I'd like to do an experience about diode to enhance the circuit's voltage. I use my laptop usb port as a ideal voltage source so i have 5v input. I set up such 10 diode in series and in series with a resistor of 10kohms and 1 white led to show there is a voltage conducting inside the circuit. However, when i measure each diode voltage, i almost get only ~180mV for each diode, and from the beginning of diode to the end of the series diode, i have total of 5V, and there is no voltage conducting thru resistor. Can any experienced person can explain that? Moreover, such what i do research online, there's a source that said when diode in parallel, they will double the current of the circuit, but when i do real experience, I still saw same amount of current when i use only one diode. Thanks

 

Yes because diodes drop 700mV, X 10 diodes that's 7V,!!!

Your Usb is only 5V. QED...

 

Expanding on Dogydave's explanation, you do not have enough voltage to make all of the diodes conduct much current. The very low current means that the diodes have a high impedance. My guess is that when you take measurements with your voltmeter your voltmeter is loading down the circuit.

If you are trying to create a stable reference voltage, you might consider these parts

TL431 (2.5V, ±1%, 1 ma, cheap)
http://www.es.co.th/detail.asp?Prod=008604535

LM385Z-1.2 (1.2V, ± 1%, 20 ua, not as cheap)
http://www.es.co.th/detail.asp?Prod=007801903
http://www.es.co.th/detail.asp?Prod=007801903

 

Just for you to grasp what is going on here, do this:

Assemble just one diode and the resistor and measure the voltage you are looking for. Then, put a second diode in series and measure again. Repeat adding successive diodes (always in series) and measuring that voltage until you see that the effect you described is happening.

Just to see the outcome, could you post with how many diodes things worked as you expected? Please also say what diodes are you using.

Enhancing voltage, you say. Are you trying to get a higher voltage using 5V at the input? Yours is not the way.

Please comment.

 

Here's how the basic diode work: Take for instance, the 1N4007 diode. It does one job - to block current from going in the reverse direction. When current flows in the forward direction it conducts nearly all the voltage across it - provided you don't exceed it's rated capabilities. If you use it in a 5 volt circuit, the typical useful voltage across that circuit - through the 1N4007 diode you'll see about 4.3 volts. The diode has a forward voltage rating; the amount of voltage it takes to make electrons cross the diode. You'd need the spec sheet to know exactly what that forward voltage is but in general it's about 0.7 volts. In other words, on a 5 volt circuit through a diode you'll get 4.3 volts.

Same is true of LED's. However, their forward voltages vary quite a bit from different type to different type. The typical red LED may have a forward voltage of 2.2 volts, meaning when you calculate the current for that circuit you have to subtract the 2.2 volts before you do the current calculation. Suppose you have a red LED and you want to run it at 15 mA (0.015 amps) on a 5 volt circuit: You would subtract 2.2 volts leaving 2.8 volts to pass through the resistor. At 15 mA on a 2.8 volt circuit (which is what you have now that the LED has sucked up that much voltage) you would need to calculate: 2.8 ÷ 0.015 and you'd need a 187 ohm resistor.

If you put two red LED's in series then you're dropping (according to my example) 4.4 volts, leaving just 0.6 volts for a resistor to use up. At 15 mA you'd need (0.6 ÷ 0.015) a 40 Ω resistor. The funny thing is (and I don't fully understand this part) is that if you were to use an LED with a forward voltage of 2.2 volts in series with another LED who's forward voltage is 3.0 volts, that's 5.2 volts. In theory you wouldn't need a resistor at all. However, you're now allowing the power supply to send as much current as it is capable of through that circuit, running the risk of overloading something. DON'T ASK ME WHY THAT IS - I DON'T KNOW. Someone else may easily explain it, and I'll be watching for just that explanation.

What you're doing is trying to light several series LED's with only 5 volts. IF they light it's because there's just enough current to overcome the forward voltage drops. They may light but they won't be near their proper brightness. I say "Proper" because they're designed to operate at a particular luminescence at a given forward voltage and a given current. You can make an LED brighter by exceeding the recommended current through the LED but then you shorten its life by burning it out prematurely. Too much current and it may last but a brief flash of light. You CAN run an LED at a lower current which will extend its life. And you can run it at a much lower current - as long as there's enough voltage to overcome the forward voltage drop for the purpose of a dimly lit indicator light.

