How to Activate Relay using Transistor?

[Svein]

If I have still errors in making the circuit as well as looking the right information in the posted data sheet, please inform me. I need to perfect this circuit for my own sake. Thank you.

Remember that you have two transistors here. Adding up two minimum values of V_be gives 0.55V + 0.55V = 1.1V.

 

[nicy12]

Remember that you have two transistors here. Adding up two minimum values of Vbe gives 0.55V + 0.55V = 1.1V.

So the 0.79 V in the exposed to light situation is not enough to turn on the transistor?

How about the negative values of currents? what is the explanation behind them?

 

[Svein]

The only problem here is the voltage in that point is 5.8 V, which is very close to the 6 V maximum rating as it shows here:

Sigh. No! The maximum voltage on the base of the first transistor is 1.54V (V_BE(on) *2). Now, if you had connected the LDR to a negative voltage, you would have had to be careful of not getting below -6V...

 

[Svein]

So the 0.79 V in the exposed to light situation is not enough to turn on the transistor?

How about the negative values of currents? what is the explanation behind them?

• 0.79V should not be enough, no. If the transistor pair ends up conducting a little, reduce the 10k resistor.
• What negative currents?

 

[nicy12]

Now, if you had connected the LDR to a negative voltage, you would have had to be careful of not getting below -6V...

Honestly, I really don't get the point. as the circuit above is concerned, the LDR is not connected to any negative voltage. The voltage on LDR is because of being in voltage divider configuration.

What negative currents?

Those in red is what "negative currents" I am talking about.

 

[Svein]

Those in red is what "negative currents" I am talking about.

How did you invent those? When the LDR has a value of 160kΩ, the voltage at the first base is below the turn-on point and no current enters the base. If you could pull a current out of the base (that is what you are implying with negative currents), then no current would flow in the transistor. It gets worse. In your two circuits you draw both sets of current with the same direction and you name them positive in one instance and negative in the other. Do you mean that the transistors suddenly start to act as voltage sources, pushing current back into the battery?

 

[nicy12]

yah

How did you invent those? When the LDR has a value of 160kΩ, the voltage at the first base is below the turn-on point and no current enters the base. If you could pull a current out of the base (that is what you are implying with negative currents), then no current would flow in the transistor. It gets worse. In your two circuits you draw both sets of current with the same direction and you name them positive in one instance and negative in the other. Do you mean that the transistors suddenly start to act as voltage sources, pushing current back into the battery?

Yahoo! Thank You Sir Svein! You clear all the doubts in my regarding this circuit (at this moment of time). The only thing to do is buy an BC547 transistors so I could make it in reality. The theoretical part was done because of you! I just need to make the circuit so I can figure out if the calculations are correct. I believed it was correct because You together with others help me to accomplish my goal. Thank You very much! Please don't hesitate to help those who like me (who is really stubborn and slow-minded). God Bless.

 

[NascentOxygen]

The only problem here is the voltage in that point is 5.8 V, which is very close to the 6 V maximum rating as it shows here:

That actually is the reverse voltage maximum for the base-emitter junction. (Notice how they refer to it as emitter-to-base?) That rating does not concern you here, it won't be reversed at all.

I can understand you wanting to pursue this circuit in the interests of learning about transistors. However, as I pointed out early in the piece, it can't be made to work reliably with even a pair of BC547 transistors. (There are products in that small signal family with gains of around 300, maybe you'd like to buy a couple and try them?)

If you eventually admit defeat, I'll return to the possibility of using a FET, it's worth a second look as I think it will manage to do what you are wanting.

 

[nicy12]

However, as I pointed out early in the piece, it can't be made to work reliably with even a pair of BC547 transistors

How about this:

.Does the circuit above won't work? The calculated voltages and currents satisfy the condition of transistors and relay. Please explain why it won't work reliably.

 

[NascentOxygen]

While the transistors are specified as β >110 it is wise to design for, say, 50, to allow a comfortable margin of safety. A design that borders on the margin of operation is not going to be reliable. HOWEVER, if you are willing to actually measure the β of whatever transistors you buy and use those with higher gain, you should be able to get the circuit going for demonstration purposes. Don't be surprised if it doesn't survive long.

