I am a transistor-noob with a question about saturation mode so it is suitable to post it here, please if anyone can enlighten me.

I understand as we increase the base voltage above ~0.7V at some point BJT will enter saturation, that is, voltage across the device will drop to it's minimum (like short circuit). In this mode voltage between the terminals is supposed to be E < B > C. Now, let's say base is connected to a +1V and E and C to -/+400V respectively, how can B become more positive than C?? Im obviously making some wrong assumption, anyone please?

 

say base is connected to a +1V and E and C to -/+400V

hi N,
Welcome to AAC.
Do you really mean -/+400v.?

E

 

Hi and yes, i seen there are BJTs available with Vceo up to 1200V.
I know Vceo stands for open circuit but im not sure of full meaning.
Is this where i am mistaken? Please explain.

 

hi,
For a regular BJT , applying a positive Vbe voltage would suggest that the transistor is a NPN type, so you would not connect a negative voltage of say
-400V to its Collector.??

E

 

No, i meant -400V to emitter. Is wrong to apply such high voltage to emitter and collector while base is not an open circuit? I want to understand the meaning of these voltage relations between active mode E < B < C and saturation mode E < B > C. Please explain.

 

hi N,
If you want to understand the BJT operation, I would recommend that you watch online video's which cover the basics of transistors.
This one is an example link.
After watching, come back with questions.

E

 

+400V applied to Collector through resistor 400k.
When you put +5V to Base through resistor 1k, you will have about +0.7V on B-E and +0.2V on C-E.
Saturation.

 

Thank you Danko, that was very helpful.

 

let's say base is connected to a +1V and E and C to -/+400V

If you put +1V on the base and -400V on the emitter, the base-emitter junction will instantly vaporise .

 

let's say base is connected to a +1V and E and C to -/+400V respectively

Normally when only one power source then we will use +V and 0V(+V/0V) to represent the power, when the power source including positive and negative then we will use +V,0V,-V to represent the power.

When you using +/-400V then we will think those are +400V,0V,400V, there are many newbies of ee always confusing and used the +V/-V to represent +V/0V, mostly when you using +V,0V,-V then you will use as two NPN or two PNP, or one NPN and one PNP, so when you just use one NPN or PNP then it is almost one power source as +V/0V.

No, i meant -400V to emitter. Is wrong to apply such high voltage to emitter and collector while base is not an open circuit? I want to understand the meaning of these voltage relations between active mode E < B < C and saturation mode E < B > C. .

When you use the +/-V power source and one BJT then it could be like the circuit below.

 

Another usefull rule of thumb, if you want to sat a bipolar transistor
force the beta to 10 by driving its base with Ib >= Ic/10.

Also, some transistor processes, inverting collector and emitter leads in circuit
will result in lower Vcesat.

Regards, Dana.

 

when you just use one NPN or PNP then it is almost one power source as +V/0V

Please elaborate on this, why is it ALMOST (as) one power source +V/0V?

 

Please elaborate on this, why is it ALMOST (as) one power source +V/0V?

ALMOST -- I meant that the most applications, you can say that it is more popular.
One transistor one power source -- Simple amplifier using one transistor.
One transistor dual power sources -- Differential Transistor Amplifiers from AAC, Differential Transistor Amplifiers from google.
---- The circuit that I posted was I saw it used in the high voltage exceed 2KV of O'Scope.
Two transistors one power source -- High side and low side, OTL Amplifier, Transistor motor bridge driver
Two transistor dual power sources -- High side and low side, OCL Amplifier.
Bipolar Junction Transistors (BJTs) -- There are two kinds of power source which including one and dual power sources, they have formulas to show how to calculate the Ic, Ib, Ie, Rc, Re, when you use this kind of circuit then you must be very carefully, otherwise it may damaged the bjt easily.

Since you said that you are a transistor-noob, I'm curious where did you get that issue(+/-400V) and the ee beginners always concentrated on the low voltage and the very basic circuit and use one power source in the circuit, because it is more difficulty with newbies, especially the one bjt circuit using dual power sources.

 

Since you said that you are a transistor-noob, I'm curious where did you get that issue(+/-400V) and the ee beginners always concentrated on the low voltage and the very basic circuit and use one power source in the circuit, because it is more difficulty with newbies, especially the one bjt circuit using dual power sources.

I chose that voltage as i need similar, bit higher voltage for my pulsed DC project. I plan to use a BiCMOS like this one. That is actually my main question, how to control this BiCMOS as a switch with a function generator. BiCMOS is MOSFET controlled so to act as a switch it needs to be in linear mode, right. Datasheet says turn on voltage is 10V and VGES is +-20V. So, to make it conduct, is it enough to pulse the gate with +10V (square wave), or do i need additional components? I don't know the values for the load right now.

 

I chose that voltage as i need similar, bit higher voltage for my pulsed DC project. I plan to use a BiCMOS like this one. That is actually my main question, how to control this BiCMOS as a switch with a function generator. BiCMOS is MOSFET controlled so to act as a switch it needs to be in linear mode, right. Datasheet says turn on voltage is 10V and VGES is +-20V. So, to make it conduct, is it enough to pulse the gate with +10V (square wave), or do i need additional components? I don't know the values for the load right now.

There is an IGBT control circuit in this site, the e of IXBH 9N140G is the common pin, you could try it.

 

See
I took the transistor model as it is, and did not adjust the parameters.

 

How good a switch are you seeking ? Eg. when on, what is the volatge drop
acoss it that you would like. This BiCMOS part at 5A has a Vcesat of 7V at
room temp, which as a switch is pretty awful.

That being said the Vcesat is speced at a Vge of 15V, so thats the pulse you
would want. To function as a switch you DO NOT want it to spend much time
in the linear region. Either on or off, thats to minimize Pdiss.The function generator
is 50 ohm (typically) and it has to drive the gate capacitance, which is 550 pF.
Gate drivers exhibit (usually) less that 1 ohm, can supply high current to charge
the gate C. But to debug a function generator fine.

Driving MOSFETs ref material -

https://www.google.com/url?sa=t&rct...57444e913f4f&usg=AOvVaw30EmxL7qVIhaoOEA610FJ-

https://www.google.com/url?sa=t&rct.../an-1067.pdf&usg=AOvVaw3X_sY-zOS-U_xBLG8QGMmZ


Regards, Dana.

 

Thank you ScottWang and Bordodynov.

Eg. when on, what is the volatge drop acoss it that you would like.

I don't know. As far as i understand it should be as low as can be in ON-mode.

This BiCMOS part at 5A has a Vcesat of 7V at room temp, which as a switch is pretty awful.

Why does Vcesat of 7V at 5A make it a bad switch? Is it heating or? Maybe you could suggest some better BiCMOS or MOSFET to switch high voltage in megahertz range.

 

BiCMOS is MOSFET controlled so to act as a switch it needs to be in linear mode,

It should be act as saturation mode and it only have OF/OFF states there is no others.

You haven't mention that what is the load and how much current will it draw, and why you need dual power sources?