So to recap: An LED has a forward voltage rating. It varies with color and manufacture. You need sufficient voltage to overcome the Vf (forward voltage) and a resistor to control the amount of current passing through the circuit. You have to subtract the Vf from the supply voltage and then calculate for the current rating you want.

Typical current ratings for LEDS can be 10 to 30 mA (for a single LED), depends on the manufacturer and the ratings for that particular LED. Typically red LED's are the lowest Vf and the lowest current rated devices and blue and white LED's are typically the highest for any single LED circuit.

When you get into LED Lights, where they have several LED's, they can be rated in much higher currents - but now you're into an animal I have no experience with and VERY little understanding.

Hope this helps.

 

Welcome to AAC!

I use my laptop usb port as a ideal voltage source so i have 5v input.

A USB port isn't an ideal voltage source. An ideal voltage source would supply infinite current with no voltage drop. A USB port will provide 100-900mA at a voltage of 4.5-5.5V.

I set up such 10 diode in series and in series with a resistor of 10kohms and 1 white led to show there is a voltage conducting inside the circuit.

You mean current, not voltage.

However, when i measure each diode voltage, i almost get only ~180mV for each diode, and from the beginning of diode to the end of the series diode, i have total of 5V, and there is no voltage conducting thru resistor.

With such a low forward bias, the diodes and LED would be conducting very little current.

For 1N4148:

The current would be less than 1uA. The voltage across the resistor would be
\( \small V = IR = 1uA*R volts \)

When discussing circuits, you should include a schematic. In your case, the circuit is simple, but we don't know what type of diodes (other than white LED) you're using or the value of the resistor.

 

The funny thing is (and I don't fully understand this part) is that if you were to use an LED with a forward voltage of 2.2 volts in series with another LED who's forward voltage is 3.0 volts, that's 5.2 volts. In theory you wouldn't need a resistor at all. However, you're now allowing the power supply to send as much current as it is capable of through that circuit, running the risk of overloading something.

You'd get the same thing that was happening with the OP's original circuit. Each device will conduct the current dictated by it's forward voltage.

If we use a pure-green and white LED with no series resistor. Using the curves below and assuming a current of 10mA (it's requires an iterative approach), the pure-green LED would have a Vf of about 2.0V and the white LED would be about 3.5V. Since you only have 5V, the LEDs would divide the voltage according to their internal resistance and current would be limited to less than 10mA; for typical diodes.

White:

 

Thanks for all of the answer above. I will improve next time which will be with the schematic diagram and more specific on which model of component i am using for the experiment. This time, i am using a zener diode to take place for the regular diode due to i have none of regular diode and its model is 1N5226. Thanks all again.

 

oreover, such what i do research online, there's a source that said when diode in parallel, they will double the current of the circuit, but when i do real experience, I still saw same amount of current when i use only one diode. Thanks

Welcome to the forum.
You do not double the current with 2 diodes in parallel, you double the current carrying capacity.
It is like making the road 2 lanes, you can have more cars driving on it at the same time. But if there is only the same number of cars on the road anyway, you do not magically make more cars.
This explanation is a bit flawed, but it was the best I could think of in a hurry.
The current is determined by the voltage (pressure) and the load + power supply source resistance. And you get a "built in" voltage drop across a diode of about 0.7V so having another one beside it will still give you a 0.7V drop. With a diode there, in fact your power supply effective voltage will now be (5 - 0.7) = 4.3V, and adding another in series makes (5 - (0.7x2)) = 3.6V. Sometimes it is a good little trick to add a diode to lower the supply volts a bit. Just make sure the current rating of the diode is up to the job.
Using a zener diode will work too. In the forward direction they are just like a normal diode but in the reverse it will be the zenner volts rating that they drop. Just be careful to observe the power dissipation rating of the zener diode as it can become significant as the current climbs.