I'm using a 222npn and bd233 npn along with bd234 pnp
all of these transistors say 5 volt max on the base via the datasheets.
but in some other research I read that its more current based.
So if it is current based, couldn't I apply 12v to the base as long as I use the correct resistor?
I'm working on a small 12volt logic gate with switching and I'm aiming for more plug and play instead of using regulators and a lot of extra things that may not me necessary.
Thanks.
The ‘reverse breakdown’ ‘Emitter Base Voltage’ VEBO is 5v, but 99.9999% of the time you will never use it in this mode.
so you're saying there's a chance hahaha
just playin
but bottom line is that I can apply 12v to the base as long as current is controlled?
You can't let the base drop more than 5V below the emitter.
If the input is 0v to 12v to the base resistor, there is zero chance of this happening.
This might be a problem in some weird bipolar situation but you will never run into this.
BTW
I have used this characteristic to create a 5v zener diode.
"maybe a really dumb transistor question..."
Nah, not asking questions is dumb.
Answers might be dumb though.
Base-Emitter acts as a diode, with a ~0.7volt forward threshold and a max reverse (breakdown) voltage.
So base of an NPN tranasistor can only be about 0.7volt higher than the emitter, because of that BE diode.
Then current starts to flow, which usually has to be limited by a resistor.
The base can also go more negative than the emitter, but has to be limited to ~5volt.
Leo..
Thank you very much.
And that seems like a cool trick to keep in ones back pocket.
It'll be a while before I get to doing things like that.
here is my schematic... might be hard to make out.
all of these transistors say 5 volt max on the base via the datasheets.
So if it is current based, couldn't I apply 12v to the base as long as I use the correct resistor?
Because the base-emitter forward voltage drop is a diode voltage drop (0.5 - 0.8V) because
a transistor consists of diode like junctions. (we can skip the details for now)
So if the forward voltage is 0.7V (typ) and as you already know , it is CURRENT based , and the
5V you referred to is actually the Breakdown Voltage (BVEBO) (the voltage at which the transistor junction is destroyed), and there is NO resistor if for example , you apply 5V to it without a resistor, the the base junction will conduct as much current as your 5V supply can deliver (800+ mA in the
case of an arduino) and the base junction will fry.
So what should you do if , say, you wanted to drive the transistor with 5V ?
The 2N2222 collector current max is 600 mA and hfe 300 so if for example your load
was 500mA (which is really pushing it for this transistor) then the base current should be:
b = Ic/hfe
= 500mA (0.5A)/300
= 1.66 mA (0.00166A)
and the base resistor Rbase = (Vin - 0.7V)/Ibase
= 5V-0.7V/1.66 mA (0.00166A)
= 4.3V/0.00166A
= 2590 ohms (2.4k ohms to 2.6k)
hfe is the transistor small signal gain, which is generally a spec used for small signal
applications where you are not using saturating the transistor as you would driving a relay .
If you want to use it as a driver instead of amplifier you would want to saturated it and would
look at the base=emitter saturation voltage, which for this transistor is specified for a collector
current of 500 mA and a base current of 50mA, in which case the base resistor would be:
the base resistor Rbase = (Vin - 0.7V)/Ibase
= 5V-0.7V/50 mA (0.05A)
= 4.3V/0.05A
= 86 ohms (91 ohms)
Now if instead of 5V, you wanted to drive it with 12V, then you would simply
follow the above but use 12V for Vin , yielding :
Ibase = (VIN-0.7v)/Rbase
or if your calculating the resistor value,
Rbase = (VIN-0.7v)/Ibase
so, for 12V,
for a base current of 2mA,
Rbase = (VIN-0.7v)/0.002A
= (11.3V-0.7V)/0.002A
= 5650 ohms
or if using it as a relay driver:
the base resistor Rbase = (Vin - 0.7V)/Ibase
= 12V-0.7V/50 mA (0.05A)
= 11.3V/0.05A
= 226 ohms (220 ohms)
My guess is that you want to use the transistor as a driver and therefore want to saturate it
IbaseSat = 10 times the small signal current , so depending on the collector current
probably 20 to 40 mA
raschemmel:
and the 5V you referred to is actually the Breakdown Voltage...
