The growing popularity of 5V plug packs and Li-ion power banks with USB A connectors provides a convenient source of power for some projects, and a USB-A to 5.5/2.1mm DC cable is a possible connection option.
Scouring eBay turned up some sources, but one can never assess the quality of the things because usually there are no meaningful specification offered, and lets face it, they are Chinese.
Above are two sample 1m cables that I purchased, the left one for about A$1, and the right for about A$3.50 (posted).
Loop resistance of the cables was measured with Kelvin probes to assess their current carrying capacity from a voltage drop perspective.
DC loop resistance of the one on the right was 0.16Ω, so the maximum current for a 5% voltage drop is 5*0.05/0.16=1.6A… not quite a 2A rating.
DC loop resistance of the one on the left was 3.3Ω, so the maximum current for a 5% voltage drop is 5*0.05/3.3=0.075A… not even a 100mA rating.
This is not surprising, experience with USB-A to USB-micro cables has revealed similar variation, and an explanation why so many of these cables are hopeless in battery charging applications.
Browsing eBay for some high power LEDs for a current project created frustration in trying to wade through the stated performance figures (to they extent that they can be relied upon).
LEDs are often headlined as having some luminous intensity in candelas, but while that might seem to be a good measure of the ‘brightness’ of the LED viewed on-axis, it gives no information about the spatial distribution off-axis and the total luminous flux output or flux density.
I wrote a little online calculator that can be of assistance in finding the total luminous flux and flux density give luminous intensity and apex angle, Calculate luminous flux (lm) from luminous intensity (cd) and apex angle (°). (Note that specified luminous intensity is usually on axis and should be discounted by perhaps 20% to provide an average luminous intensity over the cone angle.)
An example, an eBay seller advertises:
Source Material: InGaN !
Emitting Colour: 0.5W 10MM HI POWER White 0.5W LED
LENS Type: Water clear
Luminous Intensity-MCD: Typ: 290,000 mcd
Reverse Voltage: 5.0 V
DC Forward Voltage: 3.2 ~ 3.4V
DC Forward Current: 100mA
Viewing Angle: 40 degree
Lead Soldering Temp: 260¡ãC for 5 seconds
Power Dissipation: 500mW
Does it appear rational? Lets calculate average luminous intensity at 80% of 290cd, 232cd. Lets assume the viewing angle is the half power beamwidth.
Above is a calculation from the specifications. Of concern is the calculated luminous efficiency of 266lm/W, it is perhaps three times or more the expected value, so it questions the accuracy of the claims. Even at 0.5W input, the luminous efficiency is unrealistically high. Continue reading Making sense of LED output figures
This article reports an experiment to evaluate the usefulness of precision GPS for the purpose of location data for automated antenna field strength surveys.
The experiment was conducted with the rover located in a fixed location 13km North of the reference station at Symonston and with very wide view of the sky, about 7:00 am 07/04/2016.
Only GPS satellites were used for the rover.
The software was RTKLIB v2.4.3b8.
The GPS was a UBLOX LEA-6T with a small patch antenna (as sold for small UAVs). The LEA-6T provides binary data as used by RTK for carrier phase measurements. Above is the GPS and a USB-RS232 adapter. Continue reading Precision GPS experiment #1
Calculate initial load line of valve RF amplifier was written as a companion to my RF power amplifier tube performance computer tool to provide a starting point for building a model, but as it turns out, the initial load line (and related values) is a very good estimate and further modelling may not be needed.
Although written for an application to valves, it is quite applicable to any active device, keeping in mind that it assumes a linear transfer characteristic.
The update provides for both single ended and push-pull configurations.
For example, the requirement is for a single ended Class C bipolar amplifier to deliver 25W from a 13.8V DC supply. What is the ratio for a broadband output transformer to 50Ω.
Above is the solution. The required Rl is 3.3Ω, and the required turns ratio is (50/3.3)^0.5=3.9. a 1:4 (turns) transformer would be selected for a prototype. Bear in mind that output power would fall to around 20W at 12V DC supply.
Another example is the common 100W 13.8V Class B push-pull design.
With a requirement for around 3Ω collector to collector (or drain to drain), a transformer with 1:4 turns ratio would be selected.