Help with fractions

Question

I have attached an image of which I have a question about. I don’t understand how you can get from equation 1 to equation 2.

Could someone please explain this?

$$nRc=\left(\frac{V_{cc}-V_{BE}}{V_o-V_{BE}}\right)R_B-R_B\tag{1}$$

$$nRc=\frac{V_{cc}R_B-V_{BE}R_B-V_oR_B+V_{BE}R_B}{V_o-V_{BE}}\tag{2}$$

Thanks.

Just don’t understand the mathematical steps taken to get from equation 1 to 2 — David777, Apr 10 '19 at 09:28
Yeah Shaun, I'm going to have to start using it at some stage, its very good. — David777, Apr 10 '19 at 20:27

score 2 · Accepted Answer · answered Apr 10 '19 at 09:22

They have done three operations in one.

The first operation is to carry out the multiplication $$ \left(\frac{V_{CC} - V_{BE}}{V_0 - V_{BE}}\right)R_B = \frac{V_{CC}R_B - V_{BE}R_B}{V_0 - V_{BE}} $$ The second thing they did was to subtract the rightmost $R_B$ from this resulting fraction, which is a two-step process.

The first step in fraction addition and subtraction is to make sure all involved fractions have the same denominator. So we rewrite $R_B$ into a fraction with $V_0 - V_{BE}$ in the denominator: $$ R_B = \frac{ (V_0 - V_{BE})\cdot R_B}{V_0 - V_{BE}} = \frac{V_0R_B - V_{BE}R_B}{V_0 - V_{BE}} $$ Then you subtract the two fractions from one another by simply subtracting the numerators and keeping the denominator unchanged (be careful with the signs in the second fraction here): $$ \frac{V_{CC}R_B - V_{BE}R_B}{V_0 - V_{BE}} - \frac{V_0R_B - V_{BE}R_B}{V_0 - V_{BE}}\\ = \frac{V_{CC}R_B - V_{BE}R_B - (V_0R_B - V_{BE}R_B)}{V_0 - V_{BE}} \\ = \frac{V_{CC}R_B - V_{BE}R_B - V_0R_B + V_{BE}R_B}{V_0 - V_{BE}} $$ And that's it.

Thanks Arthur, get it now. Much appreciated – David777 Apr 10 '19 at 20:25 — David777, Apr 10 '19 at 20:25

score 1 · Answer 2 · answered Apr 10 '19 at 09:19

1

It is $$\frac{V_{CC}-V{BE}}{V_0-V_{BE}}R_B-R_B=\frac{(V_{CC}-V_{BE})R_B-R_B(V_0-V_{BE})}{V_0-V_{BE}}$$

answered Apr 10 '19 at 09:19

Dr. Sonnhard Graubner

95,283

Thank you Dr Sonnhard, I understand now. – David777 Apr 10 '19 at 20:26

score 1 · Answer 3 · answered Apr 10 '19 at 09:20

1

$$ \frac{V_{CC}-V_{BE}}{V_O-V_{BE}}R_B-R_B= \frac{R_B(V_{CC}-V_{BE})}{V_O-V_{BE}}-R_B=\\ \frac{V_{CC} R_B-V_{BE} R_B}{V_O-V_{BE}}-R_B= \frac{V_{CC} R_B-V_{BE} R_B}{V_O-V_{BE}}-R_B\frac{V_O-V_{BE}}{V_O-V_{BE}}=\\ \frac{V_{CC} R_B-V_{BE} R_B-R_B(V_O-V_{BE})}{V_O-V_{BE}}=\\ \frac{V_{CC} R_B-V_{BE} R_B-V_O R_B+V_{BE} R_B}{V_O-V_{BE}}. $$

answered Apr 10 '19 at 09:20

Michael Rybkin

6,646
2
11
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Thanks Michael, I understand it now. – David777 Apr 10 '19 at 20:26

Help with fractions

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