Transistor Question - Page 2 - Electronic Circuits Projects Diagrams Free

audioguru · 26th May 2009, 06:34 PM

Quote:

Originally Posted by MikeMl

Gee Carl, you must have started when most transistors had a β of 20

HFE or beta is not used when a transistor is saturated.
Simply look at the datasheet of a modern transistor like a 2N3904:
1) With a collector to emitter voltage of 1V or more and a collector current of 10mA its minimum base current is one 100th.
2) With the transistor saturated with a collector to emitter voltage of 0.2V or less and a collector current of 10mA its minimum base current is one 10th.

Sceadwian · 26th May 2009, 07:21 PM

Why calculate it out? There's a traditional rule that the base current should be 10 times the collector emitter current to act as a switch, special cases aside, go by that general rule and you'll skip about 2 years of math.

audioguru · 26th May 2009, 07:31 PM

Some transistors like the BC547 have their saturation voltage loss rated when the base current is 1/20th of the collector current.

Sceadwian · 26th May 2009, 08:00 PM

It's very nice when it's specifically stated like that. Let you relax design requirements with assurance.

I'm not an expert mind you, but everywhere I have read 1/10th is the rule of thumb, basically what you have to do is under all thermal and voltage conditions guarantee that the transistor will never come out of saturation and go into it's linear region or you'll get thermal runaway and transistor self destruction. I've heard little of safe operating area conditions but don't know much about them or how they're calculated.

MikeMl · 26th May 2009, 08:02 PM

Quote:

Originally Posted by audioguru

HFE or beta is not used when a transistor is saturated.
Simply look at the datasheet of a modern transistor like a 2N3904:
1) With a collector to emitter voltage of 1V or more and a collector current of 10mA its minimum base current is one 100th.
...

Excuse me, but the OP's collector current is only 13mA, and he is dropping about 3V across the current limiting resistor, so it doesn't matter if he drops an addional 0.8V or so across Vce. Look at the attachment from Fairchild. They are happy to rate the β at 100 to 300 near that Ic. Using a base current which gets the Vce to less than 1V is more than sufficient to turn on the LED.

There is no separate section on Fairchild's Data Sheet which differentiates between using the 2N3904 as an amplifier or as a switch, only how low can you get Vce! The test circuit where Ib=1ma is only to determine minimum switching speed, which is of no consequence to the OP who is switching an LED.

At Ic=13mA, there is no requirement that the saturation voltage of the transistor be at an absolute minimum. A base current twice that determined by βmin is more than good enough...

**Nigel Goodwin** · 26th May 2009, 08:29 PM

Quote:

Originally Posted by Sceadwian

I'm not an expert mind you, but everywhere I have read 1/10th is the rule of thumb.

A 'rule of thumb' depends on how big your thumb is!

Generally it's a question of stuffing as much current as you can down the base!.

Sceadwian · 26th May 2009, 08:45 PM

hrmm mmrmmrm m fff mmhrr mfmmffr!

Blatman Bond · 29th May 2009, 10:17 AM

Quote:

Originally Posted by MikeMl

The ratio of R3 to R2 determines at what minimum input voltage the LED turns on. Your values didn't work (the NPN never fully turned on).

Look at my values:

That you mean, my value?

It is work in real (not in simulation).
In your simulation it is turned-on when input voltage only 2V. When input reach 12V it is reach somewhat ~2V hard saturation. My value is ~1volt Vbe normal saturation (enough for driving the LED).

This also my value for 5V input: 2K2//100pf Rb+C and 680 Rbe for fast ; 4K7 Rb and 1K Rbe for slow speed.
Try it with simulation then, might be wont work, but it is work in real.

Blatman Bond · 29th May 2009, 10:27 AM

Also don't forget to look manufacture graphs about saturation characteristics (might be included in other transistor spec that specially for switch).
Normally they used β=10 for saturation value.

paul_ma7 · 29th May 2009, 09:48 PM

Quote:

Originally Posted by krich

I'm fairly inexperienced with electronics in general, but I've learned quite a bit over the last many months of self instruction (and some really good help from forums like these). I have one question on transistors whose answer has eluded my normally excellent googling skills, so I figure I would just throw out the question on this forum in hopes of getting an answer.

With a regular 2n3904 transistor, what happens if my base voltage is significantly higher than the voltage that I am switching on the Collector/Emitter? 12V on the base side, 5V on the collector/emitter side.

