# where does all the extra current go?

Discussion in 'Electronic Basics' started by anonymous, Oct 12, 2004.

1. ### anonymousGuest

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?

No.

Colin =^.^=

3. ### CFoley1064Guest

Subject: where does all the extra current go?
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

(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\$

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 www.tech-chat.de

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

4. ### dBGuest

No.
Your transformer can deliver up to 1.5A but will do so only if its

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

6. ### tempus fugitGuest

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.

7. ### Dr Engelbert BuxbaumGuest

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.