What does mean this?: The BJT doesn't work because the emitter output HAS to be lower than the base

Thread Starter

booboo

Joined Apr 25, 2015
168
Hi fellas:)
A couple weeks ago I saw this question. today I remember that I had a question about this question:

http://electronics.stackexchange.com/questions/179065/npn-transistor-for-3-volt-base-to-switch-control-5-volts

The reason that Andy pointed out is:
"The reason why a single NPN emitter follower doesn't work is because the emitter output HAS to be lower than the base (3V) to get any current passing from collector to emitter. So the emitter might be 2.5 V when base is at 3V and collector is 5V (or greater)"

I cannot get it.:(
What does mean "the emitter output HAS to be lower than the base"?
Would you please explain it to me?
 

pwdixon

Joined Oct 11, 2012
488
What it is trying to say is that the base voltage has to be higher than the emitter voltage by Vbe to turn on the transistor, I guess logically that does mean that the emitter voltage has to be lower than the base voltage therefore, it's just an odd way to say it.

The problem is that to output 5V you need the transistor hard-on but that would require the base to be 5V+Vbe and that's not good when the collector is only 5V, this is why people suggest using an PNP in this location instead.
 

Thread Starter

booboo

Joined Apr 25, 2015
168
Thanks for reply
Maybe I'm a bit confused. Yes, I know Vbe is 0.7v but what's the relation between emmiter and base? if we change the circuit to this, will this work?:



Hi ericgibbs
Thanks for reply.
In fact that was just a question to me. I'm trying to know what's going on in a transistor.:)
another question:
Could we just connect a PNP to the pin of the first MCU(3.3v) to turn on or off another one(5v MCU)?
 

korchoi

Joined Jun 5, 2015
59
What do you mean PIC?
and why that's a bad idea?
By PIC, he meant The typical brand of microcontrollers that appears out there, much like you call conserved meat "spam", or that noisy motorcycle "ducati", or that tool "kleenex".
It's a bad idea because there is a voltage drop on the transistor. it will not be a short to ground. More like a forward biased diode to ground.
The drop goes to about 0.3V. For circuits running at their higher speed limits, this drop can mean work or broke. It's better to use a mechanical relay for that(it is literally two metal pieces that touch when a coil is activated). Also some other reason i forgot and only eric can answer.
 

ericgibbs

Joined Jan 29, 2010
9,574
PIC is a MCU type.

Consider what would happen to all the pins of the MCU/PIC, they would 'float' above 0V and the internal circuit of the MCU would try to pass current from the +5V supply pin thru any external circuit connected to these pins.
 

korchoi

Joined Jun 5, 2015
59
PIC is a MCU type.

Consider what would happen to all the pins of the MCU/PIC, they would 'float' above 0V and the internal circuit of the MCU would try to pass current from the +5V supply pin thru any external circuit connected to these pins.
Wouldn't all the microcontroller pins just Become open terminals? Pins leading to the air? Any design using that method should account for the high-Z state of a disabled micro, maybe implied on code or sensed through another circuit.
 

Thread Starter

booboo

Joined Apr 25, 2015
168
By PIC, he meant The typical brand of microcontrollers that appears out there, much like you call conserved meat "spam", or that noisy motorcycle "ducati", or that tool "kleenex".
Ok. I know PIC is a type of MCUs. I thought you're talking about a new phrase/idiom in EE.;) I don't know why eric tought I'm using a PIC. That's not my question. I asked it because I would like to know the reason of it.
It's a bad idea because there is a voltage drop on the transistor. it will not be a short to ground. More like a forward biased diode to ground.
The drop goes to about 0.3V. For circuits running at their higher speed limits, this drop can mean work or broke. It's better to use a mechanical relay for that(it is literally two metal pieces that touch when a coil is activated). Also some other reason i forgot and only eric can answer.
And how can a drop about 0.3v cause problem?:D

Consider what would happen to all the pins of the MCU/PIC, they would 'float' above 0V and the internal circuit of the MCU would try to pass current from the +5V supply pin thru any external circuit connected to these pins.
WoW!
 

korchoi

Joined Jun 5, 2015
59
Ok. I know PIC is a type of MCUs. I thought you're talking about a new phrase/idiom in EE.;) I don't know why eric tought I'm using a PIC. That's not my question. I asked it because I would like to know the reason of it.