I make I_B1 to be 3μA. Did you factor in what is lost through the dark LDR? If the second transistor doesn't get an excess of base current, it won't saturate and may burn out.

 

[NascentOxygen]

When the light level in your application changes from "dark" to "light" will the transition be rapid (like when a room light is switched on), or might it be slow and unsteady (like the sun rising through a broken cloudy sky)?

 

[nicy12]

When the light level in your application changes from "dark" to "light" will the transition be rapid (like when a room light is switched on), or might it be slow and unsteady (like the sun rising through a broken cloudy sky)?

It is not very rapid (compared to room light is switched on). Why did you ask Sir? does it have a major role in the circuit?

While the transistors are specified as β >110 it is wise to design for, say, 50, to allow a comfortable margin of safety.

The minimum Beta of BC547 is 110. I think it should be designed in minimum because the range between maximum and minimum Beta is large (about 800).

Don't be surprised if it doesn't survive long.

Why would it be? We design the circuit according to its data sheet and without exceeding ecah of their maximum ratings.

I make IB1 to be 3μA

it will only produce 36.3 mA (using minimum Beta of 110) in the second transistor's collector side. Not enough to operate the relay.

Did you factor in what is lost through the dark LDR?

I don't get this. What factor are you saying Sir? I think that "factor" should be consider in this circuit.

If the second transistor doesn't get an excess of base current, it won't saturate and may burn out.

How and Why the second transistor might not get an excess of base current ( making it won't saturate and may burn out).

 

[NascentOxygen]

When the LDR is 1.5MΩ in the dark, there will be current through it to ground, roughly 1.2V/1.5MΩ. This is current through R_B that won't be going into the base of Q1.

While the BC547 seems to be guaranteed to have β>110 don't be surprised if you buy a couple to find one that doesn't. There are wide variations in β. So buy half a dozen and pick out the two with highest gain.

 

[Averagesupernova]

I think it is time to ask for more information here. There is concern about operation in between cutoff and saturation. The circuit is set up so the relay turns on when it gets dark. This implies that it is being used to control a light when it gets dark. And of course that implies that there will be times when the transistor is neither on nor off.

 

[Svein]

I think it is time to ask for more information here. There is concern about operation in between cutoff and saturation. The circuit is set up so the relay turns on when it gets dark. This implies that it is being used to control a light when it gets dark. And of course that implies that there will be times when the transistor is neither on nor off.

Yes. That is why I prefer using a comparator. I put a rough schematic in one of my previous answers.

 

[Averagesupernova]

I don't think this is too much to handle at all with discreet transistors. The current level of understanding by the OP presents a challenge though. With a comparator we will still have a transistor to drive the relay. If we use a darlington transistor with a little positive feedback to add in some hysteresis I think we will have a pretty reliable setup. I would advise a different LRD though.

 

[Svein]

I don't think this is too much to handle at all with discreet transistors. The current level of understanding by the OP presents a challenge though. With a comparator we will still have a transistor to drive the relay. If we use a darlington transistor with a little positive feedback to add in some hysteresis I think we will have a pretty reliable setup. I would advise a different LRD though.

I used this thread to talk about using transistor as switching elements. In real life, I have taken a couple of automatic dusk light switches apart, and they all use a comparator with enough output capability to drive a relay, and a small potentiometer for adjusting the trigger light level. They also use LDRs, not phototransistors.

There is, however, a small, interesting problem left. If the relay is used to turn on a light source which again shines on the LDR - how do you keep your circuit from deciding to turn the light off again? There are several solutions to this problem, but I will not disclose them for the moment. Treat it as a challenge - describe how you would solve it, and what it would cost...

 

[NascentOxygen]

@nicy12 Is this project still in the works? Have you reconsidered any aspect?

 

[nicy12]

@nicy12 Is this project still in the works? Have you reconsidered any aspect?

Yes Sir NascentOxygen. It is still in process. Gathering additional information to improve its design by means of my current understanding. In fact, this self-assessment project is still on scratch. I still haven't bought yet the needed materials.

I would like to thank all of you who helped me on designing this self-assessment project. I would also want to apologize because I became unfair. I've just thanked Sir Svein.

Please don't bother to help me again. Thank You All! God Bless Us!