Need some guidance on transistor workings and reading datasheets

Thread Starter

Aleksey Shurtygin

Joined Dec 21, 2018
54
Hi! I am relative beginner in electronics and want to try to understand transistors more. I do understand how it works basically but wanted to get some more practical knowledge.

So, as a challenge to myself I decided to try to build a simple "voltage" amplifier circuit. The idea would be that by making small voltage adjustments on the base pin I would expect to get full (or close enough) range on the collector or emitter. For example by varying voltage from 0-0.6 V on base pin I would get 0 to 5V output voltage (assuming I am using 5V power source).

I am currently a bit stuck on trying to read and understand datasheets. For example https://www.mouser.com/ds/2/149/BC550-888526.pdf

One question that is a bit unclear is that if transistor becomes like a closed switch when you apply some specific voltage to the base pin why does it have some collector-emitter saturation voltage - Vce(sat)? My only guess its the voltage drop when transistor is in saturation mode and it still may have some resistance. Is that what it is? If that is so, would not source voltage affect those values?

Second thing is the base-emitter voltages (on and saturation). On voltage sounds like when transistor starts conducting current on collector-emitter. Anything below will make it "off". If that is true then how come saturation voltage is pretty much the same. Does it mean these types of transistors will make a bad amplifier and may be really be used as a switch mostly?

Can someone comment whether my thinking is correct or makes sense or where I am getting a bit off?
Also, it would be nice to know if my so-called challenge is doable at all?

I have attached the basic schematics I am thinking that may work.
 

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Last edited:

Zeeus

Joined Apr 17, 2019
422
Hi! I am relative beginner in electronics and want to try to really understand transistors. I do understand how it works basically but wanted to get some more in-depth knowledge.

So, as a challenge to myself I decided to try to build a simple "voltage" amplifier circuit. The idea would be that by making small voltage adjustments on the base pin I would expect to get full (or close enough) range on the collector or emitter. For example by varying voltage from 0-0.6 V on base pin I would get 0 to 5V output voltage (assuming I am using 5V power source).

I am currently a bit stuck on trying to read and understand datasheets. For example https://www.mouser.com/ds/2/149/BC550-888526.pdf



Can someone comment whether my thinking is correct or makes sense or where I am getting a bit off?
.
Maybe read a little about transistors from a book or on this site first
 

dl324

Joined Mar 30, 2015
8,918
One question that is a bit unclear is that if transistor becomes like a closed switch when you apply some specific voltage to the base pin why does it have some collector-emitter saturation voltage - Vce(sat)? My only guess its the voltage drop when transistor is in saturation mode and it still may have some resistance. Is that what it is? If that is so, would not source voltage affect those values?
Only an ideal transistor would have a Vce(sat)=0V. The voltage depends a number of parameters; with base current and collector current being most important.

Second thing is the base-emitter voltages (on and saturation). On voltage sounds like when transistor starts conducting current on collector-emitter. Anything below will make it "off". If that is true then how come saturation voltage is pretty much the same. Does it mean these types of transistors will make a bad amplifier and may be really be used as a switch mostly?
No. When you're using a transistor as a switch, you're operating in the cutoff and saturation modes. For active mode, you don't want the transistor to saturate. The fourth mode (inverted), is seldom used as the transistor will have very low current gain in that configuration.

I have attached the basic schematics I am thinking that may work.
You don't show where you're taking the output from.

A simple experiment would be to take the output from the emitter and vary the voltage on the base. The voltage should track the base voltage minus the BE voltage drop.
 

Thread Starter

Aleksey Shurtygin

Joined Dec 21, 2018
54
Maybe read a little about transistors from a book or on this site first
I did read a lot about it online including wikipedia, this site, learn.sparkfun.com and few others. None really explain what I do not understand or at least not in clear or obvious way.
 

