# LDR

Discussion in 'General Electronics Discussion' started by vick5821, Jan 30, 2012.

1. ### vick5821

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Jan 22, 2012

Referring to this picture, the LED will switch ON in the dark condition.

This is due to that when in dark, the resistance of the LDR will be high and thus more voltage acorss the LDR, thus base current is high and switch on the transistor.

My questions are : isnt that the current across this circuit is constant value ?
and how the details current flow for this diagram ?
what is the use of the resistor connected to the base of transistor ? Is it necessary ?

Thank you

Last edited: Jul 5, 2012
2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
true

true

not quite true

true

No (but it's true that the current through R2 won't vary a great deal)

The important part is that when the LDR has a low resistance current flows through R2 and the LDR to ground, and as long as the voltage across the LDR is < 0.7V, no current flows through the BE junction of the transistor.

As the resistance of the LDR increases, the voltage across it rises until at 0.7V, current starts to flow through the BE junction of the transistor to ground. SOme also passes through the LDR.

It is better to think of the base being connected to a potential divider.

Only if you want the circuit to work and the transistor not to be fried (assuming the alternative is a link)

3. ### vick5821

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Jan 22, 2012
Why not quite true ? what is the true facts ?
and for the second bolded part, so means what ?

4. ### vick5821

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Jan 22, 2012
So, can I say that, current flow through the path with least resistance ? Is this concept true ? Let say when in light condition, the LDR less resistance compared to the transistor path which need to achieved 0.7V or more, so it flows through LDR and so the BE junction is off ?

5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
No, and No.

Current flows through ALL paths, the amount being inversely proportional to the resistance. diode junctions have a non-linear resistance, but the generalisation still holds.

Kinda. if the voltage across the LDR is less than 0.7v then there is insufficient voltage to forward bias the BE junction, therefore no base current flows, therefore the transistor is off.

6. ### vick5821

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Jan 22, 2012
does this means that the current flow through LDR is constant? and the voltage only depends on the resistance of LDR? V=IR

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
The current through the BE junction never gets very high, but it does increase above zero.

Better to think of the R2/LDR as a potential divider than R2 as a base resistor (well I think it's more helpful) because as either resistor falls to zero, the effect is more intuitively obvious.

And to touch on an earlier question when you asked about removing R2 (I considered you meant shorting it). If you meant removing it and leaving that part of the circuit open, it would remove the source of current from the base of the transistor (which would therefore *never* turn on)

8. ### vick5821

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Jan 22, 2012
I meant to short them.Means as what shown in the picture..No resistor connected to the base of the transistor.Will it be ok ?

If remove the resistor, for sure the circuit is open and not connected to the transistor

9. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
No, the current through the LDR varies from almost zero when the LDR is in the dark, to about 60uA when it has a very low resistance. However the voltage never rises above around 0.7V (due to the BE junction), so the current through R2 varies from about 60uA down to about 53uA. The difference between the current through R2 and the current through the LDR goes through the BE junction of the transistor.

In the dark, a max of 53uA flows into the base, with perhaps 1uA or less going through the LDR

In sunlight 60uA flows through the LDR and none into the base

At the level of light that the transistor starts turning on, about 50uA flows through the LDR and 1 or 2uA into the base.

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
The transistor will fail very quickly due to excessive current through the BE junction.

11. ### vick5821

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Jan 22, 2012
Means a little bit of light?

12. ### vick5821

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Jan 22, 2012
Oh..so as a precaution so to prolong the life of the transistor ?

13. ### vick5821

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Jan 22, 2012
So my conclusion from your explnation :

When the LDR achieved a voltage of around 0.7V, this activated the transistor BE junction and causes the current difference between R2 and LDR to flows through the transistor and switch ON the transistor and the LED lights up ?

14. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Yes.

Consider how this could happen if R2 was 0 ohms (hint, it couldn't).

Consider also that a transistor is current controlled (like a LED) and without a voltage source of sufficient impedance (essentially series resistance) an excessive current can flow. A voltage divider has an impedance determined by the value of the resistors in it (mostly the top one if you're sourcing current).

15. ### vick5821

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Jan 22, 2012
I thought sourcing current only apply in IC 555 ? How will it be applied in this case ?

16. ### vick5821

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Jan 22, 2012
But I do not understand one things.How is it possible to have a current difference between LDR and R2 ? why will this happens ?

17. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Current is sourced whenever conventional current flows from a point. The source supplies it, the sink absorbs it.

In this case the battery is the ultimate source for that current. The battery sources it, and the upper terminal or R2 sinks it (it's not the only sink the battery sees).

The junction of R2 and the LDR sources current to the base of the transistor (which sinks it). The junction of R2 and the LDR could also sink current, but in this circuit there is no source which would enable it to do so.

The LDR is a sink for R2, but a source of current for the -ve terminal of the battery (which is the ultimate sink in this circuit).

At any point in a circuit source means where current comes from, sink is where it goes to. As operating conditions change a source can become a sink, or vice versa, or something may suddenly become a source or a sink, being neither before (or vice versa)

In the case of the 555, the output pin could be a source or a sink depending on where you connect the load, and what state it's (the 555) in.

because some of the current flows from the junction of R2/LDR through the BE junction of the transistor. There are 2 paths to ground from that junction, one of which is only available when the voltage at that point reaches 0.7V

18. ### vick5821

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Jan 22, 2012
Oh, so means when the transistor if off, there will also be a small current flow to the trasistor from the R2/LDR junction ? then when the resistance of the LDR gets higher, the current through the base of transistor will increases ? So, the Higher potential acorss LDR indirectly will cause more current to flow to the base of transistor ?

p/s Since you mention the amount of current is inversely propotional to the resistance value ? So when LDR resistance high when dark, more current will pass to the base of transistor ? Can I say so ?

Last edited: Jan 30, 2012
19. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
When the base of the transistor is lower than 0.7V only a very small leakage current flows. For all intents and purposes it is zero.

As the voltage across the LDR rises, current starts to flow into the base of the transistor and it turns on.

Yes, higher potential across the LDR causes more current to flow, but because the BE junction is a forward biased diode, the voltage can rise very little and the transistor ends up hogging most of the current flowing through R2.

20. ### vick5821

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Jan 22, 2012

Mind to explain in details ? I am not too understand abt that

more currents to flow through LDR or transistor ?

Last edited: Jan 30, 2012