(the voltage at which the transistor junction is destroyed)
I think you understood that part wrong.
It's the reverse voltage you can put between BE, without resistor, and without anything happening.
Imagine that the BE junction is a ~5.6volt zener diode.
A ~0.65volt diode one way and a ~5.6volt zener diode the other way.
Leo..
I think you understood that part wrong.
It's the reverse voltage you can put between BE, without resistor, and without anything happening.Imagine that the BE junction is a ~5.6volt zener diode.
A ~0.65volt diode one way and a ~5.6volt zener diode the other way.
The OP said:
all of these transistors say 5 volt max on the base via the datasheets.
but in some other research I read that its more current based.
So if it is current based, couldn't I apply 12v to the base as long as I use the correct resistor?
I'm working on a small 12volt logic gate with switching and I'm aiming for more plug and play instead of using regulators and a lot of extra things that may not me necessary.
How does
5 volt max on the base
mean -5V on the base ?
And what does this have to do with applying 12V to the base ?
Reverse Breakdown Voltage means the base is 5V less than the emitter. (hence the 'reverse' part),
but why would the OP be asking this question if his intention is to use 12V on the base (which, as
you know is not going to happen because if there a resistor the voltage is going to be 12V-0.7V = 11.3V and the base voltage is still going to be 0.7V) ?
(point being there is never going to 5V or 12V on the base because it is going to be 0.7V unless the
OP turns the transistor upside down and grounds the collector and puts 5V on the emitter, but if
that's what he had in mind, why would he be asking about putting 12V on the base ?
Don't you need to know the transistor configuration before discussing any of this ?
The OP did not understand this value is the voltage measured emitter to base (not base to emitter) with the collector open.
Must have assumed (in error) the data sheet was talking about the forward biased voltage VBE.
BTW
Most silicon BJTs have a value for VEBO of about 5-6v.
I think there's a whole lot of 'assuming' going on...
The OP did not understand this value is the voltage measured emitter to base (not base to emitter) with the collector open.
That speaks to the OP's research (or lack thereof) on transistors. (via Google, which seems to know a lot about them (About 150,000,000 results (0.59 seconds) )
(we actually don't know what the OP was thinking or what he was looking at on the datasheet)
but in some other research I read
He does seem to have done some research, on what, that remains to be seen...
Assume
Makes an ‘ass’ out of ‘u’ and ‘me’.
I just came back from a 7 day holiday so I didn't want to push my luck saying that...
For more clarification....
I was asking if I wanted to switch 12v from collector emitter, am I limited to only being able to apply 5v on the base.
So I guess the issue was with my “hobbyist” education of transistors and the lack of complete understanding of data sheets.
Thank you all.
No problem, have fun!
Read Reply #8
LandonW:
I'm using a 222npn and bd233 npn along with bd234 pnpall of these transistors say 5 volt max on the base via the datasheets.
but in some other research I read that its more current based.
So if it is current based, couldn't I apply 12v to the base as long as I use the correct resistor?
I'm working on a small 12volt logic gate with switching and I'm aiming for more plug and play instead of using regulators and a lot of extra things that may not me necessary.
Thanks.
Yes. You need to know that the base-emitter voltage to turn the transistor on is about 0.7 volts.
Then, determine the base current you want. Let's say you want 20 milliamps.
So, R=E/I, R=(12-0.7)/0.020, R=565 ohms, closest standard value is 560 ohms.
So, 12 volts through a 560 ohm resistor will give you approximately 20 milliamps base current.
Make sense?
P.S. no question is "dumb". That's how we all learn.
