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Double balanced diode mixer question

Started by Unknown December 5, 2018
Could some electronics guru here pleae help ?

I want to simulate a double balanced diode mixer
with a LO frequency of 50 MHz and RF frequency of 
100 MHz. Both LO and RF signal generators have a 
source resistance of 50.0 Ohms. The simplest Schottky 
diode model has a fixed resistance Rs, in series with 
a praallel Cj/Rj pair(junction capacitance/resistance).
While Schottky diode SPICE modele provide values for 
both Rs and Cj, there is no quoted value for Rj.

So the question is what is the value of the Schottky 
diode input impedance that is require to impedance
match the impedance of the secondary winding of e.g.,
the LO signal balun ? The secondary winding is just 
an inductor with reactance Xl_LO = 2*PI*fLO*L. As
this reactance must match the Shottky diode's input 
impedance at the frequency f_LO, how do I determine
the imput impedance of the Schottky diode.

Thanks in advance.
 
dakupoto@gmail.com wrote:
> Could some electronics guru here pleae help ? > > I want to simulate a double balanced diode mixer > with a LO frequency of 50 MHz and RF frequency of > 100 MHz. Both LO and RF signal generators have a > source resistance of 50.0 Ohms. The simplest Schottky > diode model has a fixed resistance Rs, in series with > a praallel Cj/Rj pair(junction capacitance/resistance). > While Schottky diode SPICE modele provide values for > both Rs and Cj, there is no quoted value for Rj. > > So the question is what is the value of the Schottky > diode input impedance that is require to impedance > match the impedance of the secondary winding of e.g., > the LO signal balun ? The secondary winding is just > an inductor with reactance Xl_LO = 2*PI*fLO*L. As > this reactance must match the Shottky diode's input > impedance at the frequency f_LO, how do I determine > the imput impedance of the Schottky diode.
The LO isn't matched, or at least not by the mixer. To first order, the resistance of the diodes isn't a consideration. Ideally, their impedance is zero when forward biased and infinite when reverse biased. The IF load is matched to the RF source, because the diodes connect those through to each other, with the polarity changing each half-period of the LO. In practice, of course, the diodes have finite non- zero impedances, which cause losses and signal leakage between ports, both unwanted. Jeroen Belleman
Am 05.12.18 um 12:06 schrieb dakupoto@gmail.com:
> Could some electronics guru here pleae help ? > > I want to simulate a double balanced diode mixer > with a LO frequency of 50 MHz and RF frequency of
www.rubiola.org, in particular (linked from there): < https://arxiv.org/pdf/physics/0608211.pdf > A site full of treasures. Finally, a prof. who does not hide his stuff behind the IEEE paywall. cheers, Gerhard
On Wed, 5 Dec 2018 03:06:14 -0800 (PST), dakupoto@gmail.com wrote:

>Could some electronics guru here pleae help ? > >I want to simulate a double balanced diode mixer >with a LO frequency of 50 MHz and RF frequency of >100 MHz. Both LO and RF signal generators have a >source resistance of 50.0 Ohms. The simplest Schottky >diode model has a fixed resistance Rs, in series with >a praallel Cj/Rj pair(junction capacitance/resistance). >While Schottky diode SPICE modele provide values for >both Rs and Cj, there is no quoted value for Rj. > >So the question is what is the value of the Schottky >diode input impedance that is require to impedance >match the impedance of the secondary winding of e.g., >the LO signal balun ? The secondary winding is just >an inductor with reactance Xl_LO = 2*PI*fLO*L. As >this reactance must match the Shottky diode's input >impedance at the frequency f_LO, how do I determine >the imput impedance of the Schottky diode. > >Thanks in advance. >
Are you using LT Spice? You could use one of the schottky diodes in its pull-down list. Some of them would be fine at 100 MHz. -- John Larkin Highland Technology, Inc lunatic fringe electronics
On 5.12.18 13:06, dakupoto@gmail.com wrote:
> Could some electronics guru here pleae help ? > > I want to simulate a double balanced diode mixer > with a LO frequency of 50 MHz and RF frequency of > 100 MHz. Both LO and RF signal generators have a > source resistance of 50.0 Ohms. The simplest Schottky > diode model has a fixed resistance Rs, in series with > a praallel Cj/Rj pair(junction capacitance/resistance). > While Schottky diode SPICE modele provide values for > both Rs and Cj, there is no quoted value for Rj. > > So the question is what is the value of the Schottky > diode input impedance that is require to impedance > match the impedance of the secondary winding of e.g., > the LO signal balun ? The secondary winding is just > an inductor with reactance Xl_LO = 2*PI*fLO*L. As > this reactance must match the Shottky diode's input > impedance at the frequency f_LO, how do I determine > the imput impedance of the Schottky diode. > > Thanks in advance.
