45
\$\begingroup\$

I understand that in "saturation mode", a BJT functions as a simple switch. I've used this before driving LEDs, but I'm not sure I understand clearly how I got the transistor into that state.

Does a BJT become saturated by raising Vbe above a certain threshold? I doubt this, because BJTs, as I understand them, are current-controlled, not voltage-controlled.

Does a BJT become saturated by allowing Ib to go over a certain threshold? If so, does this threshold depend on the "load" that is connected to the collector? Is a transistor saturated simply because Ib is high enough that the beta of the transistor is no longer the limiting factor in Ic?

\$\endgroup\$
  • \$\begingroup\$ possible duplicate of How do BJT transistors work in a saturated state? \$\endgroup\$ – Leon Heller Apr 14 '11 at 18:21
  • \$\begingroup\$ That question is, "how does it work when it's saturated?", my question is, "how do I get it to be saturated?" \$\endgroup\$ – Mark Apr 14 '11 at 19:58
  • \$\begingroup\$ It's answered there. \$\endgroup\$ – Leon Heller Apr 14 '11 at 20:26
  • 1
    \$\begingroup\$ It's explained by the Ebers-Moll model of the transistor: ecee.colorado.edu/~bart/book/book/chapter5/ch5_3.htm \$\endgroup\$ – Leon Heller Apr 14 '11 at 20:53
  • 8
    \$\begingroup\$ That's not really helpful. I could become an expert in transistor theory, but then I wouldn't need to ask here... \$\endgroup\$ – Mark Apr 14 '11 at 20:55
12
\$\begingroup\$

Drive enough current into the base so that the base-collector junction becomes forward biased. How much current will depend on the type of transistor. 'saturation' has to do with how many of the charge carriers in the base region can make it into the collector region. Some will come from the base terminal, but many more will come into the base region from the emitter region. Beyond a certain amount of base current, there just won't be an increase in the available charge carriers that can cross the B-C junction.

\$\endgroup\$
  • 1
    \$\begingroup\$ Since saturation is a speed limiting factor with BJTs: Is forward biasing enough to negatively impact the turn-off time or should we approach $V_\rm{CEsat}$ for this to matter? \$\endgroup\$ – jpc Apr 15 '11 at 2:20
66
\$\begingroup\$

A transistor goes into saturation when both the base-emitter and base-collector junctions are forward biased, basically. So if the collector voltage drops below the base voltage, and the emitter voltage is below the base voltage, then the transistor is in saturation.

Consider this Common Emitter Amplifier circuit. If the collector current is high enough, then the voltage drop across the resistor will be big enough to lower the collector voltage below the base voltage. But note that the collector voltage can't go too low, because the base-collector junction will then be like a forward-biased diode! So, you will have a voltage drop across the base-collector junction but it will not be the usual 0.7V, it will be more like 0.4V.

Common Emitter Amplifier

How do you bring it out of saturation? You could reduce the amount of base drive to the transistor (either reduce the voltage \$V_{be}\$ or reduce the current \$I_b\$), which will then reduce the collector current, which means the voltage drop across the collector resistor will be decreased also. This should increase the voltage at the collector and act to bring the transistor out of saturation. In the "extreme" case, this is what is done when you switch off the transistor. The base drive is removed completely. \$V_{be}\$ is zero and so is \$I_b\$. Therefore, \$I_c\$ is zero too, and the collector resistor is like a pull-up, bringing the collector voltage up to \$V_{CC}\$.

A follow-up comment on your statement

Does a BJT become saturated by raising Vbe above a certain threshold? I doubt this, because BJTs, as I understand them, are current-controlled, not voltage-controlled.

There are a number of different ways to describe transistor operation. One is to describe the relationship between currents in the different terminals:

$$I_c = \beta I_b$$

$$I_c = \alpha I_e$$

$$I_e = I_b + I_c$$

etc. Looking at it this way, you could say that the collector current is controlled by the base current.

Another way of looking at it would be to describe the relationship between base-emitter voltage and collector current, which is

$$I_c = I_s e^{\frac{V_{be}} {V_T}}$$

Looking at it this way, the collector current is controlled by the base voltage.

