# Voltage Drop for Power Adaptor

Discussion in 'Electronic Basics' started by Homer, May 17, 2007.

1. ### HomerGuest

Hi All,

I am planning to light up 178 LEDs (3.4V-3.8V, 25MA) using a 18V/1A

Power Adapter says that is 18 volt but when I test it with volt meter
it shows 22V. When I connect a string of 6 LEDs (plus small resistor)
with it voltage is dropping to 21.4 (and shows 9.4ma current)

I don't have problem calculating my resistors (I am using
http://led.linear1.org/led.wiz) but it seems my 18V (or 22V) Adapter
is not really what is says (it drops).

My question is what should I put in that web site as my Source
Voltage? (22V? 18V? or something else).

Homer

2. ### ChuckGuest

The voltage rating on unregulated
adapters is generally the voltage at
full load (1 A in this case).

If you're calling out the parts on the
web site you could indicate both no-load

Chuck

3. ### John PopelishGuest

A fair approximation for your supply (based on both the
rated voltage and your experimental results) would be a 22
volt source that includes a 4 ohm series resistor. When you
connect an 18 ohm resistor across it, the 4 ohm internal
resistance will drop 4 volts and the 18 ohm resistor
resistor will drop 18 volts.

Include that internal 4 ohm resistance in your design
calculations.

4. ### HomerGuest

Thanks but what do you mean by unregulated adapters?
And also if the rating is the voltage at full load then how do we
explain the Voltage Drop with 6 LEDs (that is far less than 1A for
sure)?

5. ### John FieldsGuest

---
voltage with a change of output current.
---
---
As John Popelish explained earlier, your adapter can be modeled as a
stiff 22 volt source with an internal resistor in series with its
output, like this: (View in Courier)

22V>---[R]----><----+
|
[VOLTMETER]
|
GND>----------><----+

With only a typical high input impedance (10 megohms) voltmeter on
its output the output voltage will read 22V because the voltage
dropped across the series resistor will be very small.

Now, with what you rigged up:

22V
|
[R]
|
+->21.4V
|
[R]
|
[LED]
|
[LED]
|
[LED]
|
[LED]
|
[LED]
|
[LED]
|
GND

since your whole LED string plus the resistor is dropping 21.4V and
the current in it is 9.4mA, the string and the resistor look like:

E 21.4V
R = --- = --------- = 2276 ohms
I 0.0094A

and the circuit reduces to:

22V E1
|
[R1]
|
+-->21.4V E2
|
[2276R] R2
|
GND

Notice that there was 600 millivolts dropped across the internal
resistor because of the 9.4mA flowing through it.

That means that with that current flowing through it its resistance
must be:

E1 - E2 0.6V
R = -------- = --------- = 64.52 ohms
I 0.0093R

and the circuit will look like:

22V E1
|
[64.52R] R1
|
+-->21.4V E2
|
[2276R] R2
|
GND

That means that as more and more current is pulled through it by the
load, more and more voltage will be dropped across it until, with 1
amp flowing through it it will drop 4 volts and the voltage across
the load will be the specified 18V:

22V E1
|
[R1]
|
+-->18V E2
|
[18R] R2
|
GND

For that to happen, with 1 amp in it, its resistance must be:

E1 - E2 4V
R = -------- = --------- = 4 ohms
I 1A

and the circuit will look like:

22V E1
|
[4R] R1
|
+-->21.4V E2
|
[18R] R2
|
GND

Which is quite different from the 64.52 ohms with 9.3mA in it!

The reason, of course is that the impedance of the rectifier diodes
in the adapter decreases abruptly as the voltage increases past the
Vf knee.

6. ### ChuckGuest

Some "wall warts" contain internal
voltage regulators that keep the output
voltage very close to the stated voltage
on the adapter. If yours were regulated,
it would deliver 18V, regardless of
load, up to its rated current output of
1 ampere.

The unregulated variety have a
non-negligible internal resistance that
causes the output voltage to drop as the
load increases. In a rare gesture of
honesty, the manufacturer actually
promises the rated voltage at full load.
Well, the greater the external load, the
greater the voltage that will be
developed across the internal
resistance. This voltage will be
subtracted from the output voltage.

The voltage drop follows Ohm's law,
using the internal resistance and the

John has calculated the internal
resistance to be four ohms using (22V -
18V)/1A. That assumes a linear internal
resistance but the 0.6V drop with a
current of 0.0094A suggests an internal
resistance of ~63 ohms which would never
allow 18 volts output at 1 A. So maybe
it is non-linear (at least over part of
the output current range) or defective.

It can get complicated to work with an
unregulated supply. For less than a
dollar, you can add a simple IC voltage
regulator and some capacitors and have
your output voltage stay within 50mV of
18V at any load up to 1A.

Chuck