Your frequency selection is bad, with the signal a multiple of the oscillator frequency. The diodes in a diode mixer are overdriven hard, to function as switches, and the load on oscillator input is highly non-linear. There will be no impedance match here, especially with a diode ring. You need a matching attenuator here, if the oscillator is fussy about its load. To get useful output from the mixer, you do need a resonant circuit at the desired output frequency at the output, otherwise you'll see just a mess. Here's a LTspice model of the Heathkit HW-101 balanced modulator, mixing from baseband audio to 3.3 MHz double-sideband signal: Version 4 SHEET 1 1780 680 WIRE 752 112 608 112 WIRE 1008 112 752 112 WIRE 1200 112 1008 112 WIRE 1472 112 1360 112 WIRE 1600 112 1472 112 WIRE 608 144 608 112 WIRE 752 144 752 112 WIRE 1008 144 1008 112 WIRE 1200 224 1200 112 WIRE 1360 224 1360 112 WIRE 1472 240 1472 112 WIRE 96 272 32 272 WIRE 160 272 96 272 WIRE 352 272 240 272 WIRE 448 272 352 272 WIRE 608 272 608 208 WIRE 608 272 512 272 WIRE 656 272 608 272 WIRE 752 272 752 208 WIRE 752 272 720 272 WIRE 832 272 752 272 WIRE 1008 272 1008 224 WIRE 1008 272 912 272 WIRE 832 304 832 272 WIRE 1008 304 1008 272 WIRE 608 320 608 272 WIRE 752 320 752 272 WIRE 912 320 912 272 WIRE 96 336 96 272 WIRE 352 336 352 272 WIRE 608 432 608 384 WIRE 752 432 752 384 WIRE 752 432 608 432 WIRE 1008 432 1008 384 WIRE 1008 432 752 432 WIRE 1200 432 1200 304 WIRE 1200 432 1008 432 WIRE 1360 432 1360 304 WIRE 1472 432 1472 304 WIRE 1472 432 1360 432 WIRE 1472 464 1472 432 WIRE 96 480 96 416 WIRE 352 480 352 400 WIRE 912 480 912 384 WIRE 912 608 912 560 FLAG 96 480 0 FLAG 832 304 0 FLAG 912 608 0 FLAG 352 480 0 FLAG 1472 464 0 FLAG 1600 112 out IOPIN 1600 112 Out FLAG 32 272 in IOPIN 32 272 In SYMBOL voltage 96 320 R0 SYMATTR InstName V1 SYMATTR Value SINE(0 0.5 1k) SYMBOL voltage 912 464 R0 SYMATTR InstName V2 SYMATTR Value SINE(0 1.5 3.3Meg) SYMBOL res 256 256 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R1 SYMATTR Value 2k SYMBOL cap 512 256 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName C1 SYMATTR Value 0.2&micro; SYMBOL cap 720 256 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName C2 SYMATTR Value 10n SYMBOL diode 624 208 R180 WINDOW 0 24 64 Left 2 WINDOW 3 24 0 Left 2 SYMATTR InstName D1 SYMATTR Value 1N4148 SYMBOL diode 624 384 R180 WINDOW 0 24 64 Left 2 WINDOW 3 24 0 Left 2 SYMATTR InstName D2 SYMATTR Value 1N4148 SYMBOL diode 736 144 R0 SYMATTR InstName D3 SYMATTR Value 1N4148 SYMBOL diode 736 320 R0 SYMATTR InstName D4 SYMATTR Value 1N4148 SYMBOL res 992 128 R0 SYMATTR InstName R2 SYMATTR Value 320 SYMBOL res 992 288 R0 SYMATTR InstName R3 SYMATTR Value 320 SYMBOL cap 896 320 R0 SYMATTR InstName C3 SYMATTR Value 100p SYMBOL ind2 1184 320 M180 WINDOW 0 36 80 Left 2 WINDOW 3 36 40 Left 2 SYMATTR InstName L1 SYMATTR Value 1.55&micro; SYMATTR Type ind SYMBOL cap 336 336 R0 SYMATTR InstName C7 SYMATTR Value 40n SYMBOL ind2 1376 320 R180 WINDOW 0 36 80 Left 2 WINDOW 3 36 40 Left 2 SYMATTR InstName L2 SYMATTR Value 6.2&micro; SYMATTR Type ind SYMBOL cap 1456 240 R0 SYMATTR InstName C5 SYMATTR Value 375p TEXT 104 568 Left 2 !.tran 0 10m 2m TEXT 1232 504 Left 2 !K L1 L2 0.5 -- -TV
> > > Your frequency selection is bad, with the signal a > multiple of the oscillator frequency. > >
+1 good observation m
On Wednesday, December 5, 2018 at 7:27:38 AM UTC-5, Jeroen Belleman wrote:
