# Homework Help: Solving simple transistor circuit

1. Mar 28, 2016

### davidbenari

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

Determine IEQ ICQ IBQ

My problem is that this circuit is overdetermined and cannot be solved. (See below)

2. Relevant equations

$\beta = 75$

$I_E=76 I_B$

3. The attempt at a solution

The current IB is calculated as $I=\frac{1}{500}=.002mA$

The current $I_E$ is $I_E = 76*.002 = .152 mA$

Similarly $I_C= 75 * .002 = 0.15 mA$

The voltage $V_{CE}$ is calculated by knowing the current in the emitter and starting upwards from the -3V node, in other words

$V_{CE}=3-(-3+4.8*.152)=5.27V$

Knowing this we see that there is a contradiction. The voltage drop across the BE is not .7 .

Is this a sound way of proving that the BE voltage doesn't HAVE to be 0.7?

2. Mar 28, 2016

### phyzguy

Typically you don't know the beta of the transistor very accurately. A better assumption is probably that Vbe = 0.7 V, then calculate what the voltage is across the 4.8 kOhm resistor, then calculate IEQ, then beta. Beta appears to be significantly higher than 75.

3. Mar 28, 2016

### davidbenari

Oh okay. Its just I was given the value of beta.

What I've noticed is that BE voltage is always taken as 0.7V (or similar), however in saturation mode when both BC and BE are forward biased, BE is 0.7 while BC isn't necessarily 0.7 (I've seen 0.5 in problems). This seems weird to me since forward biased is considered as 0.7 or above. Is this normal?

4. Mar 28, 2016

### Staff: Mentor

A base-emitter voltage of 0.7 V is a handy approximation suitable for typical design and analysis work. But it is just a "rule of thumb" sort of approximation, pretty good for silicon transistors under typical operating conditions. Not so good at all for germanium or other types of transistors.

The base-emitter junction is essentially a diode, and as such its forward voltage will vary like a diode; the IV curve is not a perfect vertical line at 0.7V. You may see forward voltages between 0.6 and perhaps 1.3 or so volts under different circumstances. Look up the "diode equation" .