Why the ratio of Ic/Ib on BJT is not constant?

Thread Starter

jacob_c

Joined Oct 13, 2020
13
Hi. The schematic of amplifier using BJT and the result is attached.

What I learned is that the beta, ratio of Ic/Ib, would be constant.

However, the ratio of Ic/Ib varies by the time.

I googled for this and found out that the beta varies with temperature and Ic but does it vary that much?

Ic is in range of 1.1mA~1.55mA from the simulation. How can I explain this situation?
 

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WBahn

Joined Mar 31, 2012
26,398
Read the text regarding transistor characteristics and then take your best shot at giving an explanation. That will give us a starting point. We won't answer your homework questions for you.

Here's something to get you thinking -- let's say that we have a transistor set up so that Ib = 0.1 mA and Ic = 10 mA when Vce is 10 V. If we then keep Ib the same and decrease Vce, do you really expect Ic to remain 10 mA? Even when Vce is reduced all the way to 0 V? When it's made negative? If not, do you expect the ratio of Ic/Ib to change smoothly, or all at once at some point?
 

Thread Starter

jacob_c

Joined Oct 13, 2020
13
Read the text regarding transistor characteristics and then take your best shot at giving an explanation. That will give us a starting point. We won't answer your homework questions for you.

Here's something to get you thinking -- let's say that we have a transistor set up so that Ib = 0.1 mA and Ic = 10 mA when Vce is 10 V. If we then keep Ib the same and decrease Vce, do you really expect Ic to remain 10 mA? Even when Vce is reduced all the way to 0 V? When it's made negative? If not, do you expect the ratio of Ic/Ib to change smoothly, or all at once at some point?
Thank you for answer. But can you give me some more guideline for it? I'm trying but it's not quite easy to get.
 

LvW

Joined Jun 13, 2013
1,247
Hi. The schematic of amplifier using BJT and the result is attached.
What I learned is that the beta, ratio of Ic/Ib, would be constant.
However, the ratio of Ic/Ib varies by the time.
I googled for this and found out that the beta varies with temperature and Ic but does it vary that much?
Ic is in range of 1.1mA~1.55mA from the simulation. How can I explain this situation?
I must admit that I do not understand your problem description.
* You simply claim that this ratio would vary by time. No explanation or verification.
* You simply claim that the Ic would vary between 1.1 and 1.55 mA "from the simulation"....What did you simuiate and which parameer did you vary?
* What is the meaning of the shown sinus signal? What does it show in the context of your question?
* How did you arrive at the shown parts values? (They look rather strange).
 

Thread Starter

jacob_c

Joined Oct 13, 2020
13
I must admit that I do not understand your problem description.
* You simply claim that this ratio would vary by time. No explanation or verification.
* You simply claim that the Ic would vary between 1.1 and 1.55 mA "from the simulation"....What did you simuiate and which parameer did you vary?
* What is the meaning of the shown sinus signal? What does it show in the context of your question?
* How did you arrive at the shown parts values? (They look rather strange).
The simulation is conducted base on the schematic attached. I gave VSIN input of 10mV amplitude between 0 to 1ms to check
the beta, which is equal to collector current/base current.

With the given DC bias and VSIN input, the Ic varies btw 1.1~1.55mA.

The sinus signal shows beta(Ic/Ib). I expected this to be constant but it shows sinus signal as you can see.

The parts values are given.

Sorry for poor explanation of my question
 

LvW

Joined Jun 13, 2013
1,247
The simulation is conducted base on the schematic attached. I gave VSIN input of 10mV amplitude between 0 to 1ms to check
the beta, which is equal to collector current/base current.

With the given DC bias and VSIN input, the Ic varies btw 1.1~1.55mA.

The sinus signal shows beta(Ic/Ib). I expected this to be constant but it shows sinus signal as you can see.

The parts values are given.

Sorry for poor explanation of my question
OK - I did not recognize the bottom of the sinus plot. Now I see - the ratio Ic/Ib.
However, something must be wrong with your circuit.
With only 2 volts supply, I assume Vce=1V. Hence, the base cutrrent is app. Ib=(1-0.7)V/1k=0.3mA.
For a current gain of only 100 we would have a collector current of 30mA. And the voltage drop at the collector resistor is 30V.
Do you now understand my remark about the circuit and the parts values?
I suppose, you did not check the DC operating point prior to the TRAN simulation?
 

crutschow

Joined Mar 14, 2008
27,232
What I learned is that the beta, ratio of Ic/Ib, would be constant.
Wherever you learned that is giving you false information.
A real BJT transistor (or the model used in a Spice simulator) has a beta that varies with collector current and temperature.
In the real world, there is also a significant manufacturing variation between transistors, even of the same part number.
Just look at a typical transistor data sheet to see this.

That is why a circuit is normally designed so that its DC bias and AC gain is relatively insensitive to the actual transistor beta value.
 

LvW

Joined Jun 13, 2013
1,247
Wherever you learned that is giving you false information.
A real BJT transistor (or the model used in a Spice simulator) has a beta that varies with collector current and temperature.
Yes - of course, thats true.
However, the questioner has observed a change of app. 40% .
This huge deviation was caused by the fact that the transistor was not operated in the linear region (where the current gain beta is specified).
In contrary, the pn junction between C and B was open and he has measured - more or less - the properties of an open pn-diode.
 

Thread Starter

jacob_c

Joined Oct 13, 2020
13
OK - I did not recognize the bottom of the sinus plot. Now I see - the ratio Ic/Ib.
However, something must be wrong with your circuit.
With only 2 volts supply, I assume Vce=1V. Hence, the base cutrrent is app. Ib=(1-0.7)V/1k=0.3mA.
For a current gain of only 100 we would have a collector current of 30mA. And the voltage drop at the collector resistor is 30V.
Do you now understand my remark about the circuit and the parts values?
I suppose, you did not check the DC operating point prior to the TRAN simulation?
yep according to the same simulation above, the Vce is about 750mV and Vbe is about 650mV. Due to the input swing, there are some intervals that Vbe>Vce.

But usually it is on forward region while the large swing of beta is all around.

Since all values of parts are given, I cannot change it.
 

crutschow

Joined Mar 14, 2008
27,232
the Vce is about 750mV and Vbe is about 650mV
At that low Vce, the apparent beta will vary significantly with current since you are near the saturation point of the transistor.
Beta is normally specified at a few volts of collector-emitter voltage.

If you cannot change the circuit, then it is what it is.
 

Audioguru again

Joined Oct 21, 2019
3,202
Beta is current gain. But most transistor circuits are made to produce voltage gain.
Beta is used to DC bias a transistor to avoid clipping distortion.
But the supply voltage of only 2V is much lower than the 10V where the beta is listed on the datasheet.

The transistor is biased do that the collector voltage is swinging above and below the base voltage in this simple circuit and its voltage gain is only 10.75 times.
 

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