> dakupoto@gmail.com wrote: > > Could some electronics guru here pleae help ? > > > > I want to simulate a double balanced diode mixer > > with a LO frequency of 50 MHz and RF frequency of > > 100 MHz. Both LO and RF signal generators have a > > source resistance of 50.0 Ohms. The simplest Schottky > > diode model has a fixed resistance Rs, in series with > > a praallel Cj/Rj pair(junction capacitance/resistance). > > While Schottky diode SPICE modele provide values for > > both Rs and Cj, there is no quoted value for Rj. > > > > So the question is what is the value of the Schottky > > diode input impedance that is require to impedance > > match the impedance of the secondary winding of e.g., > > the LO signal balun ? The secondary winding is just > > an inductor with reactance Xl_LO = 2*PI*fLO*L. As > > this reactance must match the Shottky diode's input > > impedance at the frequency f_LO, how do I determine > > the imput impedance of the Schottky diode. > > > The LO isn't matched, or at least not by the mixer. > To first order, the resistance of the diodes isn't > a consideration. Ideally, their impedance is zero > when forward biased and infinite when reverse biased. > > The IF load is matched to the RF source, because the > diodes connect those through to each other, with the > polarity changing each half-period of the LO. > > In practice, of course, the diodes have finite non- > zero impedances, which cause losses and signal leakage > between ports, both unwanted. > > Jeroen Belleman
Thank you very much for pointing out that the IF load and the RF source need to be matched.
On Wednesday, December 5, 2018 at 1:03:18 PM UTC-5, Tauno Voipio wrote:
> On 5.12.18 13:06, dakupoto@gmail.com wrote: > > Could some electronics guru here pleae help ? > > > > I want to simulate a double balanced diode mixer > > with a LO frequency of 50 MHz and RF frequency of > > 100 MHz. Both LO and RF signal generators have a > > source resistance of 50.0 Ohms. The simplest Schottky > > diode model has a fixed resistance Rs, in series with > > a praallel Cj/Rj pair(junction capacitance/resistance). > > While Schottky diode SPICE modele provide values for > > both Rs and Cj, there is no quoted value for Rj. > > > > So the question is what is the value of the Schottky > > diode input impedance that is require to impedance > > match the impedance of the secondary winding of e.g., > > the LO signal balun ? The secondary winding is just > > an inductor with reactance Xl_LO = 2*PI*fLO*L. As > > this reactance must match the Shottky diode's input > > impedance at the frequency f_LO, how do I determine > > the imput impedance of the Schottky diode. > > > > Thanks in advance. > > > Your frequency selection is bad, with the signal a > multiple of the oscillator frequency. > > The diodes in a diode mixer are overdriven hard, > to function as switches, and the load on oscillator > input is highly non-linear. There will be no > impedance match here, especially with a diode ring. > You need a matching attenuator here, if the oscillator > is fussy about its load. > > To get useful output from the mixer, you do need a > resonant circuit at the desired output frequency at > the output, otherwise you'll see just a mess. > > Here's a LTspice model of the Heathkit HW-101 > balanced modulator, mixing from baseband audio to > 3.3 MHz double-sideband signal: > > Version 4 > SHEET 1 1780 680 > WIRE 752 112 608 112 > WIRE 1008 112 752 112 > WIRE 1200 112 1008 112 > WIRE 1472 112 1360 112 > WIRE 1600 112 1472 112 > WIRE 