Transistor Saturation

the

me

You need to check the rating of each transistor. In general, for small-signal transistors such as BC548 etc, limiting the base current to about 500uA to 1mA is sufficient. To turn on a BC548 from a 5V supply, (or chip), a 4.7K or 10K resistor in series with the base will do the job. The absolute minimum current required to turn on the transistor is the collector current divided by the current gain.

... Winfield

Sorry, I should have added, the typical base-emitter voltage of a small-signal transistor is about 0.6V. For a higher power transistor or a Darlington pair, this can be as high as 1-2V. This voltage is like the forward voltage drop of a diode. If the voltage applied to the base exceeds this voltage, it is a good idea to limit the input current to a known value. (5V-0.6V)/4.7K = 936uA.

... Winfield

You're missing the point.

The best transistors for this job are *designed* as switching transistors. They have low intrinsic resistance to reduce the Vce(sat).

Also, it's not about base *voltage* but base *current* !

Check out the 2N440x series.

Graham

Using the minimum Beta of the transistor (from the datasheet), calculate the required base current needed to pull the collector voltage down to the voltage at the base. Assuming a common emitter configuration, the voltage at the collector is equal to the (applied) collector voltage minus the IR drop in the collector resistor. Then multiply this current by 10, for a 10x overdrive factor as it is called in the Art of Electronics. Knowing the base voltage (5v in this case) and the required base current, you can calculate the needed base resistor to provide this current.

One way to find the base drive required for saturation is to create a load-line for the circuit with Ic on the vertical axis and Vce on the horizontal axis. Where this load line intersects the vertical axis is the saturation current of the BJT, for that circuit, and where it intersect the horizontal axis is cutoff. If you take value of Ic found on the vertical axis and divide by the beta of the transistor this will give you the value of base current needed to saturate the transistor. You will then need to find a value of base resistance needed to obtain this current at 5 volts.

The easiest way, to do essentially the same thing, is to find the collector saturation current by dividing the power supply voltage (Vcc) by the collector load resistance. Divide this current by the typical Beta or Hfe of the transistor. This give the base current necessary for saturation.

Your goal should to find the lowest base current required for saturation. The brute-force method is to use an overwhelming large base current but this is wasteful.

The best way to know if the transistor is in saturation is to measure a Vce of 0.3 volts or less.

Dorian

--- The rule of thumb is to divide the collector current by 10 and to force that current into the base.

Doing it this way will cause the transistor to have what is called a "forced beta" of 10 and will very nearly always assure that the transistor will go into saturation.

For example, assume a transistor with a collector-to-emitter saturation voltage (Vce(sat)) of 0.3V with 100mA of collector current, and also assume that the base-to-emitter diode (Vbe) will drop about 0.7V, worst case, when the transistor is in saturation.

With a 15V collector supply and the signal into the base being either 0V or 5V, your circuit will look like this: (View in Courier)

Vcc +15V | [Rl] | +----Vce(sat) | 0.3V C

0/5V>--[Rb]-----B / E Vbe(sat) | 0.7V | GND

Since, when saturated, the collector voltage will be about 0.3V, in order to get 100mA through the load we can calculate the value of the load resistor like this:

Vcc - Vce(sat) 14.7V Rl = ---------------- = ------- = 147 ohms Il 0.1A

Then, since the collector current will be 100mA, we'll want to turn on the transistor with 10mA through the base. Since we want the transistor to be turned on when the input signal is at 5V and we have a Vbe of about 0.7V, the base resistor will need to be:

Vin - Vbe(sat) 5V - 0.7V 4.3V Rb = ---------------- = ----------- = ------- = 430 ohms Ib 0.01A 0.01A

It's also possible to use the same procedure forcing a beta higher than 10 and, in the process, saveing some base current. Notice, however, that on most data sheets Vbe(sat) and Vce(sat) are given for a forced beta of 10.

-- John Fields Professional Circuit Designer

No particular designations, then? Well, it's not terribly important so long as they aren't very ancient.

Since they are NPN, I assume that you mean to use them with emitter connected to ground and the 15V load (whatever it is) between the collector and the +15V supply side.

Hobby play?

Well, saturation in the case I mentioned above is just a matter of forcing the collector voltage close enough to ground voltage (0V) so that the base-to-collector junction begins to be forward-biased. And the current through the collector needed to do that will depend on the load you have connected, not upon the transistor. So which exact type of transistor you use won't really matter on this narrow point. Do you see why?

Yes. When the current passing by way of the transistor's collector and through the load causes the load itself to drop almost all of your supply voltage, then the transistor will be "all the way on." When that happens, the load is taking up almost all of the voltage available, leaving only a smidgen left to support the collector to emitter voltage of the transistor. For design purposes, that remaining smidgen is usually estimated to be just a few tenths of a volt, perhaps somewhere between .1 and .4 or so.

So just estimate the collector current needed by taking your 15V, subtracting a small "smidgen" from it for the transistor, then dividing that by the load resistance of what you are connecting to the collector and supply. Let's say it is 1k ohm load and you want to estimate a saturation Vce of .2V, you'd calculate (15-.2)/1k as your collector current needed for saturation. As you can see, with a 15V supply in hand, changing your saturation estimate by a tenth or two, one way or another, won't seriously impact your calculated saturation current. The actual voltage for Vce that you will get will depend on other particulars (such as the exact device used and the base current), but it's not important at this stage to worry about the transistor data sheet.

