Okay, I'm stuck on this. I don't have a very good knowledge of transistor theory, so it's likely I'm doing something basic wrong.

I'm using a TIP125 to switch a load. This is for a computer and I'm switching power to an inverter for cold cathode lighting. I would like it to be able to turn on and off the CCFL.

The TIP125 is switching the power side of the circuit and the base is connected to a uC that switches ground, to turn it on and off.

The whole thing works perfectly unless I connect the inverter to the circuit, then the inverter just stays on, or just switches the power faintly for a few seconds before going full on. If I connect a regular LED to the circuit rather than the inverter, the LED switches perfectly; ON/OFF. I checked with a multimeter while the LED was connected and the TIP125 is switching 12V/0V correctly.

So, I have no idea what I am doing wrong. Switches perfectly if just an LED is connected, does not switch, when the inverter is connected.

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Mark Higgins

 

Should we guess, or are you going to attach your schematic?

The output of the uC goes up to only 5V doesn't it? The base of the TIP125 needs to be higher than 11V to turn off. How are you converting it?

If an LED turns on but the inverter doesn't then maybe the darlington transistor doesn't have enough base current.
What is the supply current of the inverter?

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Uncle $crooge

 

I thought that a schematic would be too simplistic. There's just the uC, a resistor to base, and the transistor switching the high side.

Okay, I am not certain about the supply current for the inverter. I will measure it, but I'm guessing it's about 500mA's or less.

I was under the impression that I did not need to turn 'OFF' the PNP by supplying current to the base, simply disconnecting it from ground would turn it 'OFF'. I will lower the base resistor from current 10k and see what happens.

Thanks you for your help.

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Mark Higgins

 

What is disconnecting the base of the darlington transistor from the 0V or 5V output of the uC? The base needs to go to about 11V or higher to turn off.

What is connecting the 10k base resistor to 0V?

The datasheet for the TIP125 shows it saturating well with a base current that is 1/250th of the collector current. Your 10k resistor is too high for a 500mA load.

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Uncle $crooge

 

Well the uC output can sink, source, or turn off. Do I need to have something disconnect it? The uC is connecting the resistor to GND to turn it on. Like I said, there's nothing more interesting working here.

Why do I need 11v? I just lowered the resistor and with an error in soldering it didn't actually connect. When I connect the CCFL with the base left unconnected to anything, the CCFL is off, like I would expect.

I'm not understanding why I'm having a problem turning it off, but not on.

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Mark Higgins

 

The base to emitter of the darlington transistor is two base-emitter diodes in series. The darlington has a very low input current so the base to emitter voltage is about 0.5V for each for a total of 1V. The emitter is connected to +12V so the base needs to be +11V or higher for it to turn off.

Some uC's have protection diodes on their input/outputs that stop external inputs from exceeding +5.7V. So the 10k resistor to the +11V base of the darlington has 11V - 5.7V= 5.3V across it which results in a base current of 0.53mA and is not turned off.

Let the uC drive another 10k resistor connected to the base of an NPN transistor with its emitter grounded. Then its open collector can turn the darlington on and off through the existing 10k resistor.
But the logic is inverted.

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Uncle $crooge

 

Audioguru, though I agree completely to your theory of input protection diode biasing the PNP base to +5.6V, I cannot find any good explanation why it had worked for a LED with 0V is being measured.

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L.Chung

 

Thanks. I will put an NPN between the uC and the base of the PNP.

I don't think it will help much, but here's the schema. I didn't do it before, because unfortunately when I changed computers I lost the original and full schematic for the control board I am working on.



Used an LED component, to represent the CCFL inverter.

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Mark Higgins

 

The LED is burnt out without a current-limiting resistor. That's why you measured 12V then 0V at it.
Maybe the LED is made with a built-in current-limiting resistor for 12V?
If it worked with the LED then it should also work with the inverter.

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Uncle $crooge

 

He doesn't mean, he's tried it with an LED, like he said the LED on the schematic represents the inverter.

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Exactly. When I used the LED I used a limiting resistor. When I say the LED turns on and off. I mean it actually cycles with the test pattern I programmed. It is not burned out.

It doesn't. So I guess this all goes back to square one. Why doesn't it? It's stumping me, but then like I said, my transistor theory is not very good.

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Mark Higgins

 

I'm sure it doesn't help either, but this is a picture of the board. Unforutunately the conformal coating makes the picture look kinda nasty. I put the conformal coating on after verifying that the PNP's worked with an LED. I assumed that would be fine. Looks like I jumped the gun on that one.

http://www.higginstribe.com/gallery/...troller-01.jpg
http://www.higginstribe.com/gallery/...troller-02.jpg

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Mark Higgins

 

I would put a resistor from base to emitter on the TIP125. Asking the darlington the turn off with an open base is asking for trouble.

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The manufacturers are way ahead of you.

Attached Images

TIP125.PNG (5.7 KB, 86 views)

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Uncle $crooge

 

Do you have an insulating washer under the TIP125s? It doesn't look like you do in the case-controller-02.jpg picture. If not, the collectors are shorting to the ground plane on your PCB!
I would still put an NPN between the uC and the base of the PNP. BTW, what type of uC are you using?

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