608 144 608 112 > WIRE 752 144 752 112 > WIRE 1008 144 1008 112 > WIRE 1200 224 1200 112 > WIRE 1360 224 1360 112 > WIRE 1472 240 1472 112 > WIRE 96 272 32 272 > WIRE 160 272 96 272 > WIRE 352 272 240 272 > WIRE 448 272 352 272 > WIRE 608 272 608 208 > WIRE 608 272 512 272 > WIRE 656 272 608 272 > WIRE 752 272 752 208 > WIRE 752 272 720 272 > WIRE 832 272 752 272 > WIRE 1008 272 1008 224 > WIRE 1008 272 912 272 > WIRE 832 304 832 272 > WIRE 1008 304 1008 272 > WIRE 608 320 608 272 > WIRE 752 320 752 272 > WIRE 912 320 912 272 > WIRE 96 336 96 272 > WIRE 352 336 352 272 > WIRE 608 432 608 384 > WIRE 752 432 752 384 > WIRE 752 432 608 432 > WIRE 1008 432 1008 384 > WIRE 1008 432 752 432 > WIRE 1200 432 1200 304 > WIRE 1200 432 1008 432 > WIRE 1360 432 1360 304 > WIRE 1472 432 1472 304 > WIRE 1472 432 1360 432 > WIRE 1472 464 1472 432 > WIRE 96 480 96 416 > WIRE 352 480 352 400 > WIRE 912 480 912 384 > WIRE 912 608 912 560 > FLAG 96 480 0 > FLAG 832 304 0 > FLAG 912 608 0 > FLAG 352 480 0 > FLAG 1472 464 0 > FLAG 1600 112 out > IOPIN 1600 112 Out > FLAG 32 272 in > IOPIN 32 272 In > SYMBOL voltage 96 320 R0 > SYMATTR InstName V1 > SYMATTR Value SINE(0 0.5 1k) > SYMBOL voltage 912 464 R0 > SYMATTR InstName V2 > SYMATTR Value SINE(0 1.5 3.3Meg) > SYMBOL res 256 256 R90 > WINDOW 0 0 56 VBottom 2 > WINDOW 3 32 56 VTop 2 > SYMATTR InstName R1 > SYMATTR Value 2k > SYMBOL cap 512 256 R90 > WINDOW 0 0 32 VBottom 2 > WINDOW 3 32 32 VTop 2 > SYMATTR InstName C1 > SYMATTR Value 0.2&micro; > SYMBOL cap 720 256 R90 > WINDOW 0 0 32 VBottom 2 > WINDOW 3 32 32 VTop 2 > SYMATTR InstName C2 > SYMATTR Value 10n > SYMBOL diode 624 208 R180 > WINDOW 0 24 64 Left 2 > WINDOW 3 24 0 Left 2 > SYMATTR InstName D1 > SYMATTR Value 1N4148 > SYMBOL diode 624 384 R180 > WINDOW 0 24 64 Left 2 > WINDOW 3 24 0 Left 2 > SYMATTR InstName D2 > SYMATTR Value 1N4148 > SYMBOL diode 736 144 R0 > SYMATTR InstName D3 > SYMATTR Value 1N4148 > SYMBOL diode 736 320 R0 > SYMATTR InstName D4 > SYMATTR Value 1N4148 > SYMBOL res 992 128 R0 > SYMATTR InstName R2 > SYMATTR Value 320 > SYMBOL res 992 288 R0 > SYMATTR InstName R3 > SYMATTR Value 320 > SYMBOL cap 896 320 R0 > SYMATTR InstName C3 > SYMATTR Value 100p > SYMBOL ind2 1184 320 M180 > WINDOW 0 36 80 Left 2 > WINDOW 3 36 40 Left 2 > SYMATTR InstName L1 > SYMATTR Value 1.55&micro; > SYMATTR Type ind > SYMBOL cap 336 336 R0 > SYMATTR InstName C7 > SYMATTR Value 40n > SYMBOL ind2 1376 320 R180 > WINDOW 0 36 80 Left 2 > WINDOW 3 36 40 Left 2 > SYMATTR InstName L2 > SYMATTR Value 6.2&micro; > SYMATTR Type ind > SYMBOL cap 1456 240 R0 > SYMATTR InstName C5 > SYMATTR Value 375p > TEXT 104 568 Left 2 !.tran 0 10m 2m > TEXT 1232 504 Left 2 !K L1 L2 0.5 > > -- > > -TV
Thank you very much for the helpful suggestions. However, the LTSpice netlist could not not be used, as I use HSpice(work) and Ngspice(home). I did get a 35 MHz(local oscillator) and 100 MHz(RF) double balanced mixer to work. The output is analysed with a straightforward C language Disrete Fourier Transform orogram that I wrote sometine ago. The first three highest peaks in the power spectrum are: 3 highest frequencies and peaks 4.958516e+06 Hz 89.167763 6.451290e+07 Hz 68.324520 1.339843e+08 Hz 29.739029 Neither the LO(35 MHz) nor the RF(100.0MHz) show up in the highest 3 peaks. In fact, the third highest peak is at 133.9 MHz, which is very closes to the sum of the LO and RF frequencies(135 MHz). So, this design, though not optimum is working. the
