im using BC547 transistor

That transistor has a maximum collector current of 100mA, so you can only drive 5 LEDs (with no safety margin). If you use it as a switch, power dissipation shouldn't be an issue.

 

zazas321 Try read this :
http://forum.allaboutcircuits.com/xfa-blog-category/circuit-design.38/view-entries


dl324 25 ohms / Ie - how can ohms divided by ampere can give as resistance ?? Also why you mixing a small-signal parameters with DC?
And re is not a resistance seen when we looking into the base terminal.

 

You use the current divider equation for parallel resistors. If you have 3A flowing through 0.5Ω||1Ω, 1/3 of the current will flow through the 1Ω and 2/3 will flow through the 0.5Ω.

Hmm. Arent those resistors in series and in series current is the same? so they should both have 3A?

 

Some other things to consider from the datasheet:

The saturation voltage could be as high as 0.6V at 100mA. The base-emitter voltage at saturation is 0.7V @ 10mA; I saw 0.9V @100 in another datasheet, but it was secure and I couldn't print or capture.

 

dl324 25 ohms / Ie - how can ohms divided by ampere can give as resistance ?? Also why you mixing a small-signal parameters with DC?
And re is not a resistance seen when we looking into the base terminal.

That's what I recalled, now I need to look it up...

Looking into the base, you see the resistance of the BE junction. At 20mA, the slope of the IV curve is quite steep and the resistance would be low. That resistance varies with current. So what I pulled from memory made sense when I was writing it.

 

Hmm. Arent those resistors in series and in series current is the same? so they should both have 3A?

No, they're in parallel (that's what || means).

I reduced the divider resistors to 1 and 0.5Ω and left the BE resistance as 1Ω. The ratioed LDR resistance is in parallel with the BE resistance. That's the current divider I was referring to.

 

No, they're in parallel (that's what || means).

I reduced the divider resistors to 1 and 0.5Ω and left the BE resistance as 1Ω. The ratioed LDR resistance is in parallel with the BE resistance. That's the current divider I was referring to.

Oh yes, i misunderstood.. I thought u were talking about the R1 and Rldr.

So okayy. using the equation Vout = Vin * R ldr / (R ldr + R1) i can find out what resistors i need , if i know that i need about 0,5-0,7 Vout and i know the battery voltage. this Vout or as u typed VLDR is the voltage that will go through those resistors and will go to B-E junction?

 

Vout or as u typed VLDR is the voltage that will go through those resistors and will go to B-E junction?

Essentially. We usually think of current going through resistors and voltage being across them.

 

Yeah okay. Hm and why did u say the b-e junction resistance is 1 Ohm? How did u calculate it? or u just gave and example with random number

 

Yeah okay. Hm and why did u say the b-e junction resistance is 1 Ohm? How did u calculate it? or u just gave and example with random number

As Jony130 pointed, I have a problem with units and using AC parameters for DC. I still think the approximation is close and good enough. You're looking at the resistance of a PN junction who's IV curve looks like similar to the LED curve I posted earlier. Resistance is the slope of the curve, so higher resistance at low currents and lower resistance at higher currents.

For the purposes of calculating current division, you need to have something. Maybe Jony130 will provide some clarifying details until I find time to do some research.

 

Yeah thats fine. I shouldnt even go so deep because its already a challenge to understand what i have now ) It seems strange to me, for example, if there are alot of light in the room, most of the current will go straight through the Collector and Emmiter junciot ( through the leds), but this current will be not enough to light the led? Do i understand this part right? For example if the R ldr when the room is light will be lets say 100k and the R1 is 50K ohm, so the current would be I=4.5/150K=0,03mA , so lets assume thats not enough to light LED? Please tell me im correct!

 

For example if the R ldr when the room is light will be lets say 100k and the R1 is 50K ohm, so the current would be I=4.5/150K=0,03mA , so lets assume thats not enough to light LED?

If I understand you correctly, that's not correct.

The LDR is a semiconductor whose resistance will decrease when exposed to light. For your circuit to work, the "light" resistance needs to be low enough to prevent the transistor from starting to turn on. Modern LEDs are very efficient and will emit light at very low currents (less than a mA).

The transistor will multiply it's base current by it's DC beta, which I usually estimate at 10% of the appropriate DC bata (to account for the fact that beta decreases with decreasing Vce). If you examine the IV curves for a transistor, you'll get a better idea of what I'm talking about. Unfortunately, the OnSemi datasheet I downloaded for BC547 doesn't have any...

The DC beta for BC547 at Ic=100mA and Vce=5V is typically 180. For saturation mode, I'd use 20. If you assume the BE resistance is much much less than the LDR, most of the current in the resistive divider will go into the base. When that induces a large enough collector current, the LED will emit light.

 

This idealized transistor IV curve was taken from an MIT course:

It doesn't illustrate clearly that beta generally decreases with increasing collector current.

It does show that beta drops off as Vce decreases. In saturation, ideally Vce is 0V.

 

For your circuit to work, the "light" resistance needs to be low enough to prevent the transistor from starting to turn on

Waaaitt.? I thought for our circuit to operate we need the transistor to turn on , not to prevent it from turning on

 

Waaaitt.? I thought for our circuit to operate we need the transistor to turn on , not to prevent it from turning on

Since you're using an LDR, I assumed you want the LEDs on when it's dark and off when it isn't; so you have to prevent it from turning on when it's "light".

 

Okay thanks for your help very muchh! Today was a tough day already and i think i will take a moment to understand everything what we have spoken today. Tommorow il ask my physics teacher and i will think about how this circuits works when there is light in the room and when there isnt I will see what i can find out , cya tommorow

 

By the way, little off topic, i came up with the problem finding out the actual voltage drop across the leds. I connected different coulours LEDs to the 4,2V battery and measure the voltage across the LED , and the measurements differ from what is written on the LEDs package they gave me when i bought those..Any ideas?

 

By the way, little off topic

It's allowable if it's your thread...

i came up with the problem finding out the actual voltage drop across the leds. I connected different coulours LEDs to the 4,2V battery and measure the voltage across the LED , and the measurements differ from what is written on the LEDs package they gave me when i bought those..Any ideas?

LED forward voltage is for a specific current. If it isn't mentioned, you assume the current used in the datasheet; which is typically 20mA (or 10mA for LEDs from a couple decades ago). I would assume 20mA.

The easiest way to measure a lot of LEDs is to make a 20mA constant current source.

For the same color LED, even from the same batch, you'll find that forward voltage is a range.