Hello,

I have to use some DBMs in a measurement instrument and of course have to
design the following bp filters.

I have often seen recommended that the DBMs see the resistive load of a
constant R filter (diplexer) rather that the reactive load of a more
traditional filter.

I can think of (model) the DBM mixer to the first order as follows :


.-- +/-1 at LO frequency
|
|
|
V
.------. ZG Zd .-------.
| | ___ ___ | BPF |
VIN--+------------| MULT |---|___|--|___|---+---| H(p) |--- VOUT
| | | | | |
| '------' | '-------'
| ___ .-.
`-|___|----. | |
ZG | | | ZL
.-. '-'
| | |
| |Zd |
| '-' |
I in| | |
| | |
V .-. |
| | |
| |ZL GND
'-'
|
GND
created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de


ZG being the source impedance,
Zd the diode dynamic resistance,
ZL the filter input impedance,
H(p) the BPF voltage transfer function.
The multiplier is a high Zin, null Zout device with unity gain.

I can't see any requirement in this first order model for a constant R load.

Do I miss something ? Is it needed because of second order effects ?
Any thought, any pointer to (preferably) detailed analysis is welcomed.

Thanks,
Fred.

Do I miss something ? Is it needed because of second order effects ?
Any thought, any pointer to (preferably) detailed analysis is welcomed.

John Stephensen. QEX May-June 2001

W4ZCB

Do I miss something ? Is it needed because of second order effects ?
Any thought, any pointer to (preferably) detailed analysis is welcomed.

John Stephensen. QEX May-June 2001

W4ZCB

Fred Bartoli wrote:

Hello,

I have to use some DBMs in a measurement instrument and of course have to
design the following bp filters.

I have often seen recommended that the DBMs see the resistive load of a
constant R filter (diplexer) rather that the reactive load of a more
traditional filter.

I can think of (model) the DBM mixer to the first order as follows :


.-- +/-1 at LO frequency
|
|
|
V
.------. ZG Zd .-------.
| | ___ ___ | BPF |
VIN--+------------| MULT |---|___|--|___|---+---| H(p) |--- VOUT
| | | | | |
| '------' | '-------'
| ___ .-.
`-|___|----. | |
ZG | | | ZL
.-. '-'
| | |
| |Zd |
| '-' |
I in| | |
| | |
V .-. |
| | |
| |ZL GND
'-'
|
GND
created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de


ZG being the source impedance,
Zd the diode dynamic resistance,
ZL the filter input impedance,
H(p) the BPF voltage transfer function.
The multiplier is a high Zin, null Zout device with unity gain.

I can't see any requirement in this first order model for a constant R load.

Do I miss something ? Is it needed because of second order effects ?
Any thought, any pointer to (preferably) detailed analysis is welcomed.

Thanks,
Fred.



The constant-R load is a requirement for diode-ring mixers, not
double-balanced mixers in general. The diode-ring mixer acts like four
switches that turn on and off as a function of the LO, RF and IF port
voltages. You'd like this to just be as a function of the LO, however.

By it's nature a diode-ring mixer does exactly as good a job of mixing
from the IF port to the RF port as it does going in the other direction
(this can be handy for transceiver design, by the way). So when you
terminate the IF (or any other port) with a non-constant R you get
different voltage-current relationships at the image frequency than you
do at the intended IF. These oddball voltages or currents (a) affect
the mixing, and (b) get dutifully converted into energy that appears
back at the RF port to get mixed _again_. Taken to extremes people with
very large brains can even terminate a diode-ring mixer make a
parametric amplifier from RF to IF.

My understanding is that you don't need a perfect DC to light constant-R
termination: you just need to terminate for the expected frequencies out
the IF port; primarily the IF and it's image (so RF +/- LO), but to be
thorough you should consider the LO's odd harmonics.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Fred Bartoli wrote:

Hello,

I have to use some DBMs in a measurement instrument and of course have to
design the following bp filters.

I have often seen recommended that the DBMs see the resistive load of a
constant R filter (diplexer) rather that the reactive load of a more
traditional filter.

I can think of (model) the DBM mixer to the first order as follows :


.-- +/-1 at LO frequency
|
|
|
V
.------. ZG Zd .-------.
| | ___ ___ | BPF |
VIN--+------------| MULT |---|___|--|___|---+---| H(p) |--- VOUT
| | | | | |
| '------' | '-------'
| ___ .-.
`-|___|----. | |
ZG | | | ZL
.-. '-'
| | |
| |Zd |
| '-' |
I in| | |
| | |
V .-. |
| | |
| |ZL GND
'-'
|
GND
created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de


ZG being the source impedance,
Zd the diode dynamic resistance,
ZL the filter input impedance,
H(p) the BPF voltage transfer function.
The multiplier is a high Zin, null Zout device with unity gain.

I can't see any requirement in this first order model for a constant R load.

Do I miss something ? Is it needed because of second order effects ?
Any thought, any pointer to (preferably) detailed analysis is welcomed.

Thanks,
Fred.



The constant-R load is a requirement for diode-ring mixers, not
double-balanced mixers in general. The diode-ring mixer acts like four
switches that turn on and off as a function of the LO, RF and IF port
voltages. You'd like this to just be as a function of the LO, however.

By it's nature a diode-ring mixer does exactly as good a job of mixing
from the IF port to the RF port as it does going in the other direction
(this can be handy for transceiver design, by the way). So when you
terminate the IF (or any other port) with a non-constant R you get
different voltage-current relationships at the image frequency than you
do at the intended IF. These oddball voltages or currents (a) affect
the mixing, and (b) get dutifully converted into energy that appears
back at the RF port to get mixed _again_. Taken to extremes people with
very large brains can even terminate a diode-ring mixer make a
parametric amplifier from RF to IF.

My understanding is that you don't need a perfect DC to light constant-R
termination: you just need to terminate for the expected frequencies out
the IF port; primarily the IF and it's image (so RF +/- LO), but to be
thorough you should consider the LO's odd harmonics.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Hello Fred,

A diplexer is certainly the Rolls Royce method. Elegant and impressive. But
there are others.

You could also provide a buffer amp that has a matching input impedance and the
mixer will be happy. One of my favorites is the AD603 if it fits the dynamic
range requirements, frequency range and cost limits. That gives you a db-linear
gain control of 40dB which often comes in real handy. They don't use Gilbert
cells in there but a patented voltage controlled attenuator so there isn't a
dynamic range penalty with gain reduction.

Regards, Joerg.

Hello Fred,

A diplexer is certainly the Rolls Royce method. Elegant and impressive. But
there are others.

You could also provide a buffer amp that has a matching input impedance and the
mixer will be happy. One of my favorites is the AD603 if it fits the dynamic
range requirements, frequency range and cost limits. That gives you a db-linear
gain control of 40dB which often comes in real handy. They don't use Gilbert
cells in there but a patented voltage controlled attenuator so there isn't a
dynamic range penalty with gain reduction.

Regards, Joerg.