On Wednesday, December 5, 2018 at 7:44:54 AM UTC-5, Gerhard Hoffmann wrote:
> Am 05.12.18 um 12:06 schrieb dakupoto@gmail.com: > > Could some electronics guru here pleae help ? > > > > I want to simulate a double balanced diode mixer > > with a LO frequency of 50 MHz and RF frequency of > > www.rubiola.org, in particular (linked from there): > > < https://arxiv.org/pdf/physics/0608211.pdf > > > A site full of treasures. Finally, a prof. who does not > hide his stuff behind the IEEE paywall. > > cheers, > Gerhard
Thank you very much for the link to a superb paper, whose author presents the topic in a clear straightforward way. You are right that the arxiv.org is a treasure tove of good papaers. Some years ago, I got hold of a similar good paper on the Zobel impedance matching network, and how it can be used to tackle any type of complex impedance matching case, for broadband applications. In this mixer paper, I really like a line the author has put in the introduction. "However simple it seems, every time I try to explain the double balanced mixer to a colleague, something goes wrong".
dakupoto@gmail.com wrote:
> On Wednesday, December 5, 2018 at 1:03:18 PM UTC-5, Tauno Voipio wrote: >> On 5.12.18 13:06, dakupoto@gmail.com wrote: >>> Could some electronics guru here pleae help ? >>> >>> I want to simulate a double balanced diode mixer >>> with a LO frequency of 50 MHz and RF frequency of >>> 100 MHz. Both LO and RF signal generators have a >>> source resistance of 50.0 Ohms. The simplest Schottky >>> diode model has a fixed resistance Rs, in series with >>> a praallel Cj/Rj pair(junction capacitance/resistance). >>> While Schottky diode SPICE modele provide values for >>> both Rs and Cj, there is no quoted value for Rj. >>> >>> So the question is what is the value of the Schottky >>> diode input impedance that is require to impedance >>> match the impedance of the secondary winding of e.g., >>> the LO signal balun ? The secondary winding is just >>> an inductor with reactance Xl_LO = 2*PI*fLO*L. As >>> this reactance must match the Shottky diode's input >>> impedance at the frequency f_LO, how do I determine >>> the imput impedance of the Schottky diode. >>> >>> Thanks in advance. >> >> Your frequency selection is bad, with the signal a >> multiple of the oscillator frequency.
[Snip! ...]
> Thank you very much for the helpful suggestions. > However, the LTSpice netlist could not not be used, as I use HSpice(work) and Ngspice(home). > > I did get a 35 MHz(local oscillator) and 100 > MHz(RF) double balanced mixer to work. The > output is analysed with a straightforward C > language Disrete Fourier Transform orogram > that I wrote sometine ago. The first three > highest peaks in the power spectrum are: > 3 highest frequencies and peaks > 4.958516e+06 Hz 89.167763 > 6.451290e+07 Hz 68.324520 > 1.339843e+08 Hz 29.739029 > Neither the LO(35 MHz) nor the RF(100.0MHz) > show up in the highest 3 peaks. In fact, > the third highest peak is at 133.9 MHz, > which is very closes to the sum of the > LO and RF frequencies(135 MHz). So, this > design, though not optimum is working.
Cleaning up the results of your Fourier transform program, what you see is 5 MHz, 65 MHz and 135 MHz. The last two are (100-35)MHz and (100+35)MHz, the principal mixer products. The 5MHz is actually (100-3*35)MHz, the product of the RF with the 3rd harmonic of the LO. The LO is, or should be, driven hard, so it has the harmonics one expects of a square wave. Jeroen Belleman