It's not uncommon to then blindly estimate the needed beta as 10, for saturation. That's because most modern transistors are pretty much going to be doing as much as you can expect from them as a switch when you supply 1/10th of the collector current as base drive. There are a few exceptions. Actually, it turns out that in many cases you can drive them with much less base current than this rule of thumb suggests. Maybe 1/30th or even 1/50th. But almost any transistor will get there with 1/10th. So you might not even need to look at the data sheet for even this value, if you are willing to accept 1/10th of the collector current as base drive.

Also, it's not uncommon to estimate the base voltage as .7V for this purpose. The exact base voltage will depend largely on the base current used (not entirely, as the collector current does play a role, but almost.) But since it varies up and down by roughly 60mV for a

10-fold change in base drive current, the exact value won't be all that far away. If your collector current is large, then the base drive current will be large too and the base voltage may be nearer .8V, .85V, or even .9V. But for design purposes, being off by one tenth of a volt won't kill your design. And .7 is a good place to be for many cases.

So far, no need yet for a data sheet to make a reasonable design.

(1) Compute collector current from the supply voltage, the load resistance, and a blind estimate of Vce, as (Supply - Vcesat) / load.

(2) Compute needed base drive current as 1/10th the value computed in step (1).

(3) Use .7V as a blind estimate for the base voltage. You can modify this per Vbe=25.865mV*(1+ln(Ib/Is+1)) @ 27C, where Is is a value found for the simulation model of the transistor or just taken to be 10^-14 or so, blindly. For example, if Ib (base drive current from step (2)) is about 100uA, then you could compute 25.865mV*(1+ln(1+100uA/10fA)), which works out to about 0.62V. This will be closer, probably, than the .7V. But the .7V won't steer you much wrong, anyway.

(4) Estimate the actual control voltage. You said 5V, but unless this is a power supply (and you've said it isn't), there is some internal impedance and supplying the base current calculated in step (2) will drop the voltage a little. Or, if this is a standard TTL output, you cannot always rely upon the voltage being much better than, say, 4V. What it is exactly will depend on your output type and what is connected to it other than the transistor. But let's use an estimate of 4.5V, as an example.

(5) Compute your base resistor used to limit the base current to the value you computed in (2) on the assumption that your 5V control voltage is low impedance enough that it can yield at least the voltage estimated in step (4) or 4.5V. This is done by dividing the voltage that the base resistor needs to drop by the estimated base drive current. Something like this: (4.5V - .7V)/100uA = 38k. Use 39k, as a standard value.

Now, suppose you are wrong about the base drive current using the estimator of 1/10th to compute it. Suppose the actual base drive current for the actual transistor works out to, say, 1/30th instead. Well, the voltage drop across the base resistor will be a lot less if that happened and that would mean that more of the 4.5V control voltage would appear at the base of the transistor. Just for a moment, let's assume that 1/3rd of our computed current was actually needed -- in that case, the base resistor would drop 1/3rd of the 3.8V we earlier used in our computation for the resistor value, or about

1.3V, instead. That would, we might guess, cause 3.2V to be at the base of the transistor. But that won't happen. That's because, as I mentioned before, the base voltage goes up by 60mV for each 10-fold increase in base current. So if it really were 3.2V, that could only happen if there were (3.2V - .7V)/60mV or about 4.2 orders of magnitude change in the base drive current. That would be more than 10,000 times. And if you tried to send that much current through the base resistor we computed, then the voltage drop would be impossibly high. So what really happens is that even if the saturation beta of the transistor _might_ be better estimated as 1/30th, than 1/10th, having designed things with the idea of 1/10th means that it will be very close to 1/10th, anyway. The actual transistor you use may impact just a tiny bit the actual current flowing through the base resistor, but it will be _very_ close to the designed value because of this exponential behavior of base current versus base voltage. So if you use 1/10th as your rule of thumb in your design, it will come out pretty darned close no matter what transistor you actually use.

So no data sheet, still.

You _may_ still want to take a look at the data sheet. One thing you need to be sure (and this is more important if you had a supply voltage larger than 15V) of is that the transistor can "stand off" your supply voltage when it is off. If there is no current flowing through the load, then the load doesn't drop any voltage and the collector will be exposed to the supply voltage. You want to be sure that the transistor can support the Vce difference without breaking down. For most transistors, given my hobby experience, you can usually start worrying when the supply voltage exceeds about 25V or so. But there are enough transistors with lower voltages that it's worth a look just to be sure.

Jon

Thank you all for the replies. I had a chance to play with a resistor decade box to the transistor base and applied the information I got here in this thread. Everything worked just like you said. I found these transistors (assorted and all T0-92 pkg) very forgiving and work great when using what I learned here. And they only cost 4 cents each! Thanks again everybody, Bart

On a final note, you may or may not have noticed that one of the things you are doing, is in effect desiging the circuit so that its performance is dependant on the commponents connected to the transistor, rather than the transistor itself.

For example, if you were using the transistor in the active mode instead of cutoff and saturation, it would (most likely) be desirable that the gain depend on the resistors used in the circuit. The reason for this is that the defining parameters of the resistor are much easier to control than the intrinsic paramaters of the transistor, such as the Beta.