Can't figure out this voltage drop

Thread Starter

Brianaala

Joined Feb 15, 2019
62
Hi folks,
I'm relatively new to electronic circuits, so this might be stupid but here goes: I'm trying to build a circuit that will light 5 LEDs (in parallel) but even when I try this circuit with a single ("flickering" 3v LED) it has a massive voltage drop (input is 3.3v and output measured at the LED junction is ~2.5v). This drop reads the same if I hook up a second (identical) LED at the junction. But the drop is NOT there when no LEDs are connected. I presume this has something to do with the load, or maybe the transistor? Any help would be great! Here is the schematic: Photoresistor to LED in dark.jpg
 

dl324

Joined Mar 30, 2015
8,877
Thanks for including a schematic.

Which node (net) do you mean by junction?

What is the "dark" resistance of the LDR?
 

Thread Starter

Brianaala

Joined Feb 15, 2019
62
Hi Dl324, thank you for responding! Ok, sorry for my lack of information: by "junction" I meant that in reality it's just where the wires from the circuit and the wires to the LEDs are twisted together. I'm honestly not sure what the resistance of the LDR is in the dark, but it drops the voltage down to 0v when covered and when not under load the voltage of the circuit is at 3.24v under load (of any or all of the LEDs) is around 2.5.
 
You have a number of problems. LEDs in parallel never good. Parallel combinations of a LED+series resistor is OK.

LEDS have a nominal current requirement. Lets' say 20 mA. You used 330 ohms. If you could get 20 mA, that drop would be I*R or 0.02*330.
= 6.6 V; That's not going to work.

The color of the LED determines the maximum voltage drop. red is about 2.4 Volts.

A bipolar transistor has a BE voltage drop and an EC voltage drop. We assume 0.6V or 0.7V if we don;t know. That voltage drops by about 10 mV/deg C. The datasheet will have a value for Vce(sat) and that's the value when fully turned on. Your not operating at saturation. Vce(sat) is a minimum value.

I did not look at the 2222a transistor, but I think they have Hfe (current gain bins) based on a suffix. If you want all the (example) RED leds to be of the same brightness, then you need to match Vf (The voltage across the LED) or get them close.

Your supply voltage is about 2.4+0.7 = 3.1 V; Everything is way to tight to work. The resistor isn't included.

So, different color LEDS have different voltage drops and
 
Last edited:

Picbuster

Joined Dec 2, 2013
982
Hi folks,
I'm relatively new to electronic circuits, so this might be stupid but here goes: I'm trying to build a circuit that will light 5 LEDs (in parallel) but even when I try this circuit with a single ("flickering" 3v LED) it has a massive voltage drop (input is 3.3v and output measured at the LED junction is ~2.5v). This drop reads the same if I hook up a second (identical) LED at the junction. But the drop is NOT there when no LEDs are connected. I presume this has something to do with the load, or maybe the transistor? Any help would be great! Here is the schematic: View attachment 180440
if the Vce is to high a problem occurs vdd-Vce should be higher than led's v forwrd
One volt from from spec sheet;
VCE (sat)* Collector-Emitter Saturation VoltageIC=500mA, IB=50mA 1V
VBE (sat)* Base-Emitter Saturation VoltageIC=500mA, IB=50mA2

Picbuster
 

dl324

Joined Mar 30, 2015
8,877
by "junction" I meant that in reality it's just where the wires from the circuit and the wires to the LEDs are twisted together.
It would be helpful if you put the voltages you're measuring on the schematic and tell us what you're referencing the voltage to. Voltages are usually referenced to ground unless it's described as a voltage across something. Since you're new, you wouldn't be expected to know the conventions.

I'm honestly not sure what the resistance of the LDR is in the dark, but it drops the voltage down to 0v when covered
Circuits like the one you're using are typically used to turn on something when it's dark. A potential problem with your circuit is that base current for the transistor is limited by the 100K resistor. If the LDR dark resistance was infinite (which it isn't), the base current to the transistor would be a maximum of about 26 microamps; not enough to saturate the transistor.
when not under load the voltage of the circuit is at 3.24v under load (of any or all of the LEDs) is around 2.5.
Where are you measuring this voltage?
 

