8
\$\begingroup\$

I have a very simple circuit which I have measured the current to be only 3mA but should be in the range of 18mA.
I have a simple circuit using 2 AA batteries (NiMH) at 1.3V each in series for a total of 2.6V and a 220Ohm resistor along with my basic red LED.

schematic

simulate this circuit – Schematic created using CircuitLab

I = E/R

Using Ohm's law I calculate that I should get approx. 11mA of current .011 = 2.6 / 220

Double-Checked Resistor Value

I double-checked my resistor with meter and they are at 215 Ohms but that would mean I would get slightly more current anyways.

Double-Checked Voltages

I also measured total voltage in circuit and it does measure at 2.6V. I measured the voltage drop across the LED and it was at approx 1.775 which seems correct also.

Why might I be getting less current than expected (3mA versus 11mA)? Is there something I'm not calculating properly for?

\$\endgroup\$
15
\$\begingroup\$

The diode has a forward volt drop of 1.775 volt hence the current that this circuit drives might be only (2.6 - 1.775)/220 = 3.75 mA.

You don't have the full 2.6 volts across the resistor hence 2.6/220 is meaningless.

\$\endgroup\$
  • \$\begingroup\$ Someone else was explaining forward voltage to me also and that is why I was investigating this so I would understand it better. So I see that my calculation should've been (V - Vf) / R. Since I wasn't doing the additional calc I was calculating improperly. Thanks very much. One thing I don't understand is => do all components cause this Vf also or just some like LED? Seems like Resistors are different and Ohms law works but in some cases I need Kirchhoff's law? Can someone point me in correct direction? \$\endgroup\$ – raddevus Jan 12 '18 at 13:20
  • 1
    \$\begingroup\$ @raddevus PN junctions (present in diodes) have this forward voltage drop because it takes a certain amount of voltage for the silicon to "turn on" (allow current to flow). Transistors also have PN junctions between certain pins, so depending on what you're measuring and how things are connected, they will have this drop as well. Any other components that consist of diodes and transistors will show this same effect, though the value of Vf will differ. \$\endgroup\$ – DerStrom8 Jan 12 '18 at 13:53
  • 1
    \$\begingroup\$ @DerStrom8 Thanks so much for explaining that. It gives me more info to go on, along with this great answer which I will mark as the answer for this question. Great info from everybody, very helpful. \$\endgroup\$ – raddevus Jan 12 '18 at 13:58
  • 3
    \$\begingroup\$ @raddevus +1 to you for asking a good question and having a good attitude towards those who answered! \$\endgroup\$ – DerStrom8 Jan 12 '18 at 13:59
  • 1
    \$\begingroup\$ @amI Voltage being dropped across a resistor is not the same as an intrinsic voltage drop 'Vf'. 'Vf' stands for Voltage forward (i.e. forward voltage) that is conventionally only an attribute of a semiconductor device, and represents the voltage required to break down the PN junction. Voltage being dropped across a component that is dependent of the current flowing through it is a different concept \$\endgroup\$ – DerStrom8 Jan 12 '18 at 19:41
5
\$\begingroup\$

If you are dropping 1.775V across the LED, you only have 0.825V across the resistor and that equates to 3.75mA. I = (0.825/220) using Ohm's law, so what you are measuring is right.

\$\endgroup\$
  • \$\begingroup\$ Thanks for adding another layer of understanding to this. The way you're explaining it, it's almost as if I can think of the LED/Resistor as one component which drops Voltage to .825. So maybe calculating total voltage drop is a way forward also? This is interesting and giving me different ways to understand. \$\endgroup\$ – raddevus Jan 12 '18 at 13:26
  • 2
    \$\begingroup\$ @raddevus -- it's a series circuit; the voltages across the individual components in the circuit will always add up to the supply voltage. \$\endgroup\$ – Pete Becker Jan 12 '18 at 14:49
  • \$\begingroup\$ @PeteBecker That's kind of what was confusing me. I was thinking that 2.6V is total voltage and 220 Ohms resistance (I=E/R) would give me @ 11mA. I believe I was thinking kind of backwards and needed to think (V Total - V drop) / Resistance = (2.6 - 1.775)/ 220. \$\endgroup\$ – raddevus Jan 12 '18 at 15:10
  • \$\begingroup\$ @raddevus It's interesting to look at the datasheet and see how the forward voltage changes with current flowing through the device and how the intensity of the LED varies with that current. You can hopefully see how the resistor value needs to change for different current and how that changes the voltage drop across the LED and the intensity of the LED. \$\endgroup\$ – DiBosco Jan 12 '18 at 15:50
0
\$\begingroup\$

Another way to think of it is like a battery in reverse in place of the diode (or emitter base junction in a BJT), equal to Vf. Rule of thumb only, as the value of Vf will increase slightly as current increases! For this reason, an LED Vf value is usually spec'd at a given current.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.