MintyBoost! - Small battery-powered USB charger
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However, there's one thing about 9V's that I've learned (from lots of bad experiences). One is that they don't have a lot of amp-hours: that is, how much current (amps) they can provide and for how long (hours). A duracell 9V provides -about- 500mAh over its lifetime. That's 500 mA (or .5A) for one hour or 100mA for 5 hours. That number is somewhat idealized but its a good starting point.
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雖然你也可以使用一個9伏的電池加上一個7805來做.
但是,我有一件對於9伏的電池的事,必須要講.
一是它(9V)並沒有很多的amp-hours.它是指一個電池可以在一個小時內,提供多少電流
(假設一個電池,它是500mAh .它就是指500mA for on hour or 100mA for 5 hours.)
這是一個好的問題開端
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Another problem is that they don't like to supply a lot of current, because they have high internal resistance (~2ohms), but basically that just means that if you want a lot of current (say to resuscitate a drained device) the 9V wont provide all 500mAh, but maybe more like 400. (Say you're drawing 250mA, then .25A*2ohm = 0.5V lost to internal resistance. For more info on 9V, read the duracell datasheet )
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另一個問題,就是它們.並不能真正地去提供這麼多的電流出來.
**因為它(9V)電池內部,有較高的內阻(~2ohoms) ,基本上它就是指,若你想要很多的電流時,
那個9V電池並不會真正地提供500mAh,大約400左右.
(且因為內阻的關係,假使你要250mAh時,你會損失.2.5 X 2 =0.5 ,在內阻上)
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Another problem with the 9V+7805 scheme is that a 7805 is a linear regulator. That means if you want 100mA at 5V (basically, USB power) then you're taking 100mA at 9V and then losing the 4V*100mA = 400mW (.4W) difference as heat.
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因一個主要的問題是, 當使用9V電池加上7805的組合時,因為7805是個線性的regulator.
而其"linear regulator"就是指說 : 若你想要100mA 5V的電源時,對於電池來說是提供了
100mA 及 9V給 7805. 而你會損失9-5 *100mA=.4W
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As the battery wears down to 7V the heat loss goes down to (7-5V)*100mA=.2W but you're still getting bad efficiency. At best the efficiency is 72% (5V/7V) and at worst its 55% (5V/9V) That means you're losing about a third of the battery power to heat!
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若其電池為7伏特電壓時,它仍然會有7-5 *100=0.2W的損失.
故得到最好效率為 5/7=72% ; 5/9 =55% .
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I'll also throw out that the 7805 itself has a quiescent current of about 5mA so you're always losing 5% (5mA/100mA) efficiency just for regulation! (& that's at least since if you're trickle charging the battery at 50mA then the 5mA quiescent is 10%)
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而其7805本身中又有大約5mA的 quiescent current .所以你在轉換電壓時,會有5%(5/100)的
損失.
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OK so basically the 7805+9V solution works but the efficiency is startlingly low, say 60% or so, and provides only 300mAh at 5V.
(故,若以7805+9V的解決方法,是可以work的.但其效率是很低的.大約只有60%)
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跟據經驗,我認為AA 2號電池是比較好的.因為不但便宜'功率高,低內阻.及容易取得.
以其功率比較時,
一個9V 伏電池 : 9 * 500mA=4.5W ;
而二個 AA電池可以提供 2*1.5*300mA=9W
有大楖二倍的功率產生.但有一個主要問題是 2個AA電池只有3V伏的電壓值.
但我們的輸出,是需要5V.伏特電壓.當你使用9伏的電池做為source你可以使用linear regulator
進行轉換.但有3伏特時,不可以用此方法.故我們需改以使用boost regulator
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設計目的:
為了設計對手機及其它的設備進行充電.
設計楖論:
參考資料:
Boost_converter
http://en.wikipedia.org/wiki/Boost_converter
原始網頁
http://www.instructables.com/id/MintyBoost!---Small-battery-powered-USB-charger/step4/The-Process-Enclosure-selection/
--------------------------------------
However, there's one thing about 9V's that I've learned (from lots of bad experiences). One is that they don't have a lot of amp-hours: that is, how much current (amps) they can provide and for how long (hours). A duracell 9V provides -about- 500mAh over its lifetime. That's 500 mA (or .5A) for one hour or 100mA for 5 hours. That number is somewhat idealized but its a good starting point.
