Using LT5560 Active Mixer - Page 1
Forum section: electronics / projects

author: Digibin
author_rank: Regular Contributor
timestamp: May 30, 2015, 09:57:06 pm
content: I'm looking for some advice regarding Linear Tech'sLT5560. I've been working on a low frequency superhet receiver design and until now I've been using theSA612A, but I'm shifting the design to 3.3V so I need something else for frequency mixing. The LT5560 seems like a good (more modern!) alternative. My system so far is all single-ended and my problem is that I'm not sure how best to interface to its differential inputs.Page 24 of the datasheet gives some info regarding low frequency applications, and I'm basing my design so far on Figure 22. My superhet frequencies are:Stage 1 upconvert:Input frequency f_in = 20kHz to 100kHz1st LO = 220kHz to 300kHzIF = 200kHzStage 2 downconvert:Input is IF = 200kHz2nd LO = 205kHzOutput frequency f_out = 5kHzThis is for a handheld device so I'm looking for the simplest solution with a small footprint - performance can be sacrificed for this.1) First off, is the LT5560 suitable?Is it a good alternative or is there a better device more suited to my application?2) LO connection:The datasheet talks about single ended LO input but from a 50 Ohm source - my LO is generated via DDS and buffered via an op-amp voltage follower. So do I need the 160 Ohm resistor (R2) to ground as shown in Fig 22, or can I just inject it via the 10nF cap alone?3) Input connection:My input signal (20kHz to 100kHz) is again single ended and low source impedance - so I can ground pin 4 and inject my signal at pin 3 through a cap, but pin 3 also needs a DC path to ground, so use a pull-down of what, 200R, 1k, 10k?4) Mixer 1 output to mixer 2 input (200kHz):I don't need any IF filtering so since the output is differential I'll feed it straight into the input of the second mixer, via a cap on each line. According to the Fig 22 I should add pull-ups of 200R to each output line, but on the input side should I simply just do the same but pull-down?5) Mixer 2 output:I need this to be single ended so could simply implement the op-amp output configuration given in Fig 22.Cheers.

author: Odysseus
author_rank: Regular Contributor
timestamp: May 31, 2015, 03:55:14 am
content: 1) It's a gilbert cell, with the twist that the input port is common-base. Should work just fine.2) Resistor not needed.3) For simplicity, ground the unused input, and provide the DC path on the other with an appropriate inductor/choke instead of a resistor. Couple your signal in with a cap. This forms an RLC high-pass filter. Set the lower frequency to 20k or so, and size the L and C so that in combination with the input R (28 ohms) it is critically damped. 2*R=sqrt(L/C)4) Here, a transformer with two center taps (connected to vcc on 1st mixer, and ground on 2nd mixer) would be the simplest solution.5) Yes.

author: Digibin
author_rank: Regular Contributor
timestamp: May 31, 2015, 12:30:57 pm
content: Thanks for the response.So for critical damping L must be 56x greater than C. This condition results in C=1pF and L=62pH for a corner frequency of 20kHz - clearly these aren't suitable since the reactance of 1pF at 20kHz is huge. So would a pull-down resistor be more suitable?Perhaps I'm searching for the wrong terminology but I can't seem to find a small 1:1 PCB transformer with a centre tap on both sides. I've looked at chip baluns but these of course are only centre-tapped on the balanced side. Could you point me towards something suitable?

author: dom0
author_rank: Super Contributor
timestamp: May 31, 2015, 12:46:36 pm
content: Got some ISDN S0 interface transformers laying around? Those should be in the right ballpark.Quote from: graymp89 on May 31, 2015, 12:30:57 pmThe reactance of 1 pF at 20 kHz is huge?Also, R=28 ?, L=62 pH, C=1 pF is not critically damped. L=100 nH, C=33 pF is and is a lot more practical (try to build a 62 pH inductor

author: Digibin
author_rank: Regular Contributor
timestamp: May 31, 2015, 02:48:52 pm
content: Good point - apparently I can't do maths. However L=100nH and C=33pF would give a cut off in the GHz region (unless my maths fails me again). For a 20kHz cut-off [fc=1/(2*pi*sqrt(LC))] the values would be L=4.4mH and C=1.4uF - no?

author: dom0
author_rank: Super Contributor
timestamp: May 31, 2015, 02:49:56 pm
content: Good point - apparently I can't read full postings

author: Digibin
author_rank: Regular Contributor
timestamp: May 31, 2015, 03:11:45 pm
content: Quote from: dom0 on May 31, 2015, 12:46:36 pmUnfortunately not. I've looked at RS and Farnell and haven't found one, problem is I don't really know what to search for or which category of Transformers to look under (have looked through most). I would have thought this would be a fairly common component and would be available on RS/Farnell?

author: Digibin
author_rank: Regular Contributor
timestamp: May 31, 2015, 03:15:13 pm
content: Perhaps I could use two baluns back to back..?Such as this one:http://uk.rs-online.com/web/p/telecom-transformers/8319324/Or better yet this one with both sides centre-tapped:http://uk.rs-online.com/web/p/lan-ethernet-transformers/8319295/

author: moffy
author_rank: Super Contributor
timestamp: May 31, 2015, 11:54:52 pm
content: You could just use digital switches e.g. ADG611, 74HC4066 as your modulators and allow your filters to get rid of the extra components. It's attractive because your frequencies are so low, they are very low power, cheap and readily available.

author: Odysseus
author_rank: Regular Contributor
timestamp: June 01, 2015, 06:14:48 pm
content: Quote from: graymp89 on May 31, 2015, 02:48:52 pmClose. That results in 2kHz cutoff. You want 440uH and 140nF.Quote from: graymp89 on May 31, 2015, 03:15:13 pmThe inductance of that transformer is 151uH. In combination with the input resistance of the 2nd mixer (2x 28 Ohms = 56 ohms) this forms a high-pass filter at 56 kHz. Since your IF is 200kHz, this part would work out okay. Since your freqs are so low, something like these may be even better:digikey search for CT pulse xfmrs

author: Digibin
author_rank: Regular Contributor
timestamp: June 06, 2015, 10:56:33 am
content: Quote from: Odysseus on June 01, 2015, 06:14:48 pmThe inductance of the transformer is in parallel with the input resistance - I thought a parallel RL circuit had no (voltage) filtering effect?Also - since the output impedance is much larger (1.2k) than the input impedance (28R) I'll need some buffering or impedance matching between the two yes?

author: Odysseus
author_rank: Regular Contributor
timestamp: June 08, 2015, 03:36:57 pm
content: Quote from: graymp89 on June 06, 2015, 10:56:33 amIt does when your source impedance is essentially a current source.Quote from: graymp89 on June 06, 2015, 10:56:33 amYes, you can choose a turns ratio to maximize the use of available voltage swing on the output of the mixer. The high pass frequency will still be set by the secondary inductance and load resistance.