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I know that is only one output per a time is active but I just tell you that is just a capture
Voltage on the LED(Y/EW) +4.99
four way traffic light - help
- Thread starter Eng Ranya Salem
- Start date
Voltage on the LED(Y/EW) +4.99
Please Tell me only one thing
The circuit that I will try to draw it is wrong ?? is there something not found on it ?
I want to finish it and see the result to do the next step of my work !!
it takes much time of me !!
Does that make sense? You said the voltage on the diode was 4.77V. How can there be a negative voltage drop across the diode?
There are a number of things "wrong" with your circuit.
The pace for progress is largely determined by you. The fact that you need to be asked questions several times before answering them means that this will be a slow process.
No one will simply give you the answer. Even if this wasn't homework, most of us won't do the work for you. We already know how to do it. The point is for you to learn how to do it.
but it is not my main Specialty to know every thing in this circuit
simple step I could do , going to any Electrical engineer and request the circuit to see the result without trying to do anything
but I wanted to know how it works by my self , how can I draw it ..
but in general you are right , but please don't ask questions I don't know it
ok let me learn , explain it " If you want "
We don't know what you know or don't know. That's why we ask questions.
You still haven't answered my last question.
I don't know
For a diode or LED to conduct, they require a minimum voltage across the anode and cathode terminals; with the anode voltage being higher than the cathode.
In the diode/LED example we're talking about, you said the voltage at the diode was 4.77V. I assume this was the voltage on the anode because you later said the voltage on the anode of the LED connected to the diode's cathode was at 4.99V. That gives a voltage drop across the diode of -0.22V, which is not possible.
how to measure it ??
I draw a voltmeter between Led AK
However you want to do it. If you ground the negative terminal on the voltmeter, all voltages are relative to ground.
The other option is to measure the voltage drop across each component.
The Volt of the A of Diode D1 is+5
the Volt of the A of Led Y/EW is+5
the Volt of the A of Led Y/EW is+5
There are no Y/EW LEDs in series with D1.
The two Y/EW LEDs are in series with Q9 which doesn't have diodes.
Yes right
and on A of LED G/NS ( with diode ) is +4.77
So the voltage drop on D1 is 0.23V and the diode is barely conducting. Are you able to measure current in the LED? With 4.77V across it, it should be fully on with a forward voltage closer to 2.5-3V.
Are you certain all of the wires are connected? When I was learning Multisim, I spent a lot of time looking for broken connections.
The Current of A Of LED (Green /Ns ) is +0.04
I draw the connections as the design
but at least one led fire if there is failed connections ( The led without Diode )
I draw the connections as the design
but at least one led fire if there is failed connections ( The led without Diode )
If the units are Amps, 0.04A = 40mA and the LED should be on. I don't think your simulator is smart enough to burn out the LED for you. Most LEDs won't tolerate much more than 20mA without burning up.