DickCappels

Joined Aug 21, 2008
5,882
The chapter on transistors is located at https://www.allaboutcircuits.com/textbook/semiconductors/chpt-2/bipolar-junction-transistors/

The reason there is a collector-emitter saturation voltage is that even though the transistor is ready to conduct the maximum amount of current, it still has resistances.
You wrote:
Second thing is the base-emitter voltages (on and saturation). On voltage sounds like when transistor starts conducting current on collector-emitter. Anything below will make it "off". If that is true then how come saturation voltage is pretty much the same. Does it mean these types of transistors will make a bad amplifier and may be really be used as a switch mostly?

Don't know about that but what you are seeing is IR drop across the bulk emitter resistance (RE) It is a form of negative feedback that limits the maximum gain out of one stage.Whether this makes for worse or better switching characteristics I don't know. You could compare those parameters between switching and amplifying transistors.

Your test circuit certainly will work, but you will get a lot more knowledge using it if you move R1 from being in series with the pot to being in series with the base and just connect your pat across the 5V supply. This is because looking at collector current vs base current gives more predictable results than comparing voltages. The view that transistors are current operated devices is the conventional view and is a lot more straight forward than the idea that it is a transconductance amplifier (change of input voltage in vs change in output current).

 

BobTPH

Joined Jun 5, 2013
2,024
You cannot nake a useful DC amplfier with a single transistor, certainly not one that can output 0 to 5V with a 5V supply.

Look at the circuit of an op amp to see what it takes to make a DC amplifier.

An AC amplifier is actually much easier and a single transistor can do a reasonable job, but again, not with an output range equal to the supply voltage.

Bob
 

Thread Starter

Aleksey Shurtygin

Joined Dec 21, 2018
54
Thanks everyone for all replies. I really do appreciate it.

Only an ideal transistor would have a Vce(sat)=0V. The voltage depends a number of parameters; with base current and collector current being most important.

No. When you're using a transistor as a switch, you're operating in the cutoff and saturation modes. For active mode, you don't want the transistor to saturate. The fourth mode (inverted), is seldom used as the transistor will have very low current gain in that configuration.

You don't show where you're taking the output from.

A simple experiment would be to take the output from the emitter and vary the voltage on the base. The voltage should track the base voltage minus the BE voltage drop.
Thanks! At this point I am not concerned about where to take an output from, be that collector or emitter. I do understand that if I'd be taking output from collector, voltage would be inverted. Higher base current/voltage - higher collector current - lower collector voltage. And reverse if I'd be taking output from emitter.

On the simple experiment. Wouldn't collector current (when voltage applied to collector) affect the emitter voltage? If so, then it would not be "base voltage minus the BE voltage drop", would it? Or there are some assumptions at play?

These posts look interesting. I'll read up on them. Thanks! I appreciate it!

Your test circuit certainly will work, but you will get a lot more knowledge using it if you move R1 from being in series with the pot to being in series with the base and just connect your pat across the 5V supply. This is because looking at collector current vs base current gives more predictable results than comparing voltages. The view that transistors are current operated devices is the conventional view and is a lot more straight forward than the idea that it is a transconductance amplifier (change of input voltage in vs change in output current).
Yes, I've seen those examples as well and I did build one of those circuits. I made up my design under assumption that voltage would govern how much transistor is open and was looking at it as a simple voltage divider. This may be incorrect way to look at it. In one of the posts that @Jony130 listed someone mentioned to think more in terms of current and not voltage when dealing with transistors. So, I will read up some more based on other responses and see if it will make more sense to me then.

Thanks! These cover mostly datasheets on a high level and do not get into specifics on transistors. I do know what those different sections mean and what they generally are for but I am having problems applying those transistor specific values in practice. This however gave me an idea to search for pages on datasheets specific to transistors. Hopefully I can dig up something useful.

