Hallo,

I have made a simple LED circuit with two leds in series and a resistor to limit the current. What would be the voltage at Anode of D2? 12V or 6-7V ? simulation shows 12V but actually I got 6-7V? what could be the reason?

Thanks,
Regards,
Surjit

 

It must be 12V,,,

there will be about 1.8 to 2.3V across each led, depending on their colour and current drain, both leds will have the same current.

 

The most probable reason for the low voltage is the voltage source has high internal resistance. High internal resistance voltage sources are often partially discharged batteries, some wallwarts, low current power supplies, power supplies with a series resistance, etc.

To verify; put the voltmeter leads directly across the supply, disconnect the load and watch the voltage jump up, reconnect the load and watch the voltage jump down.

 

Assuming a 2 volt drop across each diode - I said "Assuming", your circuit is going to be (8÷270) about 30 mA (0.0296 A). That means that 8 volts are being dropped across the resistor. Therefore in my circuit, measuring from ground I should see 8 volts at the top of the resistor, 10 volts from ground at the junction between the two LED's and of course, 12 volts across the entire circuit.

Depending on the LED's used in the SIM, you will get that particular voltage across each LED. If it's different from the one you're using on your test bench then the LED's you're using do not have the same rating.

I don't remember off hand what the TYPICAL forward voltages are but they're about 1.8 volts for red LED's, about 2 volts for yellow and green and can be as high as 3.3 volts for LED's that are blue, white and ultraviolet. But as I said - "I don't remember" what those voltages are. You have to know what the LED's you have are in order to get an accurate measurement. OR you can build a simple circuit and limit the current to a test LED, limit it to 10 mA (30 mA is pretty high) and measure the voltage across that particular LED. Even LED's of the same color will have slightly different forward voltages. That's why you're supposed to have individual resistors on each LED. Otherwise in a parallel circuit where you have all the LED's tied into one resistor, the one with the lowest forward voltage will take most of the current and likely fail. That leaves more current for the others, and the next one fails even faster. More current available means the rest will fail seemingly instantaneously.

Your circuit is a series circuit and is perfectly valid. But when calculating the resistance for your LED circuit you need to start with the supply voltage and subtract the forward voltage of the LED (or LED's) and then calculate for current based off of the new voltage. My example assumed a forward voltage of 2 volts for each LED, dropping a solid 12 volts down to 8 volts. At 8 volts, I would have shot for no more than 15 mA total current so I would have divided 8 by 0.015, which would require a resistor value of 533.3••• ohms. In my real circuit I'd use as close to that as possible but not below. Not unless I understood what a lower resistance would mean to the circuit. Maybe a 470Ω and a 100Ω resistor for a total of 570Ω and a current of 14 mA. Very close to my target current.

In short, what you're using on your test bench and what is in the sim are different. That's why you're finding a different voltage. As for why the sim is showing 12 volts at the anode of D2? Maybe you didn't specify any particular LED. Try changing the LED in the sim and see what happens.

 

Just did a sim of your circuit. Found 12 volts at the cathode of D2, 11.25 volts at the anode of D2 and 10.5 volts at the anode of D1. When I picked an LED (PT-121) I got 9.6 volts and 7.2 volts at those same points. Specifying what diode you're using in your sim will make a big difference. An LED is not just an LED. It's a specific device with specific characteristics. You need to tell your sim what you're using. If you don't know then find something with a similar forward voltage to the ones you're using on your bench. Remember, start with a regulated power supply and use resistance set for 10 mA (so you don't burn something out) and then measure the forward voltage across that LED. Don't put two LED's in series for your forward voltage testing, it could throw your readings off. Also measure the voltage drop across the resistor. Different LED's can change the current going through the circuit. In my first sim (not specifying any LED's) I got 38.9 mA. Using the specified LED's I got 26.75 mA without changing anything other than specifying the PT-121 type LED. That's where a spec sheet would be your best friend.

{oh, and I didn't set any internal resistance in my supply}

 

Found 12 volts at the cathode of D2, 11.25 volts at the anode of D2

The anode of D2 is connected directly to the 12V supply so the voltage reading there should be 12V.

 

The anode of D2 is connected directly to the 12V supply so the voltage reading there should be 12V.

I knew that. Just wanted to show all points except ground.

In the first sim the forward voltage of the generic LED was 0.75 volts. Hence, 12 - 0.75 = 11.25 volts. That minus another 0.75 volts was 10.5 volts.

 

It must be 12V,,,

there will be about 1.8 to 2.3V across each led, depending on their colour and current drain, both leds will have the same current.

About 3.4V for white LEDs - somewhere between 6 - 7V for a series pair would make sense.

 

Hallo,

I have made a simple LED circuit with two leds in series and a resistor to limit the current. What would be the voltage at Anode of D2? 12V or 6-7V ? simulation shows 12V but actually I got 6-7V? what could be the reason?

Thanks,
Regards,
Surjit

I agree with ramancini8 -- you are probably using a voltage source that is too weak to provide the 30 mA needed to drive the circuit while maintaining 12 V across its terminals. If the LEDs have about 2 V across them, then the current you are getting is about 10 mA, give or take. The supply is dropping about 6 V to supply this, so it's internal resistance, at this current draw, is about 600 Ω, give or take.

Are you by chance powering this with a coin cell battery?

 

D1 & D2 need to have a little more assigned. You also have a tran time of 12 seconds? The voltage at the anode of D2 will be the 12 volt supply voltage. D1 and D2 each drop about 0.75V because you never assigned them actual part numbers, the drawing looks very LtSpice. The circuit current is about 0.0389 Amp. D1 & D2 combined in series have a forward voltage drop of about 1.5 Volts but again they are unrealistic LEDs without a Vfwd or Ifwd called out. If you could provide the LED data sheet(s) this would go much better and make more sense.

I based my numbers on a LtSpice drawing without assigning any values to D1 & D2.

Ron

 

The TS is saying that the simulation shows 12 V at the anode of the top LED but that the ACTUAL voltage is between 6V and 7V. That implies that the TS is talking about an actual physical circuit (but the TS really needs to confirm this).