BobaMosfet

Joined Jul 1, 2009
782
Hi folks,
I'm relatively new to electronic circuits, so this might be stupid but here goes: I'm trying to build a circuit that will light 5 LEDs (in parallel) but even when I try this circuit with a single ("flickering" 3v LED) it has a massive voltage drop (input is 3.3v and output measured at the LED junction is ~2.5v). This drop reads the same if I hook up a second (identical) LED at the junction. But the drop is NOT there when no LEDs are connected. I presume this has something to do with the load, or maybe the transistor? Any help would be great! Here is the schematic: View attachment 180440
3.3V input isn't enough for this setup (even for one LED). Your BJT is taking approximately 0.7V in the BE junction so your CE is 3.3-0.7 = 2.6V which is why you're seeing the voltage drop as ~2.5V at the output.

You need to understand the difference between parallel and serial connections and how that affects voltage and current.

I suggest this book:

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3

Good luck!
 

Thread Starter

Brianaala

Joined Feb 15, 2019
62
Wow! You folks are awesome! OK I clearly have a lot to learn. Ok I will try to figure this out one bit at a time (sorry, I'm really trying my best here).
Starting with KISS; so first-off yes the LEDs that I'm using are 20mA 3v (I'm using 3 blue, that are not yet in the circuit but I have the specs for, the other two are the "flicker" LEDs from the $1 store and I do not have their specs). If I understand correctly; I should wire these in serial with a lower resistor to each or just one resistor?
As far as the "maximum voltage drop" does that mean how much the individual LED pulls out of the circuit or the maximum it can tolerate?
Beyond that I think I understand what you're saying about the transistors; that they cause an additional voltage drop (~0.7v) and I should account for that as well.
OK on to Picbuster I think that you are also showing that the Vce is being dropped too low? Honestly, that got really technical and I'm struggling a bit to comprehend.
Dl324, sorry you're right I didn't know the convention. IK I'm measuring all of these voltages from the (+) and (-) leads that would then go to the LEDs (in the schematic that would be on either side of the LED). It is indeed a circuit to turn these LEDs on when it gets dark (man, you guys are good!). So, I think you're saying that I should reduce the resistance from 100k to allow more mA to the transistor or to connect the 100k resistor after the base in the circuit?
Finally to BobaMosfet (possibly the greatest name ever!): I think you're saying that the initial voltage to the circuit should be increased just because the 2222 is pulling it down regardless of all the rest. Also, that I have to do some basic skills research! :)
Thank you all for taking the time to help me out! I will try to tackle these issues one by one and keep you posted!
 

MrChips

Joined Oct 2, 2009
19,270
Basically, you need to increase your supply voltage, perhaps 6-12V.
Then you need to adjust the resistances for the higher voltages.
 

Thread Starter

Brianaala

Joined Feb 15, 2019
62
OMG! MrChips you just made my life a billion times easier! Thankyouthankyouthankyou!!!
 

dl324

Joined Mar 30, 2015
8,877
the LEDs that I'm using are 20mA 3v (I'm using 3 blue, that are not yet in the circuit but I have the specs for, the other two are the "flicker" LEDs from the $1 store and I do not have their specs). If I understand correctly; I should wire these in serial with a lower resistor to each or just one resistor?
Blue LEDs have a forward voltage of 3V or more. The forward voltage is typically specified at a current 20mA; sometimes 10mA. They will emit light at a lower voltage, but they won't be bright.
Beyond that I think I understand what you're saying about the transistors; that they cause an additional voltage drop (~0.7v) and I should account for that as well.
The 0.7V drop is for the base-emitter junction of the transistor. But, when a transistor operates in saturation mode, the collector-base junction becomes forward biased, so the voltage from collector to emitter can be less than 0.7V; and usually is.
I'm measuring all of these voltages from the (+) and (-) leads that would then go to the LEDs (in the schematic that would be on either side of the LED). It is indeed a circuit to turn these LEDs on when it gets dark (man, you guys are good!).
We'd describe that as the voltage across the LED.
I think you're saying that I should reduce the resistance from 100k to allow more mA to the transistor or to connect the 100k resistor after the base in the circuit?
If the intent is to turn the LEDs on as brightly as possible, you want the transistor to be saturated. In saturation mode, we use a beta (transistor current gain) of 10 for 2N2222. If you want 20mA in an LED, you need to provide a base current of more like 2mA. You can choose less, but that wouldn't be a conservative design that would insure saturation mode operation.