Download the pdf version of this, very handy.

https://archive.org/details/GE_-_Transistor_Manual_1964]


Regards, Dana.
Thanks Dana! This does seem to be handy. Only down side it maybe a bit too much for my level. But I will take some time to look at it and see if it will make sense to me.

You cannot nake a useful DC amplfier with a single transistor, certainly not one that can output 0 to 5V with a 5V supply.

Look at the circuit of an op amp to see what it takes to make a DC amplifier.

An AC amplifier is actually much easier and a single transistor can do a reasonable job, but again, not with an output range equal to the supply voltage.

Bob
Thanks Bob! I am not really looking for anything useful at this point and definitely not building anything specific. As long as it is very simple and helps me understand transistor better I am perfectly fine with whatever it is.
 

dl324

Joined Mar 30, 2015
8,918
On the simple experiment. Wouldn't collector current (when voltage applied to collector) affect the emitter voltage? If so, then it would not be "base voltage minus the BE voltage drop", would it? Or there are some assumptions at play?
You didn't give values for R2-4, so I'd've made R4 sufficiently small to not affect the emitter voltage for the desired voltage range. R2 would be large enough to give the desired range, and R3 would be chosen to limit the current to a reasonable range for the transistor, which was also unspecified.
 

Thread Starter

Aleksey Shurtygin

Joined Dec 21, 2018
54
You didn't give values for R2-4, so I'd've made R4 sufficiently small to not affect the emitter voltage for the desired voltage range. R2 would be large enough to give the desired range, and R3 would be chosen to limit the current to a reasonable range for the transistor, which was also unspecified.
My bad! This schematic was meant to be only as a pseudo-schematic with no actual R values. I've put it together on the phone and for some reason could not clear resistance for that particular one. I forgot to clarify this.

Also as suggested earlier maybe it would be better to add a base resistor. So its quite likely schematic may not even be a complete one.
 

Thread Starter

Aleksey Shurtygin

Joined Dec 21, 2018
54
Hmm, let me switch gears a bit. Let me know if my thinking is correct.

So, lets assume I have a circuit with a transistor having collector connected to positive thru a resistor and emitter connected directly to ground and I have a 5V power source connected.

According to the datasheet Vce(sat) at 10mA is typically 90mV. To calculate collector resistor that acts as a current limiter I can neglect the collector-emitter resistance. This would give me resistor of about 500 Ohms (5V/0.01A). So, if transistor is fully saturated I can expect about 90 mV on the collector. But if transistor is in the cut-off mode the voltage I am going to get will be a bit under 5V.

However with higher collector current I will get higher voltage drop. If I put 50 Ohm resistor to get 100 mA I can expect 250 mV on collector in fully saturated mode.

Does what I say make sense?
 

Zeeus

Joined Apr 17, 2019
422
Connect 5v directly to the base without resistor?

Please why 250mv ?..... Wish me was home *-#)
 

Thread Starter

Aleksey Shurtygin

Joined Dec 21, 2018
54
Connect 5v directly to the base without resistor?

Please why 250mv ?..... Wish me was home *-#)
Obviously connecting base directly to +5V will probably fry it in this configuration or at least will be dangerously close to that. But for the sake of the example how one would connect base is irrelevant. Unless that may dramatically change the picture. The point is that transistor is in saturated or cut-off mode.

250mV is Vce(sat) with a condition of Ic=100 mA from BC550 datasheet.
 
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DickCappels

Joined Aug 21, 2008
5,882
The graph below should make it clear why transistor gain is usually expressed in amps out/ amps in instead of amps out/ volts in.
upload_2019-5-5_13-35-24.png

(300K is about 80°F)
Salient points:
- As base current increases the base-emitter voltage approaches being flat at some low voltage.
-For a given base current, the base voltage varies at about 2 mv/degree C (yes this can be made to measure temperature).

That's why you need a base resistor in a common-emitter circuit in analog applications.
 

