What Happens to Circuit Potentials When a Transistor is in Saturation Mode?

[Bassalisk]

Hello,

So this thread is about using transistor as a current source. I understood the concept and all. But the main point of this thread is to finally seal my knowledge of transistor in saturation mode.

We did an lab exercise about current sources, by biasing transistor with voltage divider.

Here is the scheme.

[PLAIN]http://pokit.org/get/7573eb06b9996b33837a1cc9841f1d8a.jpg

This Ropt, you can look at that as variable resistor that goes up to 1k.

I did a simulation at National Instruments, after doing a Real life exercise, and everything is ok.

Current is constant always at this Ropt, until transistor hits saturation. This is where it gets interesting.

First question(check up) is: When I am moving potentiometer, current has to be the same at collector, so the base current "automatically" changes itself to a need value in order to have the constant collector current?But when voltage across the Ropt is high enough, transistor goes into saturation(voltage across the transistor is below ~0,2 V.

Question: What happens with potentials in this circuit? Each point: Vbb, voltage across R2 etc?

I also noticed that base current is very large, in saturation mode. As expected. Where does this current go? To emitter? Is a consequence of saturation, to have a forward biased diode B-E and it has large current?

P.S. U=V at this picture, in my country the convention for voltage is U.

 

[Bob S]

This is a real current source. The output current is independent of load and input (source) voltage. The output current is determined by the value of resistance R1.

 

[Bassalisk]

Yes but the exercise was to determine the properties of the transistor, as a current source. I got everything, just the part where it goes into saturation, ergo stops being constant current source is a lit dim to me.

Not the theory of saturation itself just, the variables that are taken into consideration, while saturation is happening.

I need someone just to answer my questions, so I can write my report.

 

[vk6kro]

Saturation is not a property of the transistor. It is a property of the circuit.

I have a power transistor that could easily have 15 amps of collector current, but if it had a 1000 ohm load, it would be "saturated" at a collector current of 12 mA from a 12 Volt supply.

No extra base current could force the transistor to take more than 12 mA collector current because it is not possible to have more current flowing in a 1000 ohm resistor from a 12 volt supply.

If the collector resistor was 12 ohms, then 1 amp could flow and then the transistor would be saturated.

If the transistor had enough base current for 100 mA to flow in the 12 ohm resistor, then the voltage drop across the 12 ohm resistor would be V = I * R according to Ohm's Law.
So, V = (0.1 amps * 12 ohms) or 1.2 volts, and the remaining 10.8 volts would appear across the transistor.

Base current flows from the base to the emitter.

 

[jim hardy]

"""First question(check up) is: When I am moving potentiometer, current has to be the same at collector, so the base current "automatically" changes itself to a need value in order to have the constant collector current?""""

where is the potentiometer? I'm not sure what it is you are adjusting.
....

in your thinking take these steps:

look first at Ub, how many volts is it?
subtract from that 0.6 for one diode drop
and that'll be voltage across Re
so current through Re is known by ohm's law
and that current is sum of base-emitter current plus collector-emitter current.

so your question becomes 'how does the transistor allocate that emitter current between base and collector? '

answer lies in its datasheet
quick answer is they are in ratio of hfe,



but hfe is a function of both collector-emitter voltage and collector current
so the more detailed answer is they are in ratio of hfe at its operating condition.
there's usually a graph in datasheet for hfe vs current at various c-e voltages

and when Vce drops below a few tenths of a volt
hfe becomes very low, the transistor can no longer support high collector current for want of collector-emitter voltage;
so base's share of emitter current becomes larger

and that's saturation - circuit runs out of collector-emitter voltage


does that help?

 

[Bassalisk]

I made a mess without giving resistor values.

R1=12k
R2=2k7
Re=100
Ropt=220, 330, 470, 680, 1000

From measurements I got: when I put 1000 ohms for Ropt, I get that the Uce=0,06 V(it is in the saturation).

I get that Ub gets lower than it usually is.

For all other resistances(220-680) it is 2,45 V, but when its in saturation it gets to 2,1 V.

I also measured voltage across Re: Ue=1,42 V. (for 1000 ohms)
For all other resistances (220-680) Ue=2,45 V.

