Voltage Biased Transistor

Hello all,
I am a Beginner to this field, these days i am studying voltage biased common emitter Transistor configuration. The book is showing a pic of this configuration and do not describes the name or number of the transistor. so if i want to build that circuit, how could i with the number missing for the transistor. and my multimeter offers hfe measuring ability. i put 2N3904 transistor and the meter showed 173 hfe of the transistor and the manual of my meter says that i shows shows hfe at 10uA base current with Vce=2.8v. Now i am damn! confused about Beta. can i change value of Beta according to my requirements? i know Beta is dc current gain and hfe is Ac current gain. i know how to use it in solving for dc analysis. But still Confused about it. do i have to use the Beta or hfe given in datasheet? that is only 2 or 3 values for specified Ic. i am confused about it. please make me clear.
Thanks

ejector said:

Hello all,
can i change value of Beta according to my requirements?

Click to expand...

You can't change the beta of the transistor.

maybe this will help.

https://alltransistors.com/crsearch...150%B0C&ft=300MHz&cc=6&hfe=110/220&mnf=&caps=

When you bias the base of a transistor with a voltage divider, then since its hFE and base-emitter voltage are a little different for each transistor and vary when the temperature changes then a series emitter resistor is used so that the transistor can be biased properly.

The emitter resistor helps to automatically adjust the current in the transistor. If the collector to emitter current is too high because the transistor has a low base-emitter voltage requirement or is hot, or because it has a high hFE then the excessive current causes the emitter voltage to increase which reduces the base-emitter voltage which reduces the current. The opposite happens when the current is too low.

Hi,

You can change the beta of the transistor somewhat by changing the DC operating point, but usually you just design the stage to have a gain less than that which the beta can achieve in the design so that it will always work correctly.

Ok let suppose i need Ic=5mA so the datasheet does not supply hfe for this Ic then how to choose hfe?

ejector said:

Ok let suppose i need Ic=5mA so the datasheet does not supply hfe for this Ic then how to choose hfe?

Click to expand...

You don't choose hfe. It is what it is. You design the circuit so that it will work for the minimum value of hfe. If you look at typical bias circuits for a BJT, they use negative feedback (usually an emitter resistor) to stabilize the circuit gain and minimize the effect of beta variations on circuit operation.

Firstly, the gain of a transistor changes considerably according to the collector-emitter current.
Secondly the gain changes according to the type of circuit (the "class").
It also changes slightly due to temperature.
A transistor with a gain of 220 to 450 will have a gain of about 75 in a "self-biased" stage.
That's why you have to build the circuit and measure the results.

colin55 said:

That's why you have to build the circuit and measure the results.

Click to expand...

No.
You don't need to build and measure if you design properly for "worst case" which is the lowest spec'd current gain. Then any transistor with any amount of current gain will work in your circuit.

Ejector,
In the circuit I posted in Reply #4 there is a voltage divider to bias the base of the transistor. The current in the voltage divider is designed to have about 10 times as much current as the base current of a "typical" transistor. Then the base voltage does not change much if the transistor has a low current gain or a high current gain.
There is an emitter resistor so that the base voltage set by the voltage divider determines the collector and emitter current, not the current gain of the transistor.

Maybe you are trying to bias the transistor with a single resistor from the positive supply voltage that should NEVER be done:

I said "self-bias." I did not say to place a resistor from base to positive rail.

colin55 said:

I said "self-bias."
I did not say to place a resistor from base to positive rail.

Click to expand...

I know what you said.
The OP does not understand how to bias a transistor. He does not know about "self-bias".
I showed a simple circuit that perhaps he is using that DOES NOT ALWAYS WORK.

Hey! Thanks all,
i am using the voltage divider bias network with four resistors. the point where i am confuse is practical side, and the confusing is about β,Ic, and Vce.

hFE∗ DC Current Gain
IC = 0.1 mA VCE = 1 V 60
IC = 1 mA VCE = 1 V 80
IC = 10 mA VCE = 1 V 100
IC = 50 mA VCE = 1 V 60
IC = 100 mA VCE = 1 V 30

Now, these are the hfe Values for 2N3904 Transistor, given in its datasheet. if i choose hfe=100, i choose Ic to be 10mA. am i fixing Ic? or the load can consume the Current from it as per load's requirement?
point-2 That's why i was asking that if i want Ic=5mA then how to choose hfe.
This my confusion Thanks

If you are using a "Bridge Arrangement" with 4 resistors, the gain of the stage will heavily depend on the ratio of the collector resistor to emitter resistor. For instance, if the collector load is 10k and the emitter resistor is 470R, the gain of the stage will be 20.

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Ejector,
You DID NOT list the hfe (AC current gain) values of your 2N3904 transistor. Instead you listed the MINIMUM hFE (minimum DC current gain) values only when the transistor chip is at room temperature. Many 2N3904 transistors have much more gain perhaps 3 times more and a graph in the datasheet shows that the hFE changes a lot when the temperature changes.

The range of hFE from 100 to 300 is listed only when the collector current is 10mA. So you can design at 10mA with a "typical" hFE of 200.
Then the average base current is 50uA and the current in the voltage divider feeding the base can be 250uA to 500uA.
Higher or lower collector and base current ratios will be similar.

If you select an emitter resistor value to have a couple of volts across it then it adds some negative feedback to cancel differences in hFE values of different transistors and cancels most of the base-emitter voltage differences of different transistors.
The emitter resistor also adds AC negative feedback that reduces the voltage gain and reduces distortion. Adding a bypass capacitor across the emitter resistor increases the AC gain.

Yes, you're worrying too much about hFE.

When designing an amp as above, the only time I use hFE is to determine the voltage divider resistors. As Audio said, the current through the voltage divider needs to be about 10 times the max Ib.

So choose the collector current you need (this depend upon what the amp is driving) & calculate Ib using the min hFE in the region of the collector current. eg. if you want Ic to be about 5 mA, then use hFE nearest, eg. if the data sheet gives it at 10 mA, then that is near enough.

Then choose the emitter voltage (Audio suggested 2 Volt which is reasonable) add 0.7 to give you the approx base voltage & then design a voltage divider that gives you approx that voltage & has a current approx 10 * Ib through the resistors.

Then you can calculate the emitter resistor to give you about 2 Volt & the collector resistor to set the quiescent collector voltage about mid way between cut-off & saturation.

eg. If you have a 10 Volt supply & about 2 Volt at the emitter, then about 6 Volt is reasonable.