## transistor IC-Vce

hi
in an transitor IC-Vce graph(as it apear on the third page of this site - (first line and right graph) www.onsemi.com/pub_link/Collateral/2SB1204-D.PDF)

why are the Vce and Ic changing when the Ib is staying the same?
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 can you try phrase it in a simple language ? i dont speak fluent english and wikipedia use to add examples and phrases from all the levels of physics

## transistor IC-Vce

The collector current, Ic, is determined mainly by the base current,Ib, and the current gain of the transistor. Once the value of Vce is greater than about 0.5 to 1.0V any further increase in Vce has little effect. The transistor is said to be 'saturated'.
The values of Ic for different values of Ib are 'more or less' horizontal showing that Vce is not having a major effect.
From the table given for this transistor you can see that the current gain (hfe) is somewhere between about 30 and 70. You can also see that the value of Vce to cause saturation is about 250mV.
hope this helps
 The collector diode is reverse biased. But still there is a little leakage current that can be modeled as a resistor. This is refer to be something like the other resistance ( I don't want to get into the detail of the physics or exact model as that can be long). In normal BJT, the output resistance ro is in 10MHΩ or higher easily. So if you vary the Vce, you see the collector curve slant up with increase Vce even the Ib is constant. Take two point on the straight portion of the curve ( saturation region), find the ΔIc vs ΔVce $$r_o=\frac{\partial I_c}{\partial V_{CE}}$$ Don't challenge me on the exact equation, this is only the simplified explanation.
 the published curves for the transistor in this post give dIc/dVce of about 0.5/7 = 0.07A/V = 14ohms. (I used Ib = -15mA as the average) Do you have any references where the exact modelling and the exact equation can be followed up?

 Quote by truesearch the published curves for the transistor in this post give dIc/dVce of about 0.5/7 = 0.07A/V = 14ohms. (I used Ib = -15mA as the average) Do you have any references where the exact modelling and the exact equation can be followed up?
I should have said that the equation is only approx AT the saturation region.....The flat straight line portion of it. That's the portion that if you extrapolate to the negative voltage, it intercept the x axis at the Va that called early voltage.

There are other effect, but for common emitter ( ground emitter), this is a good approximation. If you use this straight line and calculate of the change of collector current vs Vce, you get the output resistance in range of Mega ohms. This is to take two points on the flat line and find the ΔIc vs ΔVce to get the ro. I edit my original post already.
 Transistors in amplifiers are usually operated in the saturated region (with Vce greater than about 500mV) once in this region Vce has little effect on Ic (the lines on the graph are ALMOST horizontal. The major effect on Ic is through Ib. In practical circuits with a collector resistor it is safe to assume that Vce will be about 1V (saturation voltage) in the same waythat it is afe to assume that Vbe is about 0.7V for silicon transistors. The collector resistor can be shown on the Ic against Vce graph as a straight line through the origin with the appropriate slope. It is called the load line and its intersection with the Ic~Vce curves gives the operating points of the transistor.
 Recognitions: Science Advisor In practical circuits with a collector resistor, Vce is arranged to be about half of the supply voltage. This allows the voltage to swing equally in either direction with changes in base current. For undistorted output, operating the amplifier in saturation must be avoided and it should not be driven into the cut-off region either. The load line is not drawn through the origin. It is drawn from the supply voltage point on the Vc axis to the short circuit current point on the Ic axis.
 That is what I meant VK but didn't say it correctly. Load line is from Ic axis to Vce axis. My mistake to say through origin.

 Quote by truesearch Transistors in amplifiers are usually operated in the saturated region (with Vce greater than about 500mV) once in this region Vce has little effect on Ic (the lines on the graph are ALMOST horizontal. The major effect on Ic is through Ib. In practical circuits with a collector resistor it is safe to assume that Vce will be about 1V (saturation voltage) in the same waythat it is afe to assume that Vbe is about 0.7V for silicon transistors. The collector resistor can be shown on the Ic against Vce graph as a straight line through the origin with the appropriate slope. It is called the load line and its intersection with the Ic~Vce curves gives the operating points of the transistor.
OP ask why why the Vce and Ic changing even Ib is constant. This is represented by the top right graph on page 3. Even in the saturation region, the lines are not horizontal. In fact if you extend all the lines towards the left side, they should all meet at the same point on the x-axis at a voltage called Early Voltage. From the slope of each line, you can find the output resistance of the transistor. That's the ro I posted in the first post. At the saturation region, the Ic still change when you change Vce, that's the output resistance effect. And this is with Ib held constant. That has nothing to do with load line.

At Vce between 0V to the saturation region, the curve is non linear and output resistance really don't mean a whole lot.
 thanks every one but i think it is really to profound explanation.. i dont understan (and i read comments) why should the vce change?- the ib is fixed.. how can it change?

Recognitions:
 Quote by reonem hi i have an exam next sunday and i really have no clue about this thing.. in an transitor IC-Vce graph(as it apear on the third page of this site - (first line and right graph) www.onsemi.com/pub_link/Collateral/2SB1204-D.PDF) why are the Vce and Ic changing when the Ib is staying the same?
This is just a graph of collector current vs collector voltage with different fixed base currents.

So, the base current is fixed and you can vary the collector voltage and current.

In practice, you would actually vary the base current and this would cause a change in collector current.
Because of the load resistor, this causes a change in collector voltage.

From a design point of view, you would look at this graph and see where you could place a load line which would give approximately equal changes in output voltage for the changes in base current. This would give a relatively undistorted output.

In this case, you might put a loadline from 10 volts to 4 amps, so the resistor would be 2.5 ohms, but the output would be fairly distorted because of the uneven spacing of the curves.
 the resistor is the divide yea? if i will put x amper in the base.. how can i know, through the graph, the collector current? thanks allot
 Recognitions: Science Advisor If there is no load, you can read it straight off the graph. Suppose you put 15 mA in the base, then you can read the 15 mA curve to get the collector current for different collector voltages. However if you had a 2.5 ohm load in series with the collector, you would do this: The base current is 15 mA with no signal, so the operating point is at the orange dot on the load line. The collector current is about 2.4 amps and the collector voltage is about 3.9 volts. Notice that the collector resistor is dropping about 6 volts. 2.4 amps * 2.5 ohms = 6 volts. Take this away from the 10 volt supply and you get the 4 volts on the collector. Now, if you vary the base current from 10 mA to 20 mA, the collector current will vary from 1.75 amps to 3 amps (blue lines) and the collector voltage will vary from 5.6 volts to 3 volts (purple lines).

 Quote by vk6kro If there is no load, you can read it straight off the graph. Suppose you put 15 mA in the base, then you can read the 15 mA curve to get the collector current for different collector voltages. However if you had a 2.5 ohm load in series with the collector, you would do this: The base current is 15 mA with no signal, so the operating point is at the orange dot on the load line. The collector current is about 2.4 amps and the collector voltage is about 3.9 volts. Notice that the collector resistor is dropping about 6 volts. 2.4 amps * 2.5 ohms = 6 volts. Take this away from the 10 volt supply and you get the 4 volts on the collector. Now, if you vary the base current from 10 mA to 20 mA, the collector current will vary from 1.75 amps to 3 amps (blue lines) and the collector voltage will vary from 5.6 volts to 3 volts (purple lines).
what do you mean by "load"? "If there is no load, you can read it straight off the graph."
 Recognitions: Science Advisor That means that the collector is connected directly to the supply voltage without any resistor. The load line then becomes vertical.