With a supply voltage of 3.3V, you don't have much headroom. You can't place two LEDs of any color in series at that voltage.

If you wish to use multiple LEDs, you need to connect them in parallel. Conventional wisdom says to use a current limiting resistor for each LED, but no doubt, you've seen products or circuits that skip the resistors. That's okay, but it's not considered a conservative design practice.

With a 3.3V supply, depending on current capacity, you could put LEDs directly across the voltage and not damage them. People have been making things called "throwies" that consist of LEDs taped across coin cell batteries. They attach a magnet to the assembly and "throw" them at metal objects, hoping they'll stick.
 

Thread Starter

Brianaala

Joined Feb 15, 2019
62
Thanks again Dennis for clarifying these things! I suspected that parallel was the way to go for what I was trying to accomplish; so that makes a lot of sense; so I should raise the voltage, calculate the required resistance and put one resistor at each LED (or can I get away with one resistor at the (+) lead going to all the LEDs?). I got a little lost in the transistor base saturation; as I understand for the base to become saturated (thus opening the CE gate) it needs to have an amperage of 1/10th the required current, so 0.002 Amp? How do I do that? Can resistors do that?
 

djsfantasi

Joined Apr 11, 2010
5,585
WARNING: Read the disclaimer at the end of my post. I am fully aware that I’m not addressing the TS issue...

Something which seems obvious to me, which is worrisome, is in your circuit you can reverse the function with a simple change.

Swap the positions of the LDR and the 100kΩ resistor.

Oops, confused this with another thread. I’m leaving it for informational purposes only.
 

Thread Starter

Brianaala

Joined Feb 15, 2019
62
Hi folks, ok upon further reflection I have realized several things: first that I need a seperate resistor at each LED (which if nothing else makes the math much easier), second that if I raise the supply voltage to 12v then have a 330ohm resistor at each LED that should resolve my voltage issue (12-3 / 0.20= 300). But I'm not sure how this will affect my LDR and bce issues...will that blow them up?
 

BobTPH

Joined Jun 5, 2013
2,021
If you use 12V you can put 3 LEDs in series. This is preferable because they will use only 1/3 of the power that 3 in parallel would use. This makes 9V across the 3 LEDs and leaves 3V to be dropped by the resistor. So you calculate what resistor is needed to drop 3V at 20mA:

V = I * R
3 = 0.020 * R
R = 3 / 0.020 = 150

Bob
 

Thread Starter

Brianaala

Joined Feb 15, 2019
62
Oh, ok thanks Bob! So if they are in series I presume I would just use one 150 ohm resistor at the root?
 

djsfantasi

Joined Apr 11, 2010
5,585
Hi folks, ok upon further reflection I have realized several things: first that I need a seperate resistor at each LED (which if nothing else makes the math much easier), second that if I raise the supply voltage to 12v then have a 330ohm resistor at each LED that should resolve my voltage issue (12-3 / 0.20= 300). But I'm not sure how this will affect my LDR and bce issues...will that blow them up?
I’m not sure that I get your math...
(12-3)/0.2
9/0.2
45Ω​
45Ω ≠ 330Ω !

So where did you get the figure of 0.2A or 200mA? I think an earlier post suggested that a nominal current for the LED is 20mA or 0.02A

Repeating the calculations, we get the following:
(12-3)/0.02
9/0.02
450Ω​
Still not 330Ω!

And while sometimes you can ignore the BE voltage drop, it’s best to include it. Thus, the calcs become:
(12-(3+0.7))/0.02
(12-3.7)/0.02
8.3/0.02
415Ω​

The closest resistor would be 390Ω. But the current would be 21 mA. Might be better to use a 470Ω resistor, which would limit current to 18 mA. Note that if the LED operating max current is 20 mA, a 330Ω resistor would be providing 25 mA. Too much. I didn’t show my calculations for the values in this paragraph, but I encourage you to do so.
 

dl324

Joined Mar 30, 2015
8,877
I got a little lost in the transistor base saturation; as I understand for the base to become saturated (thus opening the CE gate) it needs to have an amperage of 1/10th the required current, so 0.002 Amp?
For a conservative design, you need Ib=0.1Ic. You can get away with less base current, but conservative designs assume a beta of 10 so saturation mode operation is guaranteed.
How do I do that? Can resistors do that?
The 100k resistor limits the maximum base current. But, the values you can use are limited by the resistance of the LDR.
 
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