WARNING - dumb question follows - WARNING
consider yourself warned.
So I have a nice transformer that puts out 25V and 1.5A. I want to step it down to 9V with some voltage regulators. The regulators say they are rated for 1A and 37V max (given proper heat sinking).
If the output of the regulator is drawing < 100mA - do I have to worry about the remaining output from the transformer?
about
No.
Colin =^.^=
Not a dumb question -- just a newbie one.
First, I hope your transformer has a center tap on the secondary, because if it doesn't and you're using a 9V linear regulator and are drawing 100mA, you're talking about
(37V - 9V) * 0.1A = 2.8 Watts
Need a heat sink here. It might be somewhat easier to do it this way if your
25VAC secondary transformer has a center tap (view in fixed font or M$ Notepad):25.2 VCT Sec ____ | | + o-----. ,------>|--o---o---|7809|--o-----o )|( | C1|+ |____| +|C2 )|( | --- | --- ) ,---. | --- | --- )|( | | | | | )|( | | | | | o-----' '------>|--o---o-----o-----o-----o | | - | === === GND GND created by Andy´s ASCII-Circuit v1.24.140803 Beta
Use a 1000uF electrolytic for C1 and a 10uF electrolytic for C2.
This will give you about 18VDC peak on C1, which should make your 7809 a lot more comfortable. With total power dissipation less than a watt, you should be able to get away without a heat sink as long as the 7809 is in room temperature ambient.
Oh, yes -- transformers are rated for maximum current with resistive load. By definition, a transformer is usually rated so that the output voltage will be within 20% of the rated voltage for the rated current. Usually, if you're loading it lightly, the voltage will be higher than rated, and if you're loading it toward its limit, the voltage will be lower. The transformer is an on-demand machine -- it will produce the voltage as long as the current requirement is within limits., There isn't any extra to take care of.
Good luck Chris
No. Your transformer can deliver up to 1.5A but will do so only if its load demands it.
Similarly, your domestic a.c. supply can deliver up to "x" kilowatts but will only do so if you switch on enough appliances etc to draw that much power.
Or your domestic water supply, which can deliver "x" gallons per minute but will only do so if you open the tap/s sufficiently.
So, your "remaining current" doesn't go anywhere - it isn't flowing. If you are drawing 100mA from your 1A transformer it means that the transformer still has a further 900mA capacity which you aren't using
No. Current is drawn, not pushed. It's kind of like having a well full of water. You lower a bucket down there and draw what you need. The extra water in the well is not going to force itself into the bucket until it bursts.
Think of it as having 1.5A available if you need it.
about
------------ Nope, nope, nope, the transformer generates a voltage, which then ONLY pushes the current it is ALLOWED to push by the load resistance.
-Steve
-- -Steve Walz rstevew@armory.com ftp://ftp.armory.com/pub/user/rstevew Electronics Site!! 1000's of Files and Dirs!! With Schematics Galore!! http://www.armory.com/~rstevew or http://www.armory.com/~rstevew/Public
A 25 V AC transformer after rectification will give 25 V x sqrt(2) = 36 V DC, which is marginal for your regulator.
The output of your regulator is 9 V, the device is drawing 100 mA at this voltage. That is, your regulator converts (36 - 9)V x 100 mA = 2.7 W of electrical power into heat. This will require a heat sink to avoid frying the regulator.
For both reasons I would either use a transformer with lower output voltage (say, 12 V), or a switching regulator.
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