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# Calculate value of base-emitter resistor for Dark Activated Light

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Hi,

This will sound like a homework question. All I can say is that it isn't, and I'm not looking for any answers so I can pass any tests (those days are behind me, I hope!) For those interested I'm an IT professional dabbling in electronics.

I am designing this Dark Activated Light for fun and I'm having a problem understanding how to calculate the value of the resistor that goes into the Base of the Transistor.

The item is R2 (as below).

Some calculations below

I just made up a figure of 300uA, seemed small enough not to waste battery but large enough to turn on the transistor.

This is the part I am struggling with:

How do I calculate the value of R2, to give the right voltage drop of 0.7V when the LDR has a higher resistance.

I tried blocking off the 4k (say when dark) and focus on the VR1 + R2 path to give 0.7V -> 6V - 0.7V = 5.3V -> 5.3V / 200uA = 26500 ohms. I have 16000 ohms above, so I have about 10000 ohms which I plugged in to make up 26000-ish ohms.

However wouldn't this always switch on the transistor as the voltage drop across the parallel path will be equal?

I calculated the current through this leg of the circuit, determined the ESR to be 2857 ohms for the 4k || 10k  part, which was 318uA. The LDR leg had 227uA and the Base of the transistor 91uA. I checked the BC548 data sheet but couldn't find if it's enough current to switch on the transistor.. just the voltages. https://www.fairchildsemi.com/datasheets/BC/BC547.pdf

But am I on the right track, I just starve the Base of current so it can't turn on? Technically it has the voltage but not enough current.. ?    Then when the LDR gives resistance it gets the full 318uA which is enough to switch it on?

Also is this how electrical engineers do these calculations? Is there is a better way? I'm really just flying solo with these calculations, and you can probably tell

Edited by ElectronMan1
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The datasheet for the BC548 shows that it saturates properly when its base current is 1/20th its collector current. When the transistor is saturated its collector current is 15mA then its base current should be 15mA/20= 750uA. Then R1 will be 5.3V/750uA= about 7k. In room light the base voltage will be about 2.18V so the transistor will not completely turn off causing the LED to look dim.

You might be lucky to find a transistor with high gain (they are all different) then R1 can have a much higher value. You should be using a darlington transistor that is guaranteed to have very high gain.

R2 simply prevents the base current from being too high when R1 is reduced too low.

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Thank you audioguru.

Did you derive the saturation current ratio of 1/20 from the "Conditions - Base-Emitter Saturation Voltage" - collector current 10mA and base current 0.5mA from the datasheet?

I think I understand your calculations, I have redrawn to confirm that I understand you correctly.

I would like to explore the Darlington transistor - or two BC548 arranged into a Darlington pair.

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You forgot that the base of the NPN transistor cannot go higher than about 0.65V so in the light the base has some current. The 7k resistor has a current of (6V - 0.65V)/7k= 764uA. The 4k LDR has a current of 0.65V/4k= 163uA. Then the base current is 764uA - 163uA= 601uA so it is turned on a little and the LED lights a little.

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Ah hah. I see the problem, the LDR is too high at 4k to divert enough current to ground. The transistor requires too much base current to fully saturate and has too wide a 'window' in which it will operate but not saturate.

Time to research an appropriate Darlington transistor. Thank you

Edited by ElectronMan1

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