I'm trying to modifying and improve the MIT radar project

Here's the basic block diagram:

enter image description here

Input power is split by 3 dB SPLTR1 to transmitting antenna and LO input to the mixer. From what I understand, I don't need to input 13 dBm power to mixer, so I was thinking on putting, say, 10 dB splitter instead with relatively low insertion loss so I don't waste power.

I've searched for 10 dB RF power splitters on digi-key, mini-circuits, etc. but I can't find one. How should I search for them, or, are they impractical?


2 Answers 2


The splitter you currently show in your circuit appears to take in power X and distribute it to one port at X/2 and another port at X/2 i.e. there is no power loss. Just to confirm with real numbers, 16dBm enters and two lots of 13dBm leave. 13dBm is half the power of 16dBm hence two lots of 13dBm is 16dBm i.e. no power wasted.

What you appear to want doesn't make sense as being something useful. A "10dBm power splitter" would effectively (or so it seems) take an input of 16dBm and produce two outputs each being 6dBm. The total power out would be 9dBm (i.e. 3db higher than 6dBm) and there would be a power loss in the splitter of 7dBm. That loss is just heat and wasted.

Why would a manufacturer make one of these?

Stick with the 3dBm splitter and put an attenuator in line to the mixer. You should easily be able to get an attenuator that suits.

EDIT - worked example of a resistive splitter that maintains impedances at 50 ohm


simulate this circuit – Schematic created using CircuitLab

With R1 at 5 ohms, the power to the antenna is 83% (an attenuation of 0.8 dB). To match the input impedance to 50 ohm, R3 must be 500 ohm and this attenuates the signal to the mixer by about 21 dB.

You can mess with R1 and R3 to give different ratios but, to keep impedances correct: -

\$R_3 = \dfrac{1}{\frac{1}{50}-\frac{1}{50+R_1}} - R_4 \$

  • \$\begingroup\$ I want to deliver 80-90% of power to antenna and 10-20% to mixer. Right now it's 50/50. Would that be a better design? \$\endgroup\$ Feb 26, 2015 at 8:52
  • \$\begingroup\$ Just feed the antenna the full power and leach off a little to the mixer. If the amount you remove is small VSWR won't be hardly affected. \$\endgroup\$
    – Andy aka
    Feb 26, 2015 at 8:59
  • \$\begingroup\$ Yes, that's what trying to achieve by using a different power splitter but I couldn't find too many uneven filter. \$\endgroup\$ Feb 26, 2015 at 9:12

Wow, that MIT Radar project is cool!

Anyway, what you are looking for does exist, but it's called a coupler, because it has some directional response. (a 3 dB coupler is also directional though... go figure).
They are used to give an uneven split of power, for example tapping off some power from a master cable in an in-building distribution system.

Mini-circuits does make them, but the closest to 3 dB they make is a -10 / -0.6 dB coupler, Here which might be too low for your needs.

If you're happy to etch a PCB (and measure and re-etch a few times) you could also invent and build one yourself. 2.4 GHz is not hard to split, you could invent something with unequal impedance transformation that causes the uneven split you need. No need for any isolation in this application.

Digging a bit about mixers though, it looks like 13 dBm is a fairly good figure for the LO level. If you want to soup up this radar perhaps it would be easier to put a small amplifier on the output of the splitter, and get +23 dBm or so. You may run into issues with the law in your country.

  • \$\begingroup\$ Ok, a quick look at directional coupler vs power splitter reveals that latter splits both RF and DC but former outputs RF only to the mixer. I don't know if it will break the entire design. \$\endgroup\$ Feb 26, 2015 at 13:55
  • \$\begingroup\$ I don't think the DC matters to your radar. Most of the components will not pass DC anyway. It matters to the coupler users because some of them need the DC to power a amplifiers etc. \$\endgroup\$
    – tomnexus
    Feb 26, 2015 at 19:56

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