# BJT Voltage divider circuit design

#### rhy87_007

Joined Jul 27, 2010
2
Design a voltage-divider biasing small signal amplifier circuit using BJT transistor with
Vcc = 18V, Icq = 2mA ± 3%, emitter voltage VE = 10% of Vcc , and Vceq = 7V.
The expected β range is 80 to 200. Only commercial available the resistors are to be used.

I tried to calculate... and when i put the circuit on Multisim i cannot get what i want... attached is the circuit...

I got the relation that R1 =169/11 R2
using Vb=(R2/(R1+r2))Vcc where Vcc is 18V as stated above and Vb=Ve-Vbe=1.8-0.7=1.1V is this correct?

And using KVL,
-Vcc+IcqRc+Vceq+IeRe=0
and find the relation between Re and Rc. Is this correct?

If both correct how shall i find the relationship between R1 R2 and Rc Re?

Thanks a lot for your help!!
It'd be best if someone can design it for me.. thanks!! cuz i am a circuit idiot~ >.<

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#### Jony130

Joined Feb 17, 2009
5,183
Hmm
Vb = Vbe + Ie * Re = Vbe + Ve = Vbe + 0.1*Vcc =Vbe + 1.8V

#### rhy87_007

Joined Jul 27, 2010
2
what about the circuit shown? is it correct? I mean ignoring the values of resistors.

#### Jony130

Joined Feb 17, 2009
5,183
Yes schematics is correct.
And you must remember the rule of thumb for voltage divider current.
Voltage divider current must be greater than the base current.
In practice 5 to 30 times larger then Ib (typical 10 times larger).
So:
R2=Vb/(10*Ib)
R1=(Vcc-Vb)/(11*Ib)

#### peacelover1990

Joined Aug 11, 2010
1
how to start doing this kind of question if we doesnt know all the R values? can give any guide?

#### Wendy

Joined Mar 24, 2008
22,164
ß = 10 is a general rule of thumb for saturation. The input resistance of the base is closer to β Re in linear mode, and this resistance is in parallel with R2.

Since typical beta for a 2N3904 is 100, the Base to ground resistance is 910Ω X 100, or 91KΩ. Since βThis in parallel with 91KΩ║R2 is 47.6KΩ, which combined with R1 and R2 creates a base voltage of .55V. Since β is an approximation, this can vary somewhat. This is below the breakover of 0.6V for the BE junction, it can not a realistic answer, and the transistor is not in conduction.

OK, so lets get this transistor biased correctly.

Assume you want 9V on the Collector. This means a Ic of 1.91ma.

Ic ≈ Ie, so you would need a voltage of 1.74V on the emitter, which means you need a voltage of 2.34V on the base. If the Base to ground is 47.6KΩ (both transistor and R2) the current through R2║Q1 is 49.2µa, and R1 needs to drop 15.7V. This means R1 needs to be around 318KΩ.

This will put the amplifier in its linear area.

The gain is ≈ Rc/Re, or 5.

You can make this circuit a bit more predictable by dropping the R1/R2 values by 10 and recalculating. This reduces the effects the Base to ground resistance will have. In this design the base to ground resistance is a significant part of the circuit, and since transistors are extremely variable even within families, it will likely not come out as predicted.

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#### Wendy

Joined Mar 24, 2008
22,164
I just reread your original post. Vemitter is easy, they tell you what voltage they want.

The Re determines the current, and you have a target voltage. This gives you Re using Ohm's Law.

Rc should be midway between it's ranges, which is when the transistor is fully on and fully off. It's current is known, so once you have the required voltage then it becomes a simple matter using Ohm's Law.

Focus on these points first.