This is definitely confusing. It confused me for a long time. The truth is that you cannot really separate the base-emitter voltage from the base current, because they are interrelated. So both views are correct. When trying to understand a particular circuit or transistor configuration, I find it is usually best just to pick whichever model makes it easiest to analyze.

Edit:

Does a BJT become saturated by allowing Ib to go over a certain threshold? If so, does this threshold depend on the "load" that is connected to the collector? Is a transistor saturated simply because Ib is high enough that the beta of the transistor is no longer the limiting factor in Ic?

The bold part is basically exactly right. But the \$I_b\$ threshold is not intrinsic to a particular transistor. It will depend not only on the transistor itself but on the configuration: \$V_{CC}\$, \$R_C\$, \$R_E\$, etc.

\$\endgroup\$
  • 1
    \$\begingroup\$ Absolutely brilliantly written, thank you very much. \$\endgroup\$ – Mark Apr 15 '11 at 4:23
  • \$\begingroup\$ Another look at it: There is minimum Vce voltage (which is usually given in datasheet) below which increasing base current/voltage will not result in decreasing Vce further. At which base current/voltage it will be achieved depends on load conditions. \$\endgroup\$ – mazurnification Apr 15 '11 at 7:26
  • 3
    \$\begingroup\$ The perfect answer to this question. Related: With a Schottky diode parallel to the B-C-diode, the voltage at the collector will not dive > 0.4 V below the base, but only approx. 0.3 V below the base, which is the forward voltage of the Schottky diode. Thus, the diode will keep the transistor form being deeply saturated and the turn-off event will be allowed to happen much faster. This is the theory why thing work as described in this answer: electronics.stackexchange.com/questions/15056/… \$\endgroup\$ – zebonaut Jun 11 '11 at 7:26
  • \$\begingroup\$ So in saturation, current is limited by the external collector resistor, and below saturation, current is limited by the gain of the transistor times the base current? \$\endgroup\$ – endolith Jul 15 '13 at 18:34
  • 1
    \$\begingroup\$ Quote: "So both views are correct". I cannot agree because - physically spoken - only one single view is correct: The BJT is voltage-controlled! It is not a problem to proove resp. verify this statement (without going deep into charged carrier physics). \$\endgroup\$ – LvW Jul 12 '15 at 12:10
7
\$\begingroup\$

BJT transistor will be saturated the moment the Ic will not follow the linear relation of:

\$I_c = HFE * I_b\$.

Thus all we have to do is to limit the Ic from reaching this value.

Since \$I_b\$ is determined by the value of the resistor connected to the base and the driving voltage on its other end, it is easy to force \$I_b\$ to any value. When \$I_b\$ is determined calculate the theoretical \$I_c\$ and set the \$R_c\$ to reduce it (say by 5-8) in order to enter the saturation zone And preventing it to follow the linear relationship.

For example: \$R_b\$ is connected to 5V and \$R_c\$ (just to make it interesting) to 12V. Supose HFE = 50. If we set \$R_b=5K\$ then

\$I_b = (5-0.5)/5K ~= 1mA\$

This implies that the \$I_c\$ will be \$1mA * 50 = 50mA\$. Now if we set the \$R_c\$ to be about 2K this will limit \$I_c\$ to less that 6mA, a value almost 10 times less than the linear range and the transistor will be saturated.

If using the transistor as a switch it is recommended to add additional resistor (10K) between the base and the ground (for fast switching and leakage preventing, provided the BJT is NPN type)

\$\endgroup\$
2
\$\begingroup\$

Saturation is when an increase in input does not produce a increase in output. In a BJT, this would be because output has reached its maximum current conduction.

The method I design with to ensure that a switching BJT in common-emitter mode is taken into saturation when conducting is...

Find in the BJT's data sheet its Ic(max) and hFE(min).

Calculate the required base current Ib as 5 x Ic(max)/hFE(min)

The 5 x is a personal 'fudge factor', allowing additional base current to ensure the BJT is pushed fully into saturation.

