Understanding Vce Voltage in BC337 Transistor Circuit Design: Analysis and Interpretation

Question

When designing simple circuits with a transistor, should the voltage drop across the CE junction, i.e. the Vce voltage, be taken into account? If it is negligibly small, does it make no sense to include it in the calculations? For example, let there be a system in which 10 mA flows through the...

Anonymous · Accepted Answer

A larger Ic cannot flow because all Vcc has already been used, and the current is not limited by the transistor but by Rloads: I=(Vcc-Vled-Vcesat)/R ?(Vcc-Vled)/R - because Vcesat is small compared to Vcc. Incorrect scheme. LED should be in series with Rc only. RC is necessary. For the saturation state, we do not look at the beta value, but assume a much lower beta - according to the manufacturer's recommendations - p., the Collector Emitter Saturation Voltage table is given at Ic/Ib=10. From practice it is known that this value can be increased to 20 (Vcesat will increase slightly). Generally - the beta for entering the state of saturation must be taken much lower than the catalog one, i.e. Ib much higher than it would result from the catalog Ic/beta. Figure 5 - Vcesat=f(Ic), with Ic/Ib=10 - do you understand that? From this graph, Vcesat= about 0.05V, so Ic=(Vcc-Vled-Vcesat)/Rc=19.9mA. this difference of 0.1 mA matters to you? - because it's 0.5%, and the resistor has a tolerance of 5%. And even so, include the Vcesat value in your calculations. Because you don't understand the concept of saturation - you confuse current with voltage. this concept means that the transistor is saturated (with current carriers) and maximally open, so no more current can flow (it does not depend on the transistor - its control) - only on the power supply and load. This graph shows what current Ib needs to be applied to the base to drive the transistor into saturation (to get low, little changing Vcesat) - for different Ic. There you can see that the larger Ic, the smaller the beta and the more difficult it is to bring the transistor into saturation. Guaranteed parameters are given in the table, typical parameters in the charts (unless otherwise stated).

Xantix · Accepted Answer

There is. The graph is signed as Saturation Region - so it's definitely about saturation. And the saturation state itself does not mean that the maximum current flows through the collector. Wrong thinking - the collector current can increase, but not as a result of increasing the current Ib, but only as a result of a decrease in the resistance Rc. No - without a resistor, the diode will cook. Resistor Rc is needed to determine the assumed collector current at saturation. Wrong. You count it as if the transistor were an amplifier. And in this setup, it's the key. So it works two-way. If you want to put the transistor in saturation, you use the rule Ib>0.1 Ic and for this dependence you set the resistor Rb. Assuming Ic=20 mA, you must add a resistor Rb so that the current Ib is more than 2 mA. If Vcesat is negligibly small compared to Vcc then for practical and undemanding calculations you can neglect it. However, if the result is to be very precise then you must take Vcesat into account regardless of its value.

Xantix · Answer

But what arrangements? Amplifier, key/switch or something else? Depends on what you calculate and for what needs? Generally, each of the graphs interests us and each has its own application.

ety · Answer

Let's assume that we control several LEDs from a microcontroller pin. The current efficiency of the microcontroller pin is small, so we use a transistor.

Xantix · Answer

So the transistor as a key. Well, it seems to me that in this case the Vce voltage waveform itself does not interest us that much. The transistor works in two states in such a solution - it is either saturated or clogged.
In the clogged state, as you know, Vce=VCC and Ic=0
When in saturation, Vce is equal to Vcesat (assuming the transistor is fully saturated).
Therefore, for the case you have given, we are interested in graph number 4 - because it tells us how the Vce voltage behaves when the transistor is in saturation.
The value of Vcesat is important, among other things, because it determines the dissipation power dissipated in a saturated transistor (according to the formula Psat=Ic*Vcesat).
P.S. And most importantly: in a bipolar transistor there is no such junction as a collector-emitter junction - there are only base-emitter and collector-base junctions .

ety · Answer

Mhm ... Well, now I looked again at the construction of the npn transistor. The n layers are separated by a p layer, so electrically there is no direct connection (junction) between the collector and the emitter.

As for the thread. So graph 4 assumes that the transistor is saturated and a larger current Ic cannot flow? I've always understood the saturation state more or less in this way: the base current is so large that the collector current can no longer increase, the transistor is not able to pass beta-times higher current. If so...

Let's assume that we have some simplest arrangement, so completely pointless, as long as there is some basis for discussion.

Let it be like this:

Usup=12V - supply voltage
Rb - resistor regulating the base current
Rc - resistor for the LED (the question is whether it is needed?)

Rc gives it so that there will be a voltage drop across it, unnecessary voltage. Suppose this is 12V, and on the diode let's assume Uled=2V. The voltage drop across Rc should be 10V. I just don't know if it would work without it too? We're assuming we're using this BC337 and the note I linked.

