# need help with this circuit..

#### selena

Joined May 7, 2007
2 I'm in a bit of a hole with this following circuit. Can anyone please get me out? We are investigating this circuit in which the LED turns on when the light intensity falls below 100mWm-2.

We were given the following assupmtions
~ R(LDR) at 100mWm-2 equals to 25 kohms.
~The current gain of the transistor is 100
~The LED has an optimum voltage of 1.7 V and a limiting current of 20mA.
~Supply Voltage is 6V.

I have trouble trying to find out what R3's value should be in order for the LED to switch on at exactly that light intensity.

I have to find the values of R1, I(LDR), I(b), R(c) and R3
Just to make sure my answers are correct can someone please show me how they got the values for the above?

#### Attachments

• 307.1 KB Views: 39

#### selena

Joined May 7, 2007
2
Forgot to show my previous attempts of thie questions >_<

#### recca02

Joined Apr 2, 2007
1,212
i think this has also been discussed before (more than twice i believe) in this forum try some forum search,till some cape crusader comes to your aid.

#### bloguetronica

Joined Apr 27, 2007
1,540
Well, I think there should be a base resistor, so the transistor would never be over saturated. But I think that is no problem here.

Ok, lets assume that the transistor will start conducting if the potential on its base is above 0.7V. That makes sense, because there is only a base current (normally from Vcc), if the potential at the base equals or goes above 0,7V. As you know, if the transistor starts conducting, the LED will not receive enought current, since that current is being sunk by the transistor throught the common resistor Ra. If that happens, the LED will stop cunducting if the collector potential is below 1,7V.

When calculating R1, consider that R1 = 0.132 x Rldr = 0.132 x 25K = 3.3K. Of course 0.7 / (6 - 0.7) = 0.132, hence this factor. So R1 = 3.3K.

To calculate Ra we have to see what ammount of current is sunk by the transistor. For this we will need the transistor beta (beta=100), and the lowest Rldr possible, so we can calculate Ra according to that. I don't see what is the lowest Rldr you can have, so I'm missing that. Nevertheless, if we ignore the safety and say that the transistor is in the active region instead of being in the saturation region (since beta is 100 and not 20), Ic = 100 x Ildr. Ildr = (6V - 0.7V) / 25000 = 0.212mA. So Ic will be 21mA (current through the collector). Now we have to calculate Ra so the potential at the collector is 1.7V when the Ic = 21.2mA, that is, when the LDR is iluminated by the minimum light possible. We have a potential divider here.
So, lets assume that at 1.7V (little below) the LED is at the virge of stopping to conduct (it would be an ideal diode, with a fixed voltage). Its resistance will be a lot more highter (almost infinite) than R3, and so R3 won't matter for the case (we will get to that later). So, no apreciable current would get trought the LED or R3. So Ra = (6V - 1.7V) / 0.0212A = 202.8Ohm. Remember that all the current goes throught the transistor and the potential is divided by Ra and the transistor.

To calculate R3 we have to consider that the transistor is cutting, and that all the current goes through Ra, Rc and the LED. In that situation the LED would be bright having only 1.7V at the terminals (remember, it is considerer as an ideal diode, so the voltage will be fixed).
Lets calculate the total resistance needed to protect the LED. Rt = (6 - 1.7V) / 0.02A = 215Ohm. So R3 = 215Ohm - 202.8Ohm = 12.2Ohm.
I think that should do it.

My only doubt is that a transistor used like that should be in the saturation region, since here it is better used as a key (emmiter is grounded).