I'm having trouble with a BJT circuit.

BJT circuit

What we are given:

\$U_{CC} = 10\text{V}\$
\$R_C = 972 \Omega\$
\$R_B = 14\text{k} \Omega\$
\$U_{\text{BE}} = 0.7\text{V}\$ \$I_C = 12\text{mA}\$

We need to find the current amplification \$B\$.
My approach was to calculate \$U_C\$, the Voltage which drops at \$R_C\$. I read in a book that \$I_C\$ is the current we need to use Ohm's Law at \$R_C\$. So I solved the equation \$U_R = I_C \cdot R_C \Leftrightarrow U_R = 12\text{mA} \cdot 972 \Omega \Leftrightarrow U_R = 11.664\text{V}\$.

Having this done I was able to use the Mesh-Current-Law at the upper right part of the circuit which gave me the following equation \$-U_{CC} + U_C - U_B\$ where \$U_B\$ is the Voltage which drops at \$R_B\$. Filling the equation with the known values we receive \$U_B = 1.664\text{V}\$. Since we have \$R_B\$ given we can now apply Ohm's Law with the previously calculated Voltage which leads to the following value for

\$I_B = \frac {U_B}{R_B} = \frac {1.664\text{V}}{14000 \Omega} = 1.188571429x10^{-4}\text{ A}\$

or \$0.1188571429\text{ mA}\$.

Now I found out that the base current \$B\$ can be expressed by \$B = \frac {I_C}{I_B}\$.

Since we know \$I_C\$ as well as \$I_B\$ I went ahead and filled out the equation which gave me \$B = \frac {12\text{mA}}{0.1188571429\text{mA}} = 100.9615385\$ for \$B\$.

Am I on the right track?

  • 3
    \$\begingroup\$ Please study how significant figures work. \$\endgroup\$ – markrages Nov 29 '12 at 18:46

Not quite.

Richman's answer is good and normally I would say it is correct, but...

given Vcc=10V, Rc=972, Ic=12ma, we are not in the real physical world.

12ma through a 972 ohm resistor drops 11.664 volts, yet the supply is only 10 volts.

Somebody isn't telling us something...

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  • \$\begingroup\$ This is a serious topic and I double checked every value :(. However I still do not know if my approach was correct. \$\endgroup\$ – optional Nov 29 '12 at 18:40
  • \$\begingroup\$ Going with Vcc=12v as suggested in another comment, Ib=11.3/14k or 0.807ma Then Ic/Ib=15 at this working point; quite plausible for a transistor so close to saturation. \$\endgroup\$ – Brian Drummond Nov 29 '12 at 19:00
  • \$\begingroup\$ Why did you divide 11.3/14k? Is there any "tolerance value" which you subtracted from Vcc (Vbe?)? \$\endgroup\$ – optional Nov 29 '12 at 19:13
  • \$\begingroup\$ The loop is 12V - Vbe - Ib*Rb = 0V and my answer is correct. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 29 '12 at 20:19
  • \$\begingroup\$ As far as I know I cannot simply change the value of Vcc. I am forced to calculate the rather unusual circuit with the values given leaving me no other choice then using 10V as Vcc. Despite this "mistake" is the following calculation correct? Ib = (10V - 0.7V) / 14k Ohm = 0.66mA, Beta (current amplification factor) B = Ic/Ib = 12mA / 0.66mA = 18.06. \$\endgroup\$ – optional Nov 29 '12 at 20:37

not quite...

It seems you may be over-complicating your analysis with excessive decimal places and cryptic syntax.

The base current is not caused by collector current. It is determined solely by V+ (Ucc) ,Rb and Vbe(Ube) (assume 0.65V +-.05 or 0.7 as given ) If there are more than one Rb with pull-up and down, then convert to equivalent voltage and equivalent resistance.

We don't normally show a battery equivalent circuit on a schematic, but you may need to remember this path when doing loop calulations. enter image description here

optional< maybe its a typo, but Ucc needs to be 12V for this question to be practical. You can't have Ur > Ucc, also the dc leakage current of collector to base is neglible. So Ib= 12V/14k= 0.86mA while Ic was given as 12 mA so the hFE = 14.. This is a good number for a saturated switch. You have to derate Beta when VCe drops into saturation. Most transistors do not saturate unless this ratio is between 10 and 30. Higher current needs lower ratios.

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  • \$\begingroup\$ I am so sorry. I mixed up base current and current amplification. Yes, you are right the base current depends solely on V+, the resistor and Vbe(?). However in this case I had to calculate the current amplification, my bad. \$\endgroup\$ – optional Nov 29 '12 at 18:14
  • \$\begingroup\$ you are given the collector current, so if you find the base current, you can just divide the two to find the current amplification. \$\endgroup\$ – markrages Nov 29 '12 at 18:45
  • \$\begingroup\$ I found the base current which is called \$I_B$ in my topic. Just want to make sure that the way I attempted the problem is correct. I know that the values may seem a bit odd but thats how they are given I had no influence. \$\endgroup\$ – optional Nov 29 '12 at 19:08
  • \$\begingroup\$ No it is incorrect You actually dont need to be given Vbe and Ic to solve this question within 10%. Beta only is accuarte when Vce is between the supply rails (in linear region) When saturated (Vce=12-11.67...=0.33) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 29 '12 at 19:17
  • \$\begingroup\$ Ratios are always reduced in the range of 10~30 with few exceptions when Vce is low (saturated switch) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 29 '12 at 19:23

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