# BJT Matching?

Discussion in 'Homework Help' started by eng kryptonite, Feb 7, 2010.

1. ### eng kryptonite Thread Starter New Member

Jan 25, 2010
17
0
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! : )

2. ### mik3 Senior Member

Feb 4, 2008
4,846
65
What is the related theory behind this question?

3. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
657
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.

4. ### eng kryptonite Thread Starter New Member

Jan 25, 2010
17
0
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

5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
657
$I_1=I_s*(e^{\frac{V_{be1}}{V_T}}-1)$
$I_2=I_s*(e^{\frac{V_{be2}}{V_T}}-1)$
(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?

6. ### eng kryptonite Thread Starter New Member

Jan 25, 2010
17
0
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%.

File size:
13.7 KB
Views:
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7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
657
My version of Word won't open your file.

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

Last edited: Feb 9, 2010
8. ### eng kryptonite Thread Starter New Member

Jan 25, 2010
17
0
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.

File size:
178.5 KB
Views:
18
9. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
657
$\frac{I_{s1}}{I_{s2}}=e^{\frac{4.6}{25.85}}=e^{0.17795}$

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%.

10. ### eng kryptonite Thread Starter New Member

Jan 25, 2010
17
0
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.