And current of course: Ic~18 mA, for Ropt(220-680). For 1000 ohms Ic ~ 13 mA.

I understand that the transistor cannot give constant current any more in saturation. So Ic falls as expected, Ie falls as well and so does voltage drop across 100 ohms Re.

But question is, does Ub fall because, base is "drawing" more current now, and the less current goes through R2?

Does current go more into base because it is "seeing" less resistance, than 2k7?

 

[Bassalisk]

Base current flows from the base to the emitter.

This helped me a lot. Thank you.

 

[Bassalisk]

"""First question(check up) is: When I am moving potentiometer, current has to be the same at collector, so the base current "automatically" changes itself to a need value in order to have the constant collector current?""""

where is the potentiometer? I'm not sure what it is you are adjusting.
....

in your thinking take these steps:

look first at Ub, how many volts is it?
subtract from that 0.6 for one diode drop
and that'll be voltage across Re
so current through Re is known by ohm's law
and that current is sum of base-emitter current plus collector-emitter current.

so your question becomes 'how does the transistor allocate that emitter current between base and collector? '

answer lies in its datasheet
quick answer is they are in ratio of hfe,
but hfe is a function of both collector-emitter voltage and collector current
so the more detailed answer is they are in ratio of hfe at its operating condition.
there's usually a graph in datasheet for hfe vs current at various c-e voltages

and when Vce drops below a few tenths of a volt
hfe becomes very low, the transistor can no longer support high collector current for want of collector-emitter voltage;
so base's share of emitter current becomes larger

and that's saturation - circuit runs out of collector-emitter voltagedoes that help?

Of course it helps. Here is datasheet for the transistor.
http://www.st.com/internet/com/TECHNICAL_RESOURCES/TECHNICAL_LITERATURE/DATASHEET/CD00003226.pdf"
I will investigate what hfe is. It sounds that it gives a more detailed explanation why transistor goes into saturation. I assume that hfe is Beta gain, just this one is a function of collector current. I've been waiting for this for like a half a year.Is there any semiconductor explanation for this function hfe. We were taught only Beta gain. And that is constant in direct bias.

 

[jim hardy]

page 3 of that datasheet plots hfe vs current

hfe is collector/base currents
hfe and beta are so close that i have forgotten what is the fine point of difference

key point: collector and emitter currents differ only by base current.
and base current is small in comparison if the transistor is working as a transistor.

......

Interesting thing to do if you're tinkering and enjoy trivia:

lower that supply voltage to around 6 volts
and measure current gain
then swap emitter-collector and measure current gain again

transistor will show current gain either direction but it does a LOT better job in proper direction
and emitter-base junction when reverse biased for use as a collector will fail usually in 5 to 10 volt range. See [Vbr ebo max] in datasheet

glad it helped.

 

[vk6kro]

Once a transistor is saturated, the gain of the transistor becomes irrelevant.

Clearly, the collector current is fixed so if you increase the base current, then the gain just depends on what base current you choose.

In normal operation, there is DC current gain and small signal current gain.

DC current gain ( variously called large signal current gain, β or hFE (note the capital letters) ) is just the DC collector current divided by the base current. This is what is usually given in simple transistor testers in multimeters.

Small signal current gain (called hfe (small letters) ) is the change in collector current caused by a small change in base current, measured with a constant collector voltage. This is a more realistic measure of the current gain of the transistor under correct biasing conditions.

 

[Bassalisk]

Once a transistor is saturated, the gain of the transistor becomes irrelevant.

Clearly, the collector current is fixed so if you increase the base current, then the gain just depends on what base current you choose.

In normal operation, there is DC current gain and small signal current gain.

DC current gain ( variously called large signal current gain, β or hFE (note the capital letters) ) is just the DC collector current divided by the base current. This is what is usually given in simple transistor testers in multimeters.

Small signal current gain (called hfe (small letters) ) is the change in collector current caused by a small change in base current, measured with a constant collector voltage. This is a more realistic measure of the current gain of the transistor under correct biasing conditions.

I really thought I knew a lot about transistors. Back to my pile of books I guess. Thank you all.