0
\$\begingroup\$

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?

\$\endgroup\$
  • 3
    \$\begingroup\$ Please study how significant figures work. \$\endgroup\$ – markrages Nov 29 '12 at 18:46
2
\$\begingroup\$

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...

\$\endgroup\$
  • \$\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\$ – 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
0
\$\begingroup\$

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.

\$\endgroup\$
  • \$\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\$ – 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\$ – Sunnyskyguy EE75 Nov 29 '12 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.