I've looked at a few javascript simulators I've found online and it says my Ve=11.3V (allowing for the forward voltage drop b to e.). Is this correct? I was worried that the simulators might not handle this "reverse scenario" correctly.

Here's an example scenario:

I have a garage opener button in my garage just outside the back door. It operates on 12V. Close the circuit by mashing the button and the door opens/stops/closes.

I would like to hook up a MCU (Atmel) running at 5V to the button to detect when the button is pressed. All the inputs are max 5V, so I was thinking of using a transistor as a switch. But I worry I'll send ~12V to my MCU inputs, which is grossly out of spec and generally a bad thing.

So, is a transistor just the wrong part for the job if my base voltage is higher? I really want to understand when I can and when I can't use a transistor for such things. What about PNP?

I know there are other solutions like a relay or a solid state relay/optoisolator, which would protect the 5V side from the 12V side. This is more about understanding how a transistor works than figuring out this particular example. I've run into this situation on more than one occasion.

--------

your transistor will switch without any problems,
relays will switch high voltages (125ac and above),
the maximum voltage a transistor can switch is in its datasheet, also the maximum wattage,
what you need to be concerned is that the current at the base of the transistor is small, how small?
everything it's state in the datasheet

audioguru · 29th May 2009, 11:39 PM

A 2N3904 transistor is an NPN type. When its emitter is at 0V then its base can never be higher than about 1.0V because the base-emitter is like a diode. The MCU applies up to 1V at a low current to the base to turn on the transistor.

A 2N3906 transistor is a PNP type. When its emitter is at +12V then its base can never be less than 11.0V. Because the base is at a voltage that is much higher than the 5V for an MCU then a special circuit is needed in between.

The collector's load voltage might go to the base of a transistor only when its load and the transistor blow up.

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26th May 2009, 07:21 PM	#17
Sceadwian	Why calculate it out? There's a traditional rule that the base current should be 10 times the collector emitter current to act as a switch, special cases aside, go by that general rule and you'll skip about 2 years of math. __________________ "Because I be what I be. I would tell you what you want to know if I could, mum, but I be a cat, and no cat anywhere ever gave anyone a straight answer, har har." Last edited by Sceadwian; 26th May 2009 at 07:21 PM.

26th May 2009, 07:31 PM	#18
audioguru	Some transistors like the BC547 have their saturation voltage loss rated when the base current is 1/20th of the collector current. __________________ Uncle $crooge

26th May 2009, 08:00 PM	#19
Sceadwian	It's very nice when it's specifically stated like that. Let you relax design requirements with assurance. I'm not an expert mind you, but everywhere I have read 1/10th is the rule of thumb, basically what you have to do is under all thermal and voltage conditions guarantee that the transistor will never come out of saturation and go into it's linear region or you'll get thermal runaway and transistor self destruction. I've heard little of safe operating area conditions but don't know much about them or how they're calculated. __________________ "Because I be what I be. I would tell you what you want to know if I could, mum, but I be a cat, and no cat anywhere ever gave anyone a straight answer, har har." Last edited by Sceadwian; 26th May 2009 at 08:03 PM.

26th May 2009, 08:45 PM	#22
Sceadwian	hrmm mmrmmrm m fff mmhrr mfmmffr! __________________ "Because I be what I be. I would tell you what you want to know if I could, mum, but I be a cat, and no cat anywhere ever gave anyone a straight answer, har har."

29th May 2009, 10:27 AM	#24
Blatman Bond	Also don't forget to look manufacture graphs about saturation characteristics (might be included in other transistor spec that specially for switch). Normally they used β=10 for saturation value. Last edited by Blatman Bond; 29th May 2009 at 10:33 AM.

29th May 2009, 11:39 PM	#26
audioguru	A 2N3904 transistor is an NPN type. When its emitter is at 0V then its base can never be higher than about 1.0V because the base-emitter is like a diode. The MCU applies up to 1V at a low current to the base to turn on the transistor. A 2N3906 transistor is a PNP type. When its emitter is at +12V then its base can never be less than 11.0V. Because the base is at a voltage that is much higher than the 5V for an MCU then a special circuit is needed in between. The collector's load voltage might go to the base of a transistor only when its load and the transistor blow up. __________________ Uncle $crooge