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! : )
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: (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 . (I'm not going to do all the work for you). You should come up with an equation that contains . Do you recognize what that difference represents? Plug it and 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.
Are you aware that e = 2.71828..., which is the base for Naperian logarithms? Since 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.