--------------------------------
雖然你也可以使用一個9伏的電池加上一個7805來做.
但是,我有一件對於9伏的電池的事,必須要講.
一是它(9V)並沒有很多的amp-hours.它是指一個電池可以在一個小時內,提供多少電流
(假設一個電池,它是500mAh .它就是指500mA for on hour or 100mA for 5 hours.)
這是一個好的問題開端
--------------------------------
Another problem is that they don't like to supply a lot of current, because they have high internal resistance (~2ohms), but basically that just means that if you want a lot of current (say to resuscitate a drained device) the 9V wont provide all 500mAh, but maybe more like 400. (Say you're drawing 250mA, then .25A*2ohm = 0.5V lost to internal resistance. For more info on 9V, read the duracell datasheet )
--------------------------------
另一個問題,就是它們.並不能真正地去提供這麼多的電流出來.
**因為它(9V)電池內部,有較高的內阻(~2ohoms) ,基本上它就是指,若你想要很多的電流時,
那個9V電池並不會真正地提供500mAh,大約400左右.
(且因為內阻的關係,假使你要250mAh時,你會損失.2.5 X 2 =0.5 ,在內阻上)
--------------------------------
Another problem with the 9V+7805 scheme is that a 7805 is a linear regulator. That means if you want 100mA at 5V (basically, USB power) then you're taking 100mA at 9V and then losing the 4V*100mA = 400mW (.4W) difference as heat.
----------------------------------------
因一個主要的問題是, 當使用9V電池加上7805的組合時,因為7805是個線性的regulator.
而其"linear regulator"就是指說 : 若你想要100mA 5V的電源時,對於電池來說是提供了
100mA 及 9V給 7805. 而你會損失9-5 *100mA=.4W
----------------------------------------
As the battery wears down to 7V the heat loss goes down to (7-5V)*100mA=.2W but you're still getting bad efficiency. At best the efficiency is 72% (5V/7V) and at worst its 55% (5V/9V) That means you're losing about a third of the battery power to heat!
----------------------------------------
若其電池為7伏特電壓時,它仍然會有7-5 *100=0.2W的損失.
故得到最好效率為 5/7=72% ; 5/9 =55% .
----------------------------------------
I'll also throw out that the 7805 itself has a quiescent current of about 5mA so you're always losing 5% (5mA/100mA) efficiency just for regulation! (& that's at least since if you're trickle charging the battery at 50mA then the 5mA quiescent is 10%)
-------------------------------------
而其7805本身中又有大約5mA的 quiescent current .所以你在轉換電壓時,會有5%(5/100)的
損失.
-------------------------------------
OK so basically the 7805+9V solution works but the efficiency is startlingly low, say 60% or so, and provides only 300mAh at 5V.
(故,若以7805+9V的解決方法,是可以work的.但其效率是很低的.大約只有60%)
step 3The Process: Engineering a better solution
From experience, I know that AA's are great. They are cheap, have lots of power, very low internal resistance and are easily available everywhere. Whereas a 9V has 500mAh (for a total of 9*500 = 4.5Wh power) two AA's have 3000mAh each for a total of 2 * 1.5V * 3000mAh = 9Wh, about twice as much power. The only problem is that 2xAA's provide 3V and what we need is 5V. With a 9V battery we can use a linear regulator because 5V < 9V but, sadly, we cant use a linear regulator to turn 3V into 5V. Instead we will need to use a boost regulator (also known as a DC/DC switching/step-up regulator)----------------------------------
跟據經驗,我認為AA 2號電池是比較好的.因為不但便宜'功率高,低內阻.及容易取得.
以其功率比較時,
一個9V 伏電池 : 9 * 500mA=4.5W ;
而二個 AA電池可以提供 2*1.5*300mA=9W
有大楖二倍的功率產生.但有一個主要問題是 2個AA電池只有3V伏的電壓值.
但我們的輸出,是需要5V.伏特電壓.當你使用9伏的電池做為source你可以使用linear regulator
進行轉換.但有3伏特時,不可以用此方法.故我們需改以使用boost regulator
---------------------------------
設計目的:
為了設計對手機及其它的設備進行充電.
設計楖論:
參考資料:
Boost_converter
http://en.wikipedia.org/wiki/Boost_converter
原始網頁
http://www.instructables.com/id/MintyBoost!---Small-battery-powered-USB-charger/step4/The-Process-Enclosure-selection/
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