Why is that high voltage AC is more commonly found than High voltage DC? Example my battery powered fly swatter and fluorescent lamp both use high voltage AC. Why can't these devices increase the DC voltage from battery and directly use the High voltage DC?
Electronic – Why is high voltage AC more common than high voltage DC
acdchigh voltage
Related Solutions
While a wall adapter is usually isolated, it likely isn't rated to 1kV. A simple solution would be to find a DMM with logging capability and CAT III/IV 1kV isolated external power source or long battery life. Here are some candidates:
- Agilent U1271A: IR USB connectivity; 300 hour battery life
- Gossen Metrawatt Metrahit X-Tra: "bidirectional IR interface"; 200 hour battery life
- Fluke 289/287: "Isolated Optical Interface"; 200 hours battery life
Another option is to use a simple micro with an ADC, float the whole circuit at chassis voltage, then use proper isolation techniques (more than just optoisolation -- if you're not sure I suggest another question) to communicate with a PC over your preferred serial connection (UART -> opto -> RS232 -> USB -> opto -> PC, with cable shield voltage measurement and warning, would be my choice). Note that this means you can't touch anything on the floating (hot chassis) side of the widget. This way you can eliminate the power supply isolation worries by just using a battery, and still run it easily for 6 months to a year without replacing it, following thought on power consumption and sleep modes (ie: MSP430). Also note that a sputtering machine generates electrical noise, so you may need to use RS485 with error detection/correction algorithms.
Summary:
As you suggest, the Jumpstart_DC -> converter -> USB_5V is more likely to be the most efficient method, but this is not certain and depends on some extra points which need to be clarified. See below.
There are too many unknowns at present to be certain but
- I guesstimate DC-USB is likely to be in the 85% - 90% range BUT could be as low as 40% done poorly.
- I guesstimate AC-USB to be in the < 70% efficient range but could be 70% +.
To obtain a good comparison the following needs to be known.
- What is the JumpStart battery voltage?
- What is the JumpStart DC output voltage?
- Is DC out a direct battery connection or is there an internal SMPS (switching converter)?
- If a SMPS is used for DC out is the voltage selectable, and what are the voltage available?.
- Any available information on efficiencies, ratings etc.
- Knowing Jumpstart battery technology/chemistry would be useful.
Direct measurement may proide superior results to any technical guestimating
For a known AC load measure power in. If direct measurement not possible do run time tests with a known load.
For a known DC load do as above.
Lonnnng version:
Assumptions:
Update:
Mauvis says:
The Jump starter specified is this one:
The phone is an iPhone 4gs.
So - appears to be 12V lead acid battery with direct from battery 12V DC output. So comverter can be highly effiint if SMPS.
Converter shown may or many not be SMPS - should be.
Output is 500 mA.
(1) I assume that the use of an AC -> USB or DC-> USB converter means that either is powered from an output on the "jumpstart" device. ie either
Jumpstart_AC -> converter -> USB_5V
or
Jumpstart_DC -> converter -> USB_5V(2) I assume that the JumpStart uses a 12 volt lead acid battery.
Alternatives: Could use LiIon or NimH or ...(3) I assume that the Jumpstart DC output uses direct connection to its battery at a nominal 12 VDC. (Alternatives: could use an internal DC-DC converter and output voltage could be selectable).
Re " ...and depends on some extra points which need to be clarified ..."
You need to know or have an adequate guesstimate of:
The Jumpstart DC output voltage & how it is provided
How the proposed DC-DC converter converts to USB-5V and it's efficiency
Jumpstart battery to AC efficiency and converter AC-USB_5V efficiency.
DC -> USB_5V.
Depending on how everything is done efficincies DC-> USB could be in the 40% to 85% range. Near the higher end is likely but a better knowledge of the overall arrangements is needed for certainty.
If the Jumpstart DC output is a direct connection from its internal battery at say 12VDC and if the DC-USB converter uses a SMSP (SMPS = swith mode power supply = switching regulator - here probably a buck converter) then overall efficiency should be good.
i Jumpstart to DCout is effectively 100% efficient and the converter should be able to be 75% - 85% efficient. Higher efficiency is possible but not likely end to end.
ii If as in i above but the DC-USB converter uses a linear regulator then efficiency will typically be 5/12 =~ 40%
iii If as in i above but the Jumpstart battery and DC out is say 7.4V nominal LiIon then with a SMPS DC-USB is still probably about 75% - 85% but linear efficiency is now say 5/7 =~70% (varies with battery voltage)
iv If the Jumpstart uses a SMPS from its battery to its DC output (certainly not unknown) then that will add another 70-80% efficincy in the chain BUT if using a linear regulator the DC level may be set close to 5V so the linear regulator will be very efficient.
So in this case overall efficiency = say
SMPS: 0.8 x 0.8 =! 65% battery to USB. And
Linear: 0.8 x 0.8 =! 70% overall
AC -> USB_5V
This is a lot more certain in methods used.
AC out is an internal SMSP - probably around 70%-80% end to end.
AC-USB almost certainly rectifies the AC and uses a SMSP at 80%-90% overall.
An iron cored transformer using 60 Hz may be used and in this case may provide an excellent solution if well designed at say 90% efficiency.
SO:
AC-USB with smps converter ~=(0.7-0.8) x (0.8 - 0.9) =~ 55% - 75% range.
AC-USB with iron core transformer and careful design = (0.7 - 0.8) x 0.9 ~= 65% - 75% range.
So for comparison:
DC-USB
Could be as good as 85% - 90% with best case arrangements.
Could be a low as 40% worst case :-(
AC-USB
Using SMSP adaptor probably 55% - 75% range.
Using iron core supply = 65% - 75%.
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Best Answer
All it takes to make high voltage AC is low voltage AC and a transformer.
To make high voltage DC, you have to chop it into (what else) AC, run it through a transformer, and then rectify it back to DC. Quite a bit more hardware is necessary.
So, with mass produced products, there's a strong economic bias to use AC high voltage, so that's what you'll see, unless there's a compelling reason that the high voltage needs to be DC.