If you assume the transistor is saturated, ideally Vce would be 0V.

The datasheet will have saturation characteristics.

 

But isnt the V1 parralel with Vled and V3? so the voltage should be equal?

 

The C-B junction is forward biased when the transistor is saturated. So V1+Vbc=Vled+V3.

EDIT: Equation applies for all transistor modes of operation.

 

By saying forward biased you mean that when the current starts flowing from the collector to emiter? If BC has a voltage too, looking at a circuit i would say that V1=Vbc + Vled +V3, because Vbc seems to be in series with Vled and V3

 

By saying forward biased you mean that when the current starts flowing from the collector to emiter?

Yes, if the transistor isn't driven hard enough, it will be in active mode.

If BC has a voltage too, looking at a circuit i would say that V1=Vbc + Vled +V3, because Vbc seems to be in series with Vled and V3

I wrote it that way because the junction was forward biased and I was looking at it from the perspective that the transistor was saturated; that lets you see very clearly that Vled+V3=5V. Your equation better illustrates the fact that the quantities you mentioned are in parallel.

Another way to look at it is that KVL says the sum of all voltages in a loop is zero.

 

Yes, if the transistor isn't driven hard enough, it will be in active mode.

Active mode you mean that the the the switch will be "off", i mean the current will not flow through collector and emmiter?

I have circled the places that i understand to be parralel.. I cant grap why V1+Vbc=V3 +Vled. I dont understand why we have to sum V1 and Vbc. Those 2 votlages doesnt seem like connected in series..

 

Active mode you mean that the the the switch will be "off", i mean the current will not flow through collector and emmiter?

No. That would be "cutoff". In active mode, the transistor is on, but not driven so hard that it saturates.

I have circled the places that i understand to be parralel.. I cant grap why V1+Vbc=V3 +Vled. I dont understand why we have to sum V1 and Vbc. Those 2 votlages doesnt seem like connected in series..

If you apply KVL to that loop, you get V1 + Vbc + V3 + Vled = 0. You can use arthmetic to arrange them any way you want. Using this equation, you can write V1+Vbc = -V3 - Vled. If you substitute numbers for the voltage drops, it will be clearer. We know that V1=4.3V and Vbc=0.7V, so 5V is across the LED and resistor.

 

But i know only Vbe, that needs to be 0,7 for transistor to saturate. Why do u assume that the Vbc will be the same?. Yeah i saw that KVL voltage but if i measure all voltages from positive to negative and get 0 , that means that voltage on one side parallel should be positive and on other parralel side it should be negative, but how can that happen.

 

It might be clearer from this schematic:

\( \small V_{R1}+V_{DCB}=V_{LED}+V_{R3}\)
or
\( \small V_{R1}=V_{LED}+V_{R3}-V_{DCB} \)
Corrected equation.

 

But i know only Vbe, that needs to be 0,7 for transistor to saturate. Why do u assume that the Vbc will be the same?.

In saturation both junctions are forward biased.

 

Is this the same circuit you showed me?
Why in your equation there is Vdcb on both sides o
But if i add all those 4 voltages according to KVL there is no way i will get 0 .

Vbe is equal to Vbc when transistor is on yeah?

 

Is this the same circuit you showed me?
Why in your equation there is Vdcb on both sides o

My mistake; too much fiddling with tex and I didn't proofread my post after posting. I corrected the equation.

But if i add all those 4 voltages according to KVL there is no way i will get 0 .

If you start from the 5V supply and go counter clockwise starting with R1, you get (-4.3)+(-0.7)+3.0+2.0=0.

Vbe is equal to Vbc when transistor is on yeah?

Ideally, when the transistor is saturated. When biased in the active region, the C-B junction would be reverse biased.

Add studying the 4 modes of transistor operation (active, saturation, cutoff, and inverted) to your list of things to study.

 

Ok. There will be enough material for today, i just want to know how u measure the negative voltage How do i know which of them will be positive and wich one will be negative?

 

How do i know which of them will be positive and wich one will be negative?

Pick a convention and use it. I considered voltage drops in the direction of travel around the loop to be positive and I'm traveling against negative. Was that not clear from post #52? I meant for it to be...

 

Hello guys,

I deal with several types of LEDs that range in color and in power. That means i cant use a constant current supply unless i make it adjustable.

But for my smaller 20ma LEDs, i just use a 1k resistor and 5v regulated supply. That gives me a quick idea what characteristic voltage my LED is. This seems to be enough in order to characterize what resistor value will be needed to run the LED at a particular voltage. However, measuring two or three points would be better because after all LEDs act like diodes and have a voltage curve like that.

