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#11
07-18-2013, 09:29 AM
 AD633 Member Join Date: Jun 2013 Location: Ciudad de México Posts: 95

Quote:
 Originally Posted by Jony130 Yes, first assume that BJT is in active region and find Ib current. Then you need to check if your assumption about BJT active region was right. What?? Ic = ??
I know that Id=4.236mA.Id=IE,and therefore if the BJT transistor is in the ative region Ic approx equal to Ie.

I don't see another way to determine Ib.I could write:

$Vcc=330kOhm*Ib+VBE+VDG+VGS+1.2kOhm*Is$.... but there are many variables that i don't know.

Thanks
#12
07-18-2013, 10:27 AM
 t_n_k Senior Member Join Date: Mar 2009 Posts: 4,802

You know the emitter current for the BJT must equal the drain current of the FET.

So $\text{I_{\small{E}}=I_{\small{D}}=4.236mA}$

If the BJT is in the linear region you can simply apply

$\text{I_{\small{E}}=(1+\beta)*I_{\small{B}}}$

Hence

$\text{I_{\small{B}}=\frac{I_E}{$$1+\beta$$}}$

If you know IB then it's a simple matter to find the BJT base voltage and hence the emitter voltage.
 The Following User Says Thank You to t_n_k For This Useful Post: AD633 (07-18-2013)
#13
07-18-2013, 03:10 PM
 AD633 Member Join Date: Jun 2013 Location: Ciudad de México Posts: 95

Quote:
 Originally Posted by t_n_k You know the emitter current for the BJT must equal the drain current of the FET. So $\text{I_{\small{E}}=I_{\small{D}}=4.236mA}$ If the BJT is in the linear region you can simply apply $\text{I_{\small{E}}=(1+\beta)*I_{\small{B}}}$ Hence $\text{I_{\small{B}}=\frac{I_E}{$$1+\beta$$}}$ If you know IB then it's a simple matter to find the BJT base voltage and hence the emitter voltage.
Ib can be found trough $\text{I_{\small{E}}=(1+\beta)*I_{\small{B}}}
IB=(4,236mA)/(1+160)=26,31uA$

Now i can find Vb

$Vb=Vcc*(330kOhm)/(330kOhm+1.1kOhm)=24,92 V

Vd=Ve=Vb-Vbe=24.92 V-0.7 V=24.22 V

$

Is this correct?

Thanks
#14
07-19-2013, 12:13 AM
 t_n_k Senior Member Join Date: Mar 2009 Posts: 4,802

You seem to have some worrying gaps in quite basic DC circuit analysis understanding.

There's no logic in your assertion ...

$Vb=Vcc*(330kOhm)/(330kOhm+1.1kOhm)=24,92 V

$

How do you not see that the BJT base voltage is the supply voltage less the drop across the 330k base bias feed resistor?

You clearly need to strengthen your basic skills before attempting problems like this one. Otherwise you'll end up with significant difficulties in the future.

Last edited by t_n_k; 07-19-2013 at 01:07 AM.

 Tags bjt, circuit, jfet, transistors

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