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pgsabel
Joined: 04 Jun 2003
Posts: 12
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 13 May 2008 12:53 Monitoring lamps DC |
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Hi everyone,
Exists some circuit that allows to indicate when a light bulb of 24V burns in the proper circuit that it is installed?
I would be pleased if you help me here.Thanks
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atmaca
Joined: 13 Jan 2004
Posts: 351
Helped: 1
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| 13 May 2008 14:58 Re: Monitoring lamps DC |
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| Read current (by resistance) and voltage on lamp by ADC and uC. Then decide if the lamp is defected or not
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pgsabel
Joined: 04 Jun 2003
Posts: 12
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| 13 May 2008 15:25 Re: Monitoring lamps DC |
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| Exists another way without using a microcontroller?
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umery2k75
Joined: 19 Apr 2006
Posts: 207
Helped: 19
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| 13 May 2008 15:44 Re: Monitoring lamps DC |
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| You can use a voltage comparator.If bulb is operating then there must be X voltage drop on its one terminal, and when bulb burned up, then full applied voltage Y would appear, using this configuration, come up with a circuit, that would indicate when the bulb is failed.
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pgsabel
Joined: 04 Jun 2003
Posts: 12
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| 13 May 2008 16:02 Re: Monitoring lamps DC |
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I placed the circuit that I made in Proteus.
They could verify?
The circuit that I made is correct?
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umery2k75
Joined: 19 Apr 2006
Posts: 207
Helped: 19
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| 13 May 2008 17:49 Re: Monitoring lamps DC |
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If my circuit logic is correct, then this should work, more over I forget when the voltage comparator turns ON and turns OFF. So the LED might turn ON when the bulb is operating or turns OFF.The output of LM339 is open-collector, there by I use the pull up resistor on it.
Added after 1 hours 5 minutes:
I see a problem in your circuit, your circuit will not work for a long duration of time, because transistor will get over heat due to thermal runaway effect.
First, there is a short between Vcc and ground, as you can see in the below diagram.This will overheat the transistor and burnt it up very quickly, when the transistor gets on.
Secondly, if I calculate the collector current, then Ic=Vcc/RL = 24/0 = Infinity, since no resistance is present. So if I make the load line of the transistor BC238, which will look some what like this:
The collector resistance, determines what maximum current, you want through collector PIN of transistor. One point of the line is set by 24V and other point is calculated with the above formulation. So the transistor is showing this behaviour.
the maximum safe collector current is around 100mA, but you have exceed this limit.
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pgsabel
Joined: 04 Jun 2003
Posts: 12
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| 14 May 2008 13:11 Re: Monitoring lamps DC |
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| umery2k75 wrote: |
If my circuit logic is correct, then this should work, more over I forget when the voltage comparator turns ON and turns OFF. So the LED might turn ON when the bulb is operating or turns OFF.The output of LM339 is open-collector, there by I use the pull up resistor on it.
Added after 1 hours 5 minutes:
I see a problem in your circuit, your circuit will not work for a long duration of time, because transistor will get over heat due to thermal runaway effect.
First, there is a short between Vcc and ground, as you can see in the below diagram.This will overheat the transistor and burnt it up very quickly, when the transistor gets on.
Secondly, if I calculate the collector current, then Ic=Vcc/RL = 24/0 = Infinity, since no resistance is present. So if I make the load line of the transistor BC238, which will look some what like this:
The collector resistance, determines what maximum current, you want through collector PIN of transistor. One point of the line is set by 24V and other point is calculated with the above formulation. So the transistor is showing this behaviour.
the maximum safe collector current is around 100mA, but you have exceed this limit. |
Very good its explanation. It had not noticed the imperfection in my circuit.
In the case of its circuit as I would make to only use positive feeding (+12V) in the LM339? He has as?
Added after 1 minutes:
Very good its explanation. It had not noticed the imperfection in my circuit.
In the case of its circuit as I would make to only use positive feeding (+12V) in the LM339? He has as?
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umery2k75
Joined: 19 Apr 2006
Posts: 207
Helped: 19
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| 14 May 2008 17:20 Re: Monitoring lamps DC |
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Here are few modifications I made in your circuit.I have made this circuit, by keeping few parameters in mind.
By looking at the Datasheet of BC238. It has a DC gain(Beta) in range of (120-800). Since there's a large variation of Beta in it. I take the mean of Beta=(120+800)/2=460
As you can see in the figure, you can also say that the path is short, as I have mention in your diagram, but this transistor is partly ON.So even if I don't add the resistor on collector or even on emitter, the transistor will still not burnt up due to excessive current, because transistor has it's own internal resistance Rce. So I have only turned ON the transistor partly, in which it will only sink 10mA of current, thereby energizing the relay, but I don't know in your case how much Ib you were giving to your transistor, as there was a bulb of unknown specification, there was nothing mention about it. If Ib doesn't saturate the transistor in your circuit, your circuit can work, as it appears at this point. but more over I don't understand the purpose of your R3 resistor, seems like you are pulling it down to 0V, why? I don't understand, in this way your circuit cannot work, because it needs to have Vbe=0.6V,R3 resistor is trying not to bias the transistor, whereas R2 is trying to bias it, both resistors are fighting. So result is that, your transistor will not get bias, because it's pulled down to 0V. Even if your transistor gets biased. Ic that is produced due to unknown Ib can damage the transistor, if Ib is large enough to saturate the transistor.
About biasing the transistor in my circuit, I'm only giving 0.61V to bias the transistor.In this way the bulb gets the maximum voltage drop. As ideally it should have 24V, but due to the bias voltage, it drops to around 24-0.61=23.29V. The value of R2 was decided so that voltage on this node, would be 0.61V. So V=R/(96+R)*24=2.50Ohms. I didn't bias the transistor at 0.6V, because Rb will get to zero, according to this formulation Rb=(V-0.6)/I. As I have not calculated current getting into the base of the transistor through a current divider. I used the technique in which I add the resistor to convert voltage into current, because 0.61 V is at that point. I just add the resistor that had given me the the desired Ib.
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FvM
Joined: 22 Jan 2008
Posts: 1827
Helped: 332
Location: Bochum, Germany
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| 14 May 2008 20:46 Re: Monitoring lamps DC |
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For a current detection by shunt resistor, as discussed above, I would generally use a lower resistance shunt and an OP/comparator detection circuit that can operate with max. 100 mV voltage drop. Positive feedback (hysteresis) can prevent unwanted linear operation of the output driver.
As a simple, rugged current indicator, also a reed contact with a current sense winding can be used. Usual reed contacts have 10 - 50 ampere turns operate field strength.
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khajamoiz
Joined: 17 May 2008
Posts: 11
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| 17 May 2008 18:29 Monitoring lamps DC |
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--------------------------------------------------------------------------------
Hi everyone,
Exists some circuit that allows to indicate when a light bulb of 24V burns in the proper circuit that it is installed?
I would be pleased if you help me here.Thanks
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umery2k75
Joined: 19 Apr 2006
Posts: 207
Helped: 19
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| 18 May 2008 12:38 Re: Monitoring lamps DC |
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@khajamoiz
See above post.
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