"} X: ('M',2,10000,100000,1e+06) Y[0]: (' ',4,49.9998,0.0001,50.001)

Log: 0 0 0 PltReal: 1 PltImag: 1 Representation: 2 NeyeDiagPeriods: 0 }, { traces: 1 {524291,0,"V(output)"} X: ('M',1,10000,0,1e+06) Y[0]: (' ',0,0.0398107170553497,2,0.501187233627272) Y[1]: (' ',0,-88,8,-0) Log: 1 2 0 PltMag: 1 PltPhi: 1 0 NeyeDiagPeriods: 0 } }

---- Cut here for ConstZ_LPF.asc ---- Version 4 SHEET 1 1204 808 WIRE 144 112 80 112 WIRE 496 112 224 112 WIRE 640 112 576 112 WIRE -48 208 -96 208 WIRE 80 208 80 112 WIRE 80 208 32 208 WIRE 128 208 80 208 WIRE 288 208 208 208 WIRE 448 208 288 208 WIRE 496 208 448 208 WIRE 640 208 640 112 WIRE 640 208 576 208 WIRE 800 208 640 208 WIRE 288 304 288 208 WIRE 448 304 448 208 WIRE 640 304 640 208 WIRE 800 304 800 208 WIRE -96 352 -96 208 WIRE -96 496 -96 432 WIRE 288 496 288 368 WIRE 288 496 -96 496 WIRE 448 496 448 384 WIRE 448 496 288 496 WIRE 640 496 640 368 WIRE 640 496 448 496 WIRE 800 496 800 384 WIRE 800 496 640 496 WIRE 640 528 640 496 FLAG 640 528 0 FLAG 80 208 Input FLAG 800 208 Output SYMBOL voltage -96 336 R0 SYMATTR InstName V1 SYMATTR Value SINE(0 1 3Meg 0) SYMATTR Value2 AC 1 0 SYMATTR SpiceLine Rser=0 SYMBOL ind 128 128 R270 WINDOW 0 32 56 VTop 2 WINDOW 3 5 56 VBottom 2 SYMATTR InstName L1 SYMATTR Value {L} SYMBOL cap 304 368 R180 WINDOW 0 24 56 Left 2 WINDOW 3 24 8 Left 2 SYMATTR InstName C1 SYMATTR Value {C} SYMBOL res 224 192 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R1 SYMATTR Value {Z0} SYMBOL res 592 192 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R2 SYMATTR Value {Z0} SYMBOL res 464 400 R180 WINDOW 0 36 76 Left 2 WINDOW 3 36 40 Left 2 SYMATTR InstName R4 SYMATTR Value {Rshunt} SYMBOL res 48 192 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName Rsource SYMATTR Value {Z0} SYMBOL res 784 288 R0 SYMATTR InstName Rload SYMATTR Value {Z0} SYMBOL res 592 96 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R3 SYMATTR Value {Rseries} SYMBOL cap 624 304 R0 SYMATTR InstName C3 SYMATTR Value {Xload} TEXT -128 792 Left 2 !.AC DEC 20 {F0/10} {F0*10} TEXT -32 8 Left 2 ;Constant input impedance low-pass filter, e.g. for IF output from a mixer TEXT -128 584 Left 2 !.PARAM F0=100kHz Z0=50 k=0.98 a=1 TEXT -128 560 Left 2 ;Edit this to set the cutoff frequency, characteristic impedance, passband loss: TEXT -128 768 Left 2 ;Edit this to change the simulation frequency range: TEXT -128 648 Left 2 ;Edit this to experiment with varying load reactance: TEXT -128 672 Left 2 !; .STEP PARAM Xload LIST -100p 0p 100p\n.PARAM Xload 0p TEXT 488 656 Left 2 !.PARAM w0=2*pi*F0\n.PARAM L=Z0/(w0*k) C=1/(w0*Z0*k)\n.PARAM Rseries=Z0*(a-k)/k Rshunt=Z0*k/(a-k)\n.measure L param {L}\n.measure C param {C}\n.measure Rseries param {Rseries}\n.measure Rshunt param {Rshunt} TEXT 488 616 Left 2 ;Calculations and measurements:

Reply to
Clifford Heath

Glad that worked out. Yeah the bridged T should not be that lossy, the

150 ohm doesn't need to be there.
Reply to
bitrex

Of incidental interest here is a paper from back in the day (proceedings of the journal of network analysis-something-something, 1964) that shows it's impossible to build a variable attenuator with a constant output impedance out of any number of linear elements and linear pots ganged on a shaft:

Reply to
bitrex

The next paper in line looks interesting too "An elementary method for obtaining network responses to periodic, non-sinusoidal sources."

Maybe I gotta bribe my academic contact and see if I can get my hands on the full set.

Reply to
bitrex

A common method at Microdyne was to use a 3dB pad between the filter and a mismatch source or load.

Reply to
Michael Terrell

e

a mismatch source or load.

yes,

the key concept is that an ideal LC filter alone creates loss in the stop b and and transition band by REFLECTING energy back to the input and thus CAN NOT present broadband 50 Ohms. It has no other option. Ideal L C componen ts cannot dissipate energy. So to have a wideband 50 Ohm input you NEED to use a diplexer configuration of HPF with LPF with a dummy load or a pad.

m
Reply to
makolber

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