# Homework Help: BJT Current Mirror

1. Sep 20, 2015

### Zondrina

1. The problem statement, all variables and given/known data

1. Given the above circuit, derive an exact equation for $I_{out}$ in terms of $I_{REF}, V_{BE1}, V_{BE2}, \beta_1, \beta_2, R_{E1}, R_{E2}$ and $V_{CC}$. Do not make any assumptions.

2. Simplify the equation from part 1 by assuming $V_{BE1} = V_{BE2}$ and that both $\beta$s are very large.

2. Relevant equations

3. The attempt at a solution

I am unsure how to attempt this problem.

I started by trying to find an expression for $I_{REF}$, which I found to be:

$$I_{REF} = \frac{V_{CC} - [ V_{E1} + V_{BE1} ]}{R_{REF}} = \frac{V_{CC} - I_{E1} R_{E1} - V_{BE1}}{R_{REF}}$$

Then I thought about finding an expression for $I_{out}$:

$$I_{out} = \frac{V_{CC} - V_{out}}{R_L}$$

I can't really see a way to relate these equations. I'm not even sure this is the right approach.

I have seen another approach for a different kind of current mirror that uses the equation $I_C = I_S e^{\frac{V_{BE}}{V_T}}$ to obtain $V_{BE1}$ and $V_{BE2}$. Then it relates $I_{out}$ and $I_{REF}$. The only difference was there was no emitter resistor in the first transistor, but there was still an emitter resistor in the second transistor. I'm wondering if this approach should be used instead.

Any help with getting this one going would be appreciated.

2. Sep 20, 2015

### Zondrina

I was thinking of a different approach. What if I wrote a KCL equation at the $C1$ node to obtain:

$$I_{REF} = I_{C1} \left( 1 + \frac{1}{\beta_1} \right) + \frac{I_{C2}}{\beta_2}$$

Where $I_{C2} = I_{out}$.

I have been messing around with the above equation as well as other ideas, but nothing seems to stand out as the correct answer. I will continue to try though.

3. Sep 21, 2015

### Zondrina

I was given a hint:

I will try to perform the steps outlined by this hint. First the hint says to use KCL to find $I_{E1}$, so:

$$I_{E1} = I_{C1} + I_{B1} = I_{C1}( 1 + \frac{1}{\beta_1})$$

Now we need to use KVL around the bottom loop:

$$V_{BE2} + V_{E2} - V_{BE1} - V_{E1} = 0$$
$$V_{E2} = V_{BE1} - V_{BE2} + V_{E1}$$
$$I_{E2} = \frac{V_{BE1} - V_{BE2} + I_{E1}R_{E1}}{R_{E2}}$$

The hint now says to obtain $I_{C2}$ by multiplying by $\alpha_2 = \frac{\beta_2}{1 + \beta_2}$:

$$\frac{\beta_2}{1 + \beta_2} I_{E2} = \frac{\beta_2}{1 + \beta_2} \frac{V_{BE1} - V_{BE2} + I_{E1}R_{E1}}{R_{E2}}$$
$$I_{C2} = \frac{\beta_2}{1 + \beta_2} \frac{V_{BE1} - V_{BE2} + I_{E1}R_{E1}}{R_{E2}}$$

The hint then mentions base currents, which I don't have in the expression. So I guess nothing needs to be done in that sense. Now using the fact $I_{C2} = I_{out}$ and subbing in for $I_{E1}$, we get:

$$I_{out} = \frac{\beta_2}{1 + \beta_2} \frac{V_{BE1} - V_{BE2} + I_{C1}( 1 + \frac{1}{\beta_1})R_{E1}}{R_{E2}}$$

Now using the simplifying assumption $V_{BE1} = V_{BE2}$ and taking both $\beta$s to be very large, we get:

$$I_{out} = \frac{I_{C1}R_{E1}}{R_{E2}}$$

Which I hope looks okay.

EDIT: I still need to relate $I_{out}$ to $I_{REF}$. Applying KCL at the node $C1$ yields the equation given in post #2.

Now using the simplifying assumption, we see $I_{C1} = I_{REF}$ from this KCL equation.

Therefore the final result would be:

$$I_{out} = I_{REF} \frac{R_{E1}}{R_{E2}}$$

Last edited: Sep 21, 2015