There are many places we can begin looking at this.

We know that the collector current is 2 mA, and that \$V_{CE}\$ is 5V. Since the emitter is at ground, the collector must be at 5V, and so the voltage across \$R_C\$ is therefore 10V: the difference between 5V and 15V. So \$R_C\$ must be \$10V/.002A = 5K\Omega\$.

Next, \$R_2\$ and \$R_1\$ which span a voltage from \$-15V\$ to \$+5V\$ must form a voltage divider such that the top of \$R_1\$ is at 0.7V (\$V_{BE}\$). Hint: the voltage divider spans a range of 20V, and the 0.7V transistor base voltage is 15.7V above the bottom of the voltage divider.

This approach assumes that we can ignore the base current because it is small. Often when analyzing transistor circuits we can do that, but not in this case because \$R_1\$ is such a high resistor. The voltage divider is not "stiff" at all with regard to the resistance in the base circuit of the transistor (which has no emitter resistor at all).

A more exact answer requires that we account for the base current. The transistor is carrying only 2mA of current, and so is nowhere near hard saturation, and so the base current is only 0.02 mA, or 20 micro-Amperes (2mA divided by \$\beta\$).

Determine how much current is flowing through \$R_1\$ from its resistance, and 15.7 voltage. The current flowing through \$R_2\$ is the sum of the transistor base current (.02 mA) and the current through \$R_1\$. Knowing the current through \$R_2\$ and the voltage across it, we can calculate its resistance.

There is a sign error in the 1st equation. The node voltage at the base is

$$V_B = -I_B\cdot R_B$$

so the correct KVL equation is:

$$-I_B\cdot R_B = V_{BE} + I_E \cdot R_E + \left(-V_{EE}\right)$$

The 2nd equation does not have the sign error.

If you do not have the value for \$V_{EE}\$, all you can do is provide the answer in terms of \$V_{EE}\$. Deriving one of the DC bias equations, as you have done, may be all that is required.

Also can you please show me the scheme of the common base transistor
in this case?

It isn't clear to me what you're asking for since the canonical common base circuit schematic can be found in almost any transistor circuits textbook and is easily found with a Google search.

## Best Answer

The link to the relay is broken, but it's for you to look up anyway.

You already know the transistor can handle 100 mA, so the question is whether the relay will require more than that. You say the relay only requires about 21 mA, which is obviously less than 100 mA by a good margin, so I don't see what you are actually asking.

You say you'd be powering the relay from 3.8 V. The transistor can handle 45 V, so that is clearly within its capabilities. That is again so obvious it's hard to see what you are actually asking.

You have to make sure that your relay can really operate from 3.6 V (the transistor will eat up about 200 mV when on) coil voltage, and that it really does draw less than 100 mA when it does. This again should be obvious, coming directly from the relay datasheet.

Two things that are less obvious to watch out for are: 1 - make sure to put a reverse polarity diode across the relay coil. 2 - Make sure the transistor can dissipate the power. That should be a no-brainer if all the current thru it comes when it is saturated. Since that can't be more than 100 mA and saturation is probably (your job to check) 200 mV or less, it can't dissipate more than 20 mW in that case.