And how can a drop about 0.3v cause problem?:D


WoW!
0.3V may be small, but you never know...
some microcontrollers are very picky on the supply voltage.
 

djsfantasi

Joined Apr 11, 2010
6,196
Just a question. Why are you switching power to the MCU? Why not send the output of the level shifter to a digital pin. Then code it so the functions of the 5.5V MCU only execute when the pin is HIGH?
 

ErnieM

Joined Apr 24, 2011
8,019
Maybe I'm a bit confused. Yes, I know Vbe is 0.7v but what's the relation between emmiter and base?
Yes, you are confused. Vbe *is* the relation between the emitter and the base.

The 'e' in Vbe stands for "emitter." The 'b' stands for "base." 'V' is for "voltage."

Vbe is about 0.7V when the device is on. Anything much less and it is off. Anything much more and the magic smoke comes out.
 

pwdixon

Joined Oct 11, 2012
488
Yes, you are confused. Vbe *is* the relation between the emitter and the base.

The 'e' in Vbe stands for "emitter." The 'b' stands for "base." 'V' is for "voltage."

Vbe is about 0.7V when the device is on. Anything much less and it is off. Anything much more and the magic smoke comes out.
I'd like to bet the OP will now say "I know that"!
 

Thread Starter

booboo

Joined Apr 25, 2015
168
Just a question. Why are you switching power to the MCU? Why not send the output of the level shifter to a digital pin. Then code it so the functions of the 5.5V MCU only execute when the pin is HIGH?
Maybe that guy would save the energy.:rolleyes:
Yes, you are confused. Vbe *is* the relation between the emitter and the base.

The 'e' in Vbe stands for "emitter." The 'b' stands for "base." 'V' is for "voltage."

Vbe is about 0.7V when the device is on. Anything much less and it is off. Anything much more and the magic smoke comes out.
Let me to explain what I know about BJT and why I'm confused:
look at this picture:



As you can see a transistor is like a channel. when you open the basis gate, the cellector gate would open subsequently and how much you can open the cellector depends on the basis gate. no matter from where those water would come. BUT this is the point! we need 0.7v to open the cellector. no matter from where it would come. either 3.3v or 5v! but in this sentence Andy said "the emitter output HAS to be lower than the base (3V) to get any current passing from collector to emitter. So the emitter might be 2.5 V when base is at 3V and collector is 5V"
Base is 3v?o_O

I'd like to bet the OP will now say "I know that"!
You beat me pwdixon.:D
 

Jony130

Joined Feb 17, 2009
5,145
To open (turn it on) NPN transistor the voltage at base must by 0.5V...0.6V higher than the emitter voltage.
Also base current must flow into the base
 

gerty

Joined Aug 30, 2007
1,273
If the base and emitter were the same in your water analogy, when the base gate opened water would also flow up from the emitter, leaving nowhere for the collector water to flow
 

MikeML

Joined Oct 2, 2009
5,444
Booboo:

When you understand this simulation, you will have a good understanding of NPN and PNP transistors and how they are commonly used:

Note that the drive voltage V(in) (?from a microprocessor port pin, perhaps?) is the same for all four cases. Compare the four outputs.

Why do two of the cases "follow" the input voltage? Why does V(Nf) stop following V(in) when it reaches 4.2V? Why does V(Pf) stop following V(in) when it reaches 0.8V?

Why do two of the cases "square up" (amplify?) the input voltage? What is the switching threshold for these two cases?

252.gif
 
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