I was asked to look at an experiment. Very simple setup:
-9, 5mm LEDs connected in parallel
-bench power supply providing 5 vdc
-a 150 1/4 watt resistor in series with the positive voltage to the circuit
-all 9 LEDs lighted normally with the bench supply showing current being supplied is .02amps

Now,
A) why does the power supply show current being supplied is only 20ma and not 180ma?
B) why is the 5 volts, when divided between all 9 LEDs even enough to light the white LEDs at all (white LEDs require 3-3.4vdc)
C) the 150 ohm resistor reduces the voltage again furthering the voltage drop

Thanks

 

20 mA into a 150 ohm resistor = 3 V.
Hence the voltage across the resistor is 3 V.
The voltage across the LED is 5 V - 3 V = 2 V.
Each LED is drawing about 20 mA / 9 = 2.2 mA

 

I was asked to look at an experiment. Very simple setup:
-9, 5mm LEDs connected in parallel
-bench power supply providing 5 vdc
-a 150 1/4 watt resistor in series with the positive voltage to the circuit
-all 9 LEDs lighted normally with the bench supply showing current being supplied is .02amps

Now,
A) why does the power supply show current being supplied is only 20ma and not 180ma?
B) why is the 5 volts, when divided between all 9 LEDs even enough to light the white LEDs at all (white LEDs require 3-3.4vdc)
C) the 150 ohm resistor reduces the voltage again furthering the voltage drop

Thanks

Why were you asked to look at this?

Sounds like some kind of homework or lab assignment.

In any event, you will learn more if you make your best effort to answer the questions as best you can and walk through your reasoning. We can then see where you are right and where you are going off the rails. You will learn a lot more that way.

Here are some questions to help you get started.

Why would you expect 180 mA? How did you arrive at this value?

If 180 mA is flowing through a 150 Ω resistor, what is voltage going to be across that resistor?

If you have 20 mA (as shown by your power supply meter, which is probably in the ballpark, but probably not terribly accurate), what is the voltage drop across the 150 Ω resistor?

If 20 mA leaves the supply, flows through the resistor, and then splits evenly (not a great assumption, but we need to start somewhere) into the nine LEDs, how much current flows in each LED.

Are you familiar with Kirchhoff's Laws (Voltage Law and Current Law)? If not read up on them.

 

FIRST: all 9 LEDs IN PARALLEL, so the same voltage is across all the LEDS. Next, the current is only 20 Ma because that is the total current thru the 9 parallel LEDs, WHICH, since they are diodes, they are NOT LINEAR devices at all At 20 mA thru 150.00 ohms, V=I xR =0.020 A x 150.00 ohms=3 volts actross the resistor and 5-3= 2 volts across the 9 LEDs so at least one of the LEDs is biased far enough ito conduction for some current to flow.

 

FIRST: all 9 LEDs IN PARALLEL, so the same voltage is across all the LEDS. Next, the current is only 20 Ma because that is the total current thru the 9 parallel LEDs, WHICH, since they are diodes, they are NOT LINEAR devices at all At 20 mA thru 150.00 ohms, V=I xR =0.020 A x 150.00 ohms=3 volts actross the resistor and 5-3= 2 volts across the 9 LEDs so at least one of the LEDs is biased far enough ito conduction for some current to flow.

Thanks for the reply. I do understand your explanation but observing what is happening, aside from the calculations is what is happening but doesn’t seem as it should.
All the LEDs are on and full bright. The voltage for white LEDs is 3-3.4ish. But it under what they need as you have calculated. I’m not getting this ….

 

Are you sure of the value of the series resistor?

 

The truth is that LED brightness is difficult to evaluate! AND it is not linear with the electrical power input, either. I experimented with that long ago!! RED LEDs worked quite well driven by CMOS ICs that were rated at much less output current than the LED 20 mA spec described. AND the gate output was usually adequate to trigger other gates. and all of this with only the 5 volt supply , not the higher voltage that CMOS gates can handle. AND they could often be used without any current limiting resistors.