# How can I calculate the base current of a transistor to get a specific emitter-collector voltage?

For this circuit I am asked to calculate how much base current I need to supply to the transistor to get a Vce = 220.745 mV. The thing that has me stuck in this is that the problem does not provide me with a (betha) value.

Is there a way to calculate the base current without (betha)?

If not I will use (betha) as 100, since the problem tells me to use a Vce I calculated in a previous problem using (betha forward) as 100 and (betha reverse) as 0.01. As you can see I already calculated the collector current (Ic) using KVL. Since the collector-emitter saturation voltage is approximately 0.2 V, I assume that my transistor is in saturation.

When the transistor is in saturation, does Ic = (betha)(Ib) still apply?

• The beta is not provided because the transistor is in saturation meaning that beta will be much smaller compared to when the transistor is not in saturation. Due to the transistor being in saturation the base current will be much larger, thus making beta appear lower. So beta = Ic/Ib does still apply but beta has a smaller value ! And you don't know what that value is. – Bimpelrekkie Apr 1 '16 at 6:42
• Is there a way to calculate beta without the base current? Or do you know any other way of calculating the base current value? – Rodolfo David Apr 1 '16 at 21:13
• No, in saturation you need to know Ib and Ic, then you can calculate beta. Or you choose a certain beta (this is possible since we're in saturation) then from Ic you can calculate Ib and then set the base current to this value. – Bimpelrekkie Apr 2 '16 at 12:54
• I doubt that there is even a (not too complex) mathematical way to solve your question. In practice you would consult a (measured) graph in the BJT's datasheet to get an indication of behaviour in saturation. A BJT simulator model can also solve it of course but then the BJT is described by many more parameters than you were given for this assignment. So if there is a (reasonably simple) solution then I'm curious how that would work. I can't remember seeing one in the 25 years I've been dealing with BJTs. – Bimpelrekkie Apr 2 '16 at 12:59
• Are you sure that you understand the question? From your picture, the question seems to be, "Does Ic = beta Ib still apply". Your title asks how to calculate the base current, which is something elses. At one level, the answer is obvious: yes. This is because beta is defined as the ratio of collector and base currents. What you need to realize is that the value of beta depends on the current levels and Vce involved. – WhatRoughBeast Dec 7 '18 at 17:23

There exists a method for determining the saturation voltage $v_{CESAT}$ for a given current $i_B$. With a few assumptions for the forward and reverse current gain $\beta_F$ and $\beta_R$ , we can determine the current $i_B$ from the saturation voltage $v_{CE}$.

I've referenced chapter 5 page 233 in Microelectronic Circuit Design by Jaeger and Blalock ISBN 978-0-07-319163-8

Assume that both junctions are forward biased. (This is required to approximate $i_B$ and $i_C$ in order to arrive at this equation).

$v_{CESAT} = V_Tln( \frac{1}{\alpha_R} \frac{1+\frac{i_C}{(\beta_R+1)i_B}} {1-\frac{i_C}{\beta_Fi_B}})$ for $i_B > \frac{i_C}{\beta_F}$
This equation is only valid in the condition condition that $i_B > \frac{i_C}{\beta_F}$

This equation is important and highly useful in the design of saturated digital switching circuits. For a given value of collector current, [this] can be used to determine the base current required to achieve a desired value of $v_{CESAT}$.

With $V_{CE} < 0.7V$, it can be assumed operation is in the forward saturation region. For your situation with $V_{CE} = 220.745mV$, we can simplify to $6835 = \frac{1}{\alpha_R} \frac{1+\frac{i_C}{(\beta_R+1)i_B}} {1-\frac{i_C}{\beta_Fi_B}}$

But, we still need to know $\beta_R, \beta_F,$ and $\alpha_R = \frac{1}{\beta_R+1}$

Choosing arbitrarily, and within reason, $\beta_F = 50$ and $\beta_R = 1$ we can solve for
$i_B = 0.39414mA$ with $i_B > \frac{i_C}{\beta_F} = 0.39116mA$
And with a similarly reasonable $\beta_F = 100$ we find
$i_B = 0.19850mA$ with $i_B > \frac{i_C}{\beta_F} = 0.19558mA$

That is great and all, but how do we know these results make sense? If you are computing $V_{CESAT}$, you expect a small value close to zero, even $0.06V$ is sensible. Plugging the currents back into the original equation in both cases it yeilds $0.220745V$

Some methods to check the work cannot be used without a value for $I_S$. Also note that the above equation is independent of $I_S$.
Final note, the above work was done assuming npn transistor, but pnp can be computed as well.