This presumes a simple case: a small BJT in common-emitter mode switching small (say <2 A) loads a low frequency (say <50 kHz) with a capable base current source. Otherwise there are further analogue conditions to be considered, such as if saturating the BJT will give good switching performance or if a MOSFET/etc. should be used instead. (That's beyond the scope of this answer, though.)

\$\endgroup\$
  • \$\begingroup\$ Do you mean hFE(min) rather than max? \$\endgroup\$ – Kevin White Dec 26 '16 at 0:37
  • \$\begingroup\$ @KevinWhite, yes I do, or yes I should do - have corrected this. Thanks very much and Merry Christmas :-) \$\endgroup\$ – TonyM Dec 26 '16 at 0:39
1
\$\begingroup\$

I know this is an old question, but a lot of people are still viewing it.

One other way to know if your transistor is in saturation is by looking at the ratio of \$i_C/i_B\$. This parameter is called the "forced beta." Forced beta can be thought of as the beta value required for the current state of the transistor.

If you find that the value of the forced beta is lower than the value of beta (\$h_{fe}\$), then you know you are in saturation, because in the active region you would be using the "full" value of the beta.

This way is useful when you don't know the value of \$V_{BE}\$.

\$\endgroup\$
  • \$\begingroup\$ What is the use of Vbe (saturation) in all this? I think I understand the use of Vce (saturation) though \$\endgroup\$ – quantum231 Nov 13 '16 at 18:15
1
\$\begingroup\$

It is worth noting that in the "real world" saturation is NOT a single well defined state. As you apply increasing base current \$V_{CEsat}\$ will continue to drop for a given collector current.

"Long ago" I used a bipolar transistor to switch a voltage divider. The saturation voltage of the transistor affected the output voltage of the divider. I used a high gain transistor (probably BC 817-40 with a \$\beta\$ of ~= 400) and a base current drive of about ten times the collector current - i.e. "forced Beta" of 0.1. This reduced \$V_{CEsat}\$ to a few mV compared to the 10's of mV usually seen at low \$I_{C}\$.

A beta of 0.1 would seldom be useful or acceptable, but in this case it was.

Nowadays I would use a suitable low \$R_{DSon}\$ MOSFET for the switch.

\$\endgroup\$
1
\$\begingroup\$

There are two way of bringing transistor in saturation mode:

1)Use of Rc resistor: we can calculate max current (Ic) by assuming Vce = 0. Ic(max)= Vcc/Rc

you can find corresponding base current (Ib) = Ic/(beta).

Transistor will be in saturation if you apply base current greater then above calculated base current

2)By using rated saturation current (Datasheet): You can apply base current that tends to produce larger collector current then stated in datasheet

\$\endgroup\$
0
\$\begingroup\$

Saturating a transisor also depends on its datasheet. You must find a graph with a nonlinear \$h_{FE}\$ including \$V_{BEsat}\$ and \$V_{CEsat}\$ and use that \$\beta\$ for your calculations.

Now you can easily calculate the base current which is \$I_{C} \over h_{FE}\$ and the base needed resistor which is

$$ R_{B} = { (V_{B} - V_{BE}) \over I_{B} } $$

Look for another graph which shows the influence of \$I_{C}\$ upon the DC gain.

Be careful so this gain is the one you want.

\$\endgroup\$
-4
\$\begingroup\$

The NPN BJT will enter the saturation mode when Vcb will below some value. Sedra&Smith use value of 0.4V, but this will be device dependent.

Though I have no idea why you would whant to use BJT as a switch. MOSFETS are better suited for this task.

\$\endgroup\$
  • 3
    \$\begingroup\$ Because I have BJTs, and I don't have any MOSFETs. I also understand BJTs better than I understand MOSFETs. \$\endgroup\$ – Mark Apr 14 '11 at 20:13
  • 4
    \$\begingroup\$ No, not if there isnt more current in the base than in the collector divided by the current gain. And MOSFETs are not always better \$\endgroup\$ – Martin Apr 15 '11 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.