What degree of matching in emitter area is required for the base-emitter voltages of two bipolar devices with identical collector currents to have a difference of less than 4.6 mV if we use VT = 25.85 mV? How to approach this? Its quite confusing for me. Thanks! : )

 

What is the related theory behind this question?

 

Here is how I would go about it.
For a given current, the current density is inversely proportional to emitter area. Think of the diode equation and current ratio.

 

Here is how I would go about it.
For a given current, the current density is inversely proportional to emitter area. Think of the diode equation and current ratio.

If all I have are the VT & the voltage .. how would I go with the diode equation and current ratio? Can you please show some figures so I can follow up or explain a little further? I still don't understand the question

 

Start with diode equations for the two transistor base-emitter junctions:
\( I_1=I_s*(e^{\frac{V_{be1}}{V_T}}-1)\)
\( I_2=I_s*(e^{\frac{V_{be2}}{V_T}}-1)\)
(Keep in mind what I said previously about emitter area and current density.)
For reasonable current levels, you can ignore the -1 term in each equation.
Now, take the natural log of the ratio \(\small \frac{I_1}{I_2}\). (I'm not going to do all the work for you).
You should come up with an equation that contains \(\small Vbe1-Vbe2\). Do you recognize what that difference represents? Plug it and \( \small V_T\) into your equation, then take the antilog. What answer do you get?

 

Thanks Ron .. The idea became much better now and I have came up with a solution thats attached. Nothing brilliant but now this thing makes sense to me. Sounds good to you?

what I got is 1.78%.

 

My version of Word won't open your file.
Your answer is ln(I1/I2). You need I1/I2.

EDIT: ln(I1/I2)=0.178. Even if this were the correct answer, it is not equivalent to 1.78%.

 

I know its not 1.78% .. What I got is 0.178e-1 * 100 which gave me that percentage. I converted the doc to word 2003 and it should be okay now.

 

\(\frac{I_{s1}}{I_{s2}}=e^{\frac{4.6}{25.85}}=e^{0.17795}\)

Are you aware that e = 2.71828..., which is the base for Naperian logarithms?

\( \frac {I_{s1}}{I_{s2}}=1.1948\)

Since \( I_s\) is proportional to emitter area, then the areas must match within ≈20%.

 

Ooops
I missed out the "e" part
for some reason under all this stress I calculated that as (10^x) ..
Thanks a lot Ron .. your help is appreciated once again.