# Voltage not dropping in circuit

This schematic shows the circuit that I'm trying to build. At the moment I'm not using V2, as I'm having a problem with V1. From my understanding the diode should drop the voltage by about 0.7V, and the resistor by 1.1V, reducing the Voltage from 2.6 to about 0.8V. The voltage is only dropping by 0.7 V with the LED's lit, and by 0.2V without the LED's. Can someone please explain why, and what I'm doing wrong?

The reason I'm using a 2nd circuit, is that I want V2 to supply extra voltage to switch on the LED's in V1. There are other LED's in the V2 circuit, and I want them all to switch on at the same time. The LED's need about 2.4V to be fully lit. Thanks in advance.

simulate this circuit – Schematic created using CircuitLab

• What voltage(s) are you measuring and where? What voltage(s) are you expecting? What's the forward voltage drop of the LEDs? Please clarify your situation, because the question is unclear at the moment. Welcome to EE.SE, by the way. Oct 21, 2014 at 3:43
• Sorry I accidentally posted before I wrote the question. Oct 21, 2014 at 3:50
• If I measure the voltage across the LED, it's 1.9V. If I measure from after the resistor to V1- it's 2.4V. I would be expecting approx 0.8V from after the resistor to V1-. Thanks Oct 21, 2014 at 3:57

The LEDs, by your own admission, need 2.4V to operate. But you start with a voltage source that is 2.6V and then add a diode and a resistor in series. D1 alone should drop about 0.7V, so you're already below the operating voltage of the LEDs before you even get to the resistor. Add the voltage drop from the resistor and the voltage available to the LEDs is below their normal operating conditions.

So your primary problem is that V1 (and V2) are too low. You need to pick a current for the LEDs (based on the datasheet and how bright you want them) and then choose a voltage that satisfies that current. If, say, you wanted 20mA through the LEDs and you're keeping the 110Ω resistor, your voltage would be: $$V1=V_{D1} + V_{R1} + V_{LED} = 0.7V+(110Ω*20mA)+2.4V = 5.3V$$

Alternatively, you could pick a voltage (maybe 5V) and calculate a resistor that gives 20mA to the LEDs: $$R_{1} = \frac{V1-V_{D1}-V_{LED}}{20mA}=\frac{5V-0.7V-2.4V}{20mA}=95Ω$$

But let's look at what's actually happening in the circuit that you posted. As we've already concluded, there's not enough voltage to light the LEDs in their normal operating region. The only reason any amount of current is flowing at all is because the forward drop of a diode is actually a function of current. The lower the current, the lower the voltage drop. Here's the graph out of a datasheet from a 1N4148:

Let's guess 1mA is going through the circuit. So I drew a line from 1mA on the Y-axis and then down to corresponding forward voltage on the X-axis. It landed just above 0.6V. We'll just use 0.6V and see how that works out.

Let's see how much of a voltage drop we'll get from V1 to the bottom of R1: $$2.6V - 0.6V - (110Ω * 1mA) = 1.89V$$ Oh hey, look at that! That's almost exactly what you measured! Our guess was nearly correct. In fact, it was a bit high. The exact current is slightly less than 1mA. But now we know that just under 1mA is going through the whole circuit. And with two LEDs in parallel, they're getting less than 500uA each! That's not very bright, if you can see it at all.

• Thank you, it's a very good explanation of what's going on. I only have basic electronics knowledge. The overall problem is this. I have a set of LED's that are being powered from a PCB which has 5V. There are 7 Blue LEDs, and 2 red LEDs. I can light all the Blue LEDs which measure about 2.7V. If I add 1 Red LED, the Blue ones dim, and if I add 2 Red LEDs, all the Blue ones go out and the red ones turn on. I'm not sure how this PCB works because it supplies 5V, but the voltage doesn't burn out the LED. Any ideas for how I can light the red LEDs, and switch them off at same time as the blue? Oct 21, 2014 at 5:08
• I'm assuming you're putting all of the LEDs in parallel? Red LEDs have a much lower voltage drop than blue LEDs, so when you put them in parallel, the red LEDs will "steal" all of the current from the blues. Moral of the story: you can't put LEDs in parallel like that. Oct 21, 2014 at 5:38
• Yes, all in parallel. Would it work if I put the blue in parallel, and then both red in parallel, (but in series to the blue)? Oct 21, 2014 at 7:37
• As long as your source voltage was larger than the combined voltage drops of both colors of LEDs, then yes, that would work. However, there's another issue with putting LEDs in parallel that has to do with their negative temperature coefficient. Too long to explain here. Check out my explanation on this answer (5th paragraph): electronics.stackexchange.com/a/132366/43931 Oct 21, 2014 at 17:58
• Ok, I will try this. I read about the temperature problem, and if it happens, might have to try something else. Oct 21, 2014 at 23:46

In any case, in answer to to your comment - questions, if you have a 5 volt supply, 7 blue LEDs, and 2 red LEDs, you should connect each of them to its own ballast resistor and then connect the strings in parallel, as shown below.

In order to determine the value of the ballast resistor, you subtract the LED's forward voltage from the supply voltage and then divide that difference by the current you want to let through the LED.

For example, if a blue LED has a typical Vf (forward voltage) of 3.5 volts with 20 milliamperes through the LED and you're powering it from a 5 volt supply, then you'd write:

$$\ Rs = \frac{Vs -Vf(led)}{If(led)} = \frac{5V - 3.5V}{0.02A}= 75\Omega.$$

That value is shown below for the 7 blue LEDs, and the resistor value for red LEDs with a Vf of 2 volts with 20mA through them - also shown below - was worked out similarly.

• It's going to be very difficult to do that because of this kit. It's not my setup, it's something I bought. I might try the other method, if it doesn't work, I'll try this. Oct 21, 2014 at 9:45
• What just happened? I looked earlier and you had accepted my answer for 15 rep, but now they've disappeared and my answer is unaccepted and zero rep. Did you do that? Oct 21, 2014 at 10:34
• yes sorry, i thought it would work, but then I realised I couldn't put resistors in without great difficulty, so really I still am not sure what to do.also I don't know how you got to those numbers. it's not clear to me Oct 21, 2014 at 11:12
• That's kind of cheap but, anyway, I've edited my answer so that you can figure out for yourself how to determine the value of the LED ballast resistor(s). BTW, what "other way" are you going to try? Oct 21, 2014 at 12:33
• I'm not really knowing much about the point system. Just focusing on the solution here. Thanks for clarifying..didn't realise it was 3.5V. I think the LEDs are only 2.7V. See above for other method. Oct 21, 2014 at 23:43