My UPS powers one 120V AC device and a whole bunch of small 'bricks' that end up rectifying things down to 5-12V DC. Instead of using the 'bricks', my idea is to tap the battery directly, and use DC-DC converters to get down to desired voltage.
Would this result in better efficiency? Would it even work? Any potential pitfalls?
Electronic – Tapping the battery from an UPS to feed low voltage devices
batteriespower supplyups
Related Solutions
Your solution started out as bearable (5V at 100mA) but ended up completely unacceptable at 500 mA. You say that your "wall wart" is rated at 300 mA. When you supply a voltage using a linear regulator the current in is the same as the current out - the regulator drops the difference in voltage. So here if you draw 500 mA at 5V you must supply 500 mA at 12V or 24V. The transformer will be overloaded in either case.
If the ratings are as you say then a potentially acceptable solution is to use a switching regulator (SR) operating from 24V in. \$5V \times 500 mA = 2.5 W\$.
\$24V \times 5 W =~ 210 mA\$. If the SR is 80% efficient (easily achieved) that rises to 260 mA. As that is liable to be an occasional requirement the total current at 24V will probably be acceptable with a 300 mA supply - depending on how many solenoids you wish to maintain on.
If you switch only one solenoid on at once the current drain with N activated is \$20 \times N + 20 mA\$. The surge current is essentially immaterial.
If you wanted more than 3 or 4 solenoids then current drain at 5V may need to be limited.
e.g.
- 10 solenoids at 20 mA = \$200 mA\$
- Balance = \$300mA-200mA = 100 mA\$
- Available current at 5V at 80 % efficient = \$ 100 mA \times \frac{24}{5} \times 0.8 = 384 mA\$, say \$400 mA\$.
Note that when a switching regulator is used, using a higher input voltage will result in less input current drain. Hence it is better here to use the full 24V supply.
Note also that if the transformer is a genuine 24 VAC then the rectified DC will be about \$24 VAC \times 1.414 - 1.5V - \$ "a bit" \$~= 30 VDC \$
Because:
\$VDC_{peak} = VAC_{RMS} \times \sqrt{2} ~= VAC \times 1.414 ~= 34 V\$.
A full bridge rectifier will drop about 1.5V.
34 VDC is peak voltage and available DC will be slightly lower - depends on load. There will be "a bit" of ripple and wiring loss and transformer droop and ...
At 80% efficiency this gives a 24VAC to 5V DC current boost of \$ \frac{30}{5} \times 0.8 = 4.8:1 \$
e.g.
- for 48 mA at 5V you need 10 mA at 30V.
- for 480 mA at 5V you need 100 mA at 30V.
So you about get 10 solenoids plus almost 500 mA at 5V DC :-)
One solution of many:
There are many SR IC's and designs. Here a simple buck regulator will suffice. You can buy commercial units or "roll your own". There are many modern ICs but if cost is at a premium you could look at ye olde MC34063. About the cheapest switching regulator IC available and able to handle essentially any topology. It would handle this task with no external semiconductors and a minimum of other components.
MC34063. $US0.62 from Digikey in 1's. I pay about 10 cents each in 10,000 qauntity in China (about half Digikey's price).
Figure 8 in the datasheet referenced below happens to be a "perfect match" to your requirement. Here 25 VDC in, 5V at 500 mA out. 83% efficient. 3 x R, 3 x C, diode, inductor. It would work without alteration at 30 VDC in.
Datasheet - http://focus.ti.com/lit/ds/symlink/mc33063a.pdf
Prices - http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=296-17766-5-ND
- Added:
Figure 8 in the LM34063 datasheet shows ALL component values except for the inductor design (inductance only is given). We can spec the inductor for you from Digikey (see below) or wherever and/or help you design it. Basically it's a 200 uH inducor designed for general power switching use with a saturation current of say 750 mA or more. Things like resonant frequency, resistance etc matter BUT are liable to be fine in any part that meets the basic spec. OR you can wind your own for very little on eg a Micrometals core. Design software on their site.
From Digikey $US0.62/1. In stock. Bourns (ie good).