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Jony130

Joined Feb 17, 2009
4,975
I measured the Vbe (base-emitter voltage) vs Ib (base current) for BC337-40
The power supply voltage was Vcc = 10V

RB = 680kΩ....Vbe = 0.614V....Ib = 13.8µA

RB = 470kΩ....Vbe = 0.616V....Ib = 20µA

RB = 220kΩ....Vbe = 0.624V....Ib = 42.61µA

RB = 100kΩ....Vbe = 0.639V....Ib = 93.61µA

RB = 50kΩ......Vbe = 0.659V....Ib = 187µA

RB = 10kΩ......Vbe = 0.719V....Ib = 928µA

RB = 5kΩ........Vbe = 0.748V....Ib = 1.85mA

RB = 2kΩ........Vbe = 0.787V....Ib = 4.6mA

RB = 1kΩ........Vbe = 0.819V....Ib = 9.18mA

RB = 500Ω......Vbe = 0.856V....Ib = 18.29mA

RB = 200Ω......Vbe = 0.989V....Ib = 45mA

As you can see the base current change 45mA/13.8μA = 3260 times.
But the Vbe change only by 375mV.
This very small change in Vbe compared with a huge change in the base current justified the assumption that we use for Vbe voltage (Vbe = 0.6...0.7V) in hand calculations.

In reality the Vbe will go up by a 60mV or down by 60mV for every upward /down change in Ic by a factor of ten (decade change).

For example if Vbe is 0.62V for Ic = 1mA if now we increases the colector current to Ic = 10mA the Vbe voltage will rise to 0.62V + 60mV = 0.68V.
Or if Ic current drops to from 1mA to 0.1mA the Vbe will also drop to 0.56V.
Or if you double the current the Vbe will change by about 18mV.
If the Vbe voltage changes from 0.62V to 0.638V the collector current will double from 1mA to 2mA.
 

Thread Starter

Aleksey Shurtygin

Joined Dec 21, 2018
54
The graph below should make it clear why transistor gain is usually expressed in amps out/ amps in instead of amps out/ volts in.
View attachment 176620

(300K is about 80°F)
Salient points:
- As base current increases the base-emitter voltage approaches being flat at some low voltage.
-For a given base current, the base voltage varies at about 2 mv/degree C (yes this can be made to measure temperature).

That's why you need a base resistor in a common-emitter circuit in analog applications.
Would this be equivalent to saying that the base resistor is being used simply as a current limiter and this graph simply represents base-emitter resistance? For usual resistor this graph would be a linear and in base-emitter case its non-linear?
Is this graph is a characteristic of a specific transistor or its a common one?

I am not considering temperature variations at this point and concentrating on the room temperature for now. Otherwise it's adding unnecessary complexity for me at this point.
 

Papabravo

Joined Feb 24, 2006
12,405
The base resistor is there to convert a voltage source to a current source. As you must be aware, a transistor is a current controlled device -- base current controls collector current. Whatever voltage the emitter is at, the base can only be raised about 0.7 above that value. Connecting a voltage source to the base would be a bad bad thing because it might try to raise the base more than 0.7 V above the emitter with catastrophic results. The base resistor ensures whatever else happens, the current into or out of the base will be limited. Maybe not enough, but better than no resistor at all.
 

Thread Starter

Aleksey Shurtygin

Joined Dec 21, 2018
54
In reality the Vbe will go up by a 60mV or down by 60mV for every upward /down change in Ic by a factor of ten (decade change).

For example if Vbe is 0.62V for Ic = 1mA if now we increases the colector current to Ic = 10mA the Vbe voltage will rise to 0.62V + 60mV = 0.68V.
Or if Ic current drops to from 1mA to 0.1mA the Vbe will also drop to 0.56V.
Or if you double the current the Vbe will change by about 18mV.
If the Vbe voltage changes from 0.62V to 0.638V the collector current will double from 1mA to 2mA.
Does this mean that Ic will affect base current and consequently Vbe will change?
 
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