HI All,

I'm using a TIP41C transistor to control a 21 watt 12 volt bulb. How do I work out what voltage/current I need to turn the trasistor fully on?

Any help, or a link to a tutorial would be great.

Thanks

Tip41C datasheet attatched

Attached Files

TIP41C.pdf (42.9 KB, 11 views)

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Are you just using the transistor as a switch to either turn the lamp on or off or are you wanting to dim the lamp as well?

If in the dimming mode, I'd not use the transistor in an analog mode but would drive it with a 555 and use PWM -- much, much more efficient.

Dean

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NPN transistors are turn ON via current going into the base pin.

First one have to determine the load current. For a lamp of 21W at nominal 12V supply, the current is 21W/12V=1.75A. However, we should allow for extra margins so I would say about 2A.

If you than look at the datasheet graph on page 2 "DC Current Gain", it gives a current gain at this particular current to be 40. Again we allow for some margins so I would say it is about 30. Thus for a collector current of 2A, the base current will be roughly 2A/Gain = 2A/30 = 67mA.

If this is an industrial application, I would even lower the DC gain to 15 according to the figure quoted by page1 of the datasheet(Hfe Min 15 at 3A). This ensure the TIP41C will will be fully ON no matter what.

Thus in your case you need to pass a current of 67mA into the base pin to result in a collector current flow of 2A. At this current, the Vbe(sat) is about 1V and you can work out the resistor value say from a 5V drive. Base resistor = (5 - Vbe(sat) ) / 67mA = 60 ohm.

Of course in real life the lamp current will be less than 21W/12V because at 2A, the saturation voltage of the TIP41C is about 2V thus only about 10V is available to light up your lamp. Your TIP41C is dissipating 3~4W of heat so it will definitely need a heatsink.

Hopes this help.

Edited: The Vce(sat) value of 2V at 2A collector current was mis-read from the datasheet. It should be about 200mV instead of 2V. The heat dissiplation is thus also 10 times small. Sorry for the confusion. Thanks Roff for pointing this out.

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L.Chung

 

This brings up another question in my mind. The hot resistance of the bulb on the subject is 12V/1.75A or 6.8 ohms. The cold resistance of the bulb if measured is about 1/10 of the hot resistance. Since the base current determins the collector current, will the transistor limit the surge current? The surge current in a normal circuit(just a switch to turn on the lamp) would be about 17A, but the surge current is only for a fraction of a second. So with the transistor as a switch would the surge current be limited?

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Thanks eblc,

That's a great explanation. Thanks again

Neil

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I read the graph as Vce(sat)~200mV at Ic=2A. Notice the graph says this is with Ic/Ib=10 (Ib~175mA). I would recommend using this drive ratio, because beta is lower in saturation than it is with Vce=4V, which is where the beta numbers you quoted are specified.

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Ron

 

Hi L. Chung,
The hFE of a transistor is used when it has 4V across it as a linear amplifier. When the transistor is saturated then it needs a much higher base current.
The base current is shown to be 1/10th the collector current for the spec'd saturation voltage. Then the base current should be 175mA.

Hi K7,
The current gain of a transistor is a range. Some are good and others are not very good. If you design the circuit so there is enough base current for all transistors like I do then the sensitive ones will conduct much more than the weaker ones and not limit the surge current.

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Uncle $crooge

 

I think he was looking at the possibility of surge current limiting as a good thing. With 175mA base drive, you should get some limiting.
EDIT: I would look at using a MOSFET instead of a bipolar transistor.

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Ron

 

Good catch, Roff. Not aware that Y-axis was with different scale between Vce(sat) and Vbe(sat).

The suggestion by you and audioguru of using gain=10 is very true when Vce(sat) voltage is low. I have forgotten that fact.

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L.Chung

 

Roff is correct. I have been working on a circuit to limit the surge current on 12V bulbs, specifically those used as lighting in RV's. I think we all know that most bulbs burn out when first turned on because of the initial surge current as a result of the cold resistance of the lamp filament. I have been using power MOSFETS in a timed circuit to short out a surge limiting resistor that has extended the life of live of 12V bulbs a great deal. Seeing this post gave me some more insight on how stuff works. My initial thought was that in this case the surge current would exceed the maximum collector current of the transistor, but I don't think that will occur as the base current limits the collector current in this case.
I posted my surge limiting stuff on a couple of RV forums, but there wasn't much interest. Perhaps I will post it on one of the project threads of this forum.

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Or even simpler, you could slowly turn the MOSFET on using a resistor and capacitor.

I agree, use a MOSFET.

I'd also consider replacing the lamp with LEDs.

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They're not, at least on the datasheet that Fingaz posted.

Attached Images

TIP41 sat.PNG (14.4 KB, 10 views)

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Ron

 

What is wrong with semiconductor manufacturers today?
The datasheet shows an excellent typical saturation voltage but they keep selling ones that have horrible spec's. Surely they have improved their manufacuring over the years.
Those horrible ones should be discarded.

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Uncle $crooge

 

No. they are of the same scale. I don't know why I have said that.

I must have contacted the "foot in mouth" virus.

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L.Chung

 

I'm taking medication for that right now.

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Ron