I plan to make an automatic curve tracer at some point but in the mean time the 1k resistor and +5v supply seems to be enough to tell the difference between the types of LEDs i have. For example, one type of green measures close to 1.9v while the other measures close to 2.4v, so i know right away these are two fundamentally different green types. More measurments include:
Blue: 2.7v
Red: 1.8
White: 2.8
Yellow: 1.9

Also logged with the voltage is the supply voltage and resistance used, so the current at those voltages can be calculated. For example for the blue LED the resistor voltage is 5-2.7=2.3 and 2.3/1000=2.3ma. So i know both the current and the voltage at one point on the curve.
Using two different resistor values however would give me two different operating points which would be better, and then i might use the diode equation and solve for two of the constants, or measuring three points and using the diode equation to solve for three constants perhaps. But knowing just that one point helps to classify the LED as to it's characteristic voltage so i expect it to act like other LEDs with the same measurement.

Other ideas include a resistance substitution box and power supply. The subs box only has to include a few different value resistors so you could build a dummed down version of one for testing LEDs. Maybe three to five resistor values.

Of course another idea is a variable power supply and a fixed resistor.

 

I deal with several types of LEDs that range in color and in power. That means i cant use a constant current supply unless i make it adjustable.

The whole point of a current source is that it delivers a constant current regardless of load. To do that, the source voltage has to have sufficient head room.

If you wanted the circuit I posted to work with any LED, all you have to do is use a higher supply voltage. I breadboarded a PNP version of the current source and used it to test some white LEDs I had wired to work with a 6V battery. I increased the supply voltage to 9V and was able to "test" them; series 220 ohm resistor and all...

You're doing too much work. Forward voltage for modern LEDs is typically stated at 20mA. If you're testing at some other current, your forward voltage will not match manufacturer specs.

 

I dont know but this still doesnt make any sense for me , why the hell it should show the negative on one side and positive on one side.
If i measure it , i would say it will show me all positive values

 

Okay i almost understand how the KVL works after i read good article about it.
http://www.facstaff.bucknell.edu/mastascu/elessonsHTML/Basic/Basic5Kv.html

Lets say we choose the loop i made on the bottom of the paper.. Lets assume that we are moving clockwise and lets assume that we are gonna make the voltage positive at the end of every element.. So if we are moving clockwise i wrote the + and - signs on the resistor, but the guy from the website has the opposite signs at V4.. If we are moving clockwise the end of the element will be downwards, so + should be down..

 

I gave that article a quick scan and it seems correct. You're not interpreting what he said correctly.

Take the loop below:

I marked the voltage drops as they would normally be.

If I start at R1 and go clockwise, we have VR1 + VR2 + (-VR3) = 0.

There is a negative sign on VR3 because we encountered a negative sign first, so it's polarity is the opposite of the other two resistors. Rewriting gives VR1 + VR2 = VR3.

BTW, I couldn't read the equations at the top of the page. Contrast is poor and they were too small. From what I could make out, I could tell that you didn't interpret what you read correctly.

 

The whole point of a current source is that it delivers a constant current regardless of load. To do that, the source voltage has to have sufficient head room.

If you wanted the circuit I posted to work with any LED, all you have to do is use a higher supply voltage. I breadboarded a PNP version of the current source and used it to test some white LEDs I had wired to work with a 6V battery. I increased the supply voltage to 9V and was able to "test" them; series 220 ohm resistor and all...

You're doing too much work. Forward voltage for modern LEDs is typically stated at 20mA. If you're testing at some other current, your forward voltage will not match manufacturer specs.

Hello,

I am not sure why are you saying all that. Who said i needed a higher voltage? Not me

Too much work? How so?
There is a correlation between the voltage at one current level and the voltage at another current level. If we were going to worry only about the manufacturers spec at 20ma then we're in trouble if we want to run at 2ma unless we know the correlation or test for that second specific point. This is what curve tracers do.
So with my test i know one point on the voltage/current curve, and with your test you know another point, but without the correlation information (which can simply come from experience with these LEDs) we dont know what voltage it will be at any other current level. I happen to have enough experience with the different colors so that when i know one reasonable point on the v/i curve i know the characteristic of that particular LED and all i need to get that information is a single series resistor and a source that i always have around anyway.

The resistor i use is 1 percent, but i dont even need that much precision. As my example gave, one green measured 1.9v and the other 2.4v, so i can immediately tell these are different basic types of LED, and i got that information from one measurement.

A curve tracer is much nicer of course because then we have the whole curve laid out right before our eyes. If we want to run at 1ma, 2ma, 5ma, 10ma, 20ma, we know what to expect without doing any more work.