# Transistor help

Discussion in 'Electronic Basics' started by Wong, Nov 13, 2003.

1. ### WongGuest

Hi,
I just read a tutorial about biasing a transistor. It says that we
cant use the formula Ic=Beta.Ib when it is in saturation region but in
active region we can use that. But in another tutorial, I found that
the transistor is in saturation region as a switch but it applied the
above formula, so which one is true ?

2. ### John PopelishGuest

They might both be, depending on the definition of beta. If beta is
defined as the ratio of collector to base current, and you approximate
beta by a constant, then the first one is better, If you include in
the definition that beta changes as collector voltage approaches and
falls below base voltage, then the second one can still be right.
Transistors in saturation are often driven to a forced bets that is
well below the beta you see when there is plenty of collector
voltage. For instance, if a transistor exhibits a beta of about 100
over a large range of collector voltage, but you saturate it by
driving it with a base current equal to 1/10 of the collector current,
you are forcing it to the beta = 10 operating point (forcing the beta
down to 10, 1/10th of its normal beta). This is a measure of how
thoroughly saturated the transistor is.

3. ### MacGuest

When you use a transistor as a switch, you normally want to saturate it.
So you calculate the maximum base current you could possibly need, then
multiply by 10 and design for that current.

For example, suppose I have a transistor with a gauranteed minimum DC
current gain of 100. I am using it as an inverter, and I have a 1k pullup
on the collector. I know that I need to drive it down to 0.8V to get a
valid LVTTL low.

So, I can go through the calculations to find the minimum collector current:

(3.3V - 0.8V) / 1k = 2.5 mA

This means that the minimum base current is:

2.5mA / 100 = 0.025mA

So, in theory, I would be safe if I can put 0.025 mA into the base. But I
want to be really safe, so I'll shoot for 0.25mA into the base. So that is
why I use the formula, to find the minimum current, which I then multiply
by 10.

Anyway, to finish, since the base will be at about 0.8 volts, and my input
voltage will probably be 3.3 volts, I need (3.3 - 0.8) / 0.25mA = 10k in
series with the base. Since I have a ten-fold safety margin, I don't need
to worry about 5% resistors or inputs that might be substantially less
than 3.3V.

HTH

Mac
--

4. ### Watson A.Name - Watt Sun, Dark RemoverGuest

So put a milliammeter on the collector of a transistor and a
microammeter on the base. You will see that as you get closer to
saturation, it takes more and more increase in base current for the
same increase in collector current. I.e. the Ic to Ib ratio drops.
Instead of 160 for example, it drops to as low as 10 when fully
saturated.

You don't have to use a milliammeter. You can measure the voltage
across the resistors, such as a 1k which gives 1 milliamp per volt.

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5. ### GPGGuest

2SD1071, Darlington power switch ,300V, 6A, 40W, B1500, TO220

6. ### WongGuest

OK, one more question. Why we need to use as switch in saturation
region and then, amplifier in active region ? Is this related to the
"Beta Stabilization" ?

7. ### John PopelishGuest

If you try to set the base current to turn a switch on to 1 volt
collector to emitter, you will have a hard time making two units that
act the same. Some will drop 1/2 volt, and some will drop volts.
Even a change in temperature will alter the drop when the switch is
on. Double or quadruple that base current, and thousands of units
will saturate with the same (smaller) drop to within 100 mv or so of
each other. If that switch is delivering power (as in a switching
supply) it will also produce only a fraction of the heat because of
the lower voltage drop. All that for a bit more base drive.