Price: http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=SDR1005-221KLCT-ND
Datasheet: http://www.bourns.com/data/global/pdfs/SDR1005.pdf
Slightly better spec
$US0.75/1.
Surface mount.
Bourns.
Let's say your "inverter" works just like a battery backed up UPS, and that it has a pretty wide margin on its storage voltage, going from 14V down to 7V before it taps out.
That means your main capacitor in case of the low voltage option can only drop 7V.
The voltage drop across a capacitor is an integral over time of the current signal taken from it, which becomes a very nice fiddly bit of magic concerning systems of differential equations once you add in a 10W constant load and a variable converter efficiency, so I'm going to roughly ball-park it, because I'm lazy and it's evening and I have a million things to do.
(( Ref: Wikipedia page jumped to the spot where the voltage current relation of a capacitor is given ))
How will I do this?
- Coursely take 80% efficiency for a converter going from 12VDC to 120VAC (which may be seriously overestimating it in a DIY scenario, to be honest).
- Estimate the current draw to be constant, calculated at a capacitor voltage of 9.5V, rather than the exact average, which voltage I drew from my large hat of "that'll probably do". If you want to do other use cases, you can take the average, since a constant current assumption will be off any ways.
- Simplify the integral for constant current, which then becomes a simple linear equation: V = (I*t) / C.
So, the current from the capacitor will be:
I = (10W / 0.8 [=efficiency]) / 9.5V =~ 1.32A
Which then can be put into the simplified linear equation for the assumption of constant current (be aware, this is a very broad and lazy assumption):
V =~ (1.32A * t) / C
Let's say you want only ten seconds of power, with the known voltage drop of 7V across the capacitor, that becomes:
7V ~= (1.32A * 10s) / C
Which becomes:
C =~ 13.2As / 7V =~ 1.88F
Let's quickly do that for 120VDC as well:
Same assumptions, but the voltage range will be 80V to 120V, probably, so a drop of 40V is allowable, estimating the constance of current at the 90V point:
I = 10W / 0.8 / 90V =~ 139mA
with t=10s:
40V =~ 1.39As / C
C =~ 1.39As / 40V =~ 35mF --> Charged up to 120V = very, very lethal.
So, you see, I've already used a lot of assumptions about all the stuff you're not giving us about your project, and how you will personally be able to complete the electronics and it's still a lot of calculation, even though I made a very bad and broad assumption of constant current
The final choice will depend on fixing all the parameters and some will intertwine. There's no solution to that and that's what makes electronics design a difficult field.
This is just your very first, very broad ball park. But to be honest, re: "very, very lethal", if you are asking this question I don't really think you should be considering anything above 30VDC to store energy the likes of this.
Best Answer
Most small UPSes are so-called "off-line" systems. When grid power is live, they feed input AC directly through to the output. In parallel, they trickle-charge the batteries. When grid power fails, they quickly switch to inverter mode, where the batteries are discharged into an DC->AC inverter to power the downstream loads.
For your application, an off-line UPS wouldn't work for a number of reasons, with the primary risk being that the trickle charger can't keep up with a full load (now DC).
In contrast, an "on-line" UPS feeds all power through the battery bus, such that it's all converted to DC, then re-inverted to AC. On power-loss, the AC->DC front end simply powers down and the DC-AC inverter continues to operate off batteries. This might work for your application, though it's impossible to determine if the ripple on the DC bus would play nicely with your loads.
As to efficiency, the efficiency of the "on-line" UPS is less than that of an "off-line" UPS, as there is an extra AC->DC->AC conversion step at all times. Let's look at the normal power flow through each type of system:
Off-line, grid live: AC->AC(pass-thru)->DC(brick)
On-line, grid live: AC->DC(battery)->AC->DC(brick)
Direct battery tap - on-line, grid live: AC->DC(battery)->AC(DC/DC converter)->DC(DC/DC converter)->DC(load power)
To sum up, I'd expect the grid-live efficiency of an off-line UPS running DC/DC converters to run power bricks to be less efficient than an on-line UPS powering loads directly.