I'm reading the Art of electronics and trying to calculate current and voltages of the following scheme:

^{simulate this circuit – Schematic created using CircuitLab}

I wrote the current as the letters with indexes (I1-I9).

So I got the following equations:

```
I1 = I5 + I6
I2 = I3 + I4
I9 = I6 + I4
I8 = I7 + I5
```

Also we know that:

```
I1 = 5/1000 = 0.005A
I2 = 5/1000 = 0.005A
```

Then I tried calculating the first transistor's current (on the left side):

```
I7 = V/R = (5 - 0.6)/25K = 0.000176A
```

Then I used the general rule that:

```
Ie = b * Ib, where b=beta coefficient, Ie - emitter current, Ib - base current
```

So we got something like:

```
I8 = b * I7
```

Lets say b = 100, then

```
I8 = 0.000176 * 100 = 0.0176A
```

From `I8 = I5 + I7`

we get that:

```
0.0176 = 0.000176 + I5
```

giving us:

```
I5 = 0.017424A
```

I1 = I5 + I6:

```
0.005 = 0.017424 + I6
```

**I6 = -0.012424A**

And here I don't understand why its minus.

Whenever transistors and diodes comes into the circuits I can't estimate the voltages and the currents. I think that I'm missing something there.

## Best Answer

You are forgetting that you can overdrive a base so that it goes into saturation. This is desirable to get the transistor fully turned on.

Your calculation for I7 is correct:

$$ I_7 = \frac {V}{R_3} = \frac {5 - 0.6}{25k} = 0.176~mA $$

You then calculate the

maximumcollector current using \$ I_8 = \beta I_7 \$ Lets say \$ \beta \$ = 100, then \$ I_8 = 0.176 * 100 = 17.6~mA \$.Now look at R1. The maximum current through it is given by

$$ I_1 = \frac {V}{R_1} = \frac {5 - 0.6}{1k} = 4.4~mA $$

Clearly the maximum possible value for \$ I_8 \$ is this 4.4 mA + 0.176 mA from \$ I_7 \$.

The calculations indicate that we could reduce the base current into Q1. In practice we want to ensure that the drive is adequate to cover variations in \$ \beta \$ due to production spread or even transistor substitution. Drive it hard and make sure it's fully 'on' is the normal approach.