We want the collector current Ic to be 20mA. In addition, the Ube voltage drop is assumed to be 0.7V (I don't see it in the note). I guess that's all.

First we calculate Rb. In Fig.3 we have a gain graph. For Ic=20mA, the gain is about 180. So Ib=20mA/180=111uA. Let be Ib=0.1mA.

Now from Ohm's law:
Ub=12V-0.7V=11.3V
Rb=Ub/Ib=11.3V/0.1mA=113k?

Now we calculate Rc. Exactly. The voltage drop across Rc should be 10V. What about the drop between collector and emitter? Because there will be some?

Without the drop, Uce (or Vce) would be:

Rc=Uc/Ic=10V/20mA=500?

Well cool. First of all, what about Uce?
The second thing - what is with the graph you mentioned. This Fig.4. Because I don't think I'm satiated here, right? Collector current 20mA, and according to the note, the maximum is 800mA.

ety · Answer

No no. What I meant was that in this example of mine the transistor is not in saturation. Graph - known to be about saturation.

Anonymous · Answer

Why? - I'm afraid you still don't understand saturation. Read carefully what I wrote above. Do you think that Icmax is supposed to flow through the saturated transistor (for BC337 it's 800mA)? No - the current flowing through the saturated transistor depends only on Vcc and Robc. If you remove the LED from your schematic (leaving Rb and Rc), then by providing a sufficiently large Ib we saturate the transistor and for Rc=1k, Ic=12mA, and for Rc=100? - Ic=120mA. And in both cases the transistor is saturated.

ety · Answer

Where is a beta of 10 stated? You mean we should start a small beta? Was it a mistake and meant beta 100? Because maybe I'm blind, but I don't see 10 in the table ... Also not on the hFE chart. I must be missing something :D Well, now I think I see what it was. And I know where to read Vce. I know, I know, the so-called pharmacy difference. So if I want to put the transistor in saturation at Ic=20mA, I can safely calculate everything for the base current of 1mA? That's how I understand +/- from the graph. Fig4. I see. I thought that we are able to control the working point of the transistor (because I think that's what it's called) so that without the additional resistance Rc the current would be 20mA. I know it's a stupid question, but where did this rule come from? Is it just an engineering formula for quick estimation? As for wavelength 300/f[MHz]=lambda [m]. It's just supposed to be a switch, i.e. there is either a clogged state (LED is off) or a saturation state (LED is on). I'm beginning to understand. What about the active state between these two states. Theoretically, if instead of the Rb resistor, put a potentiometer? So theoretical - where would the resistance start at the calculated Rb and increase as it spins, reducing the base current? Then we would move from saturation to clogging (through active state) relatively smoothly?

Xantix · Answer

Apparently it was established empirically - in any case, I always followed this rule in calculations and never had problems with incomplete saturation of the transistor. So then the transistor would work in active state.

Anonymous · Answer

In table page 2 for the line Collector Emitter Saturation Voltage the conditions IC = 500 mA, IB = 50 mA - i.e. Ic/Ib = 10, are given. And in Figure5 the curve is described as Vcesat@Ic/Ib=10. We won't get anywhere if you don't at least try to read what I write - because I gave it too carefully. You could say. The manufacturer of the BC337 guarantees that Vcesat will be at the given value @ic with the given Ic/Ib. In other conditions - it is not known, because the transistor is not equal to the transistor. Read: Yes, but such control of the transistor is not stable and changes in conditions (changes in external temperature, transistor heating, load change, power supply) can cause the transition from the shallow saturation state to the active state (and this will either destroy the transistor or cause the load to malfunction), or going from the active state to the saturation state.

ety · Answer

Thank you very much. I haven't burned any transistor yet, and as you can see, I don't know everything yet Good thing I asked Vce about it. I understand that in theory such a system as I proposed (well, I should move the LED) would work. In practice, however, for some more serious applications, the base could be susceptible to interference. With a larger base current, a drop in this current would not significantly affect the collector current. Of course, within certain limits. That's how I understand it, from the linked materials. OK, but what about audio amplifiers with adjustable gain? Of course, I understand that when controlling via a transistor on the on/off basis, we follow slightly different rules. We are supposed to have stable two states ON and OFF. Clogging and saturation. We don't seem to be interested in the active state.

Xantix · Answer

Well, amplifiers with adjustable gain are rather much more complicated to build. And as for such a simple amplifier with one transistor, more complex transistor polarization systems are used, designed to ensure the stability of the amplifier parameters regardless of fluctuations in external factors. Basically that's how it is.

Anonymous · Answer

We are interested in the fact that everything should be done to prevent the transistor from entering this state, or even to approach it. Of course, we are talking about a system where the key works at low frequency - then you can and even should introduce the transistor into deep saturation (Ic/Ib=10-20), when the frequency is high, you need to do otherwise. It is the high base current that protects against this. The base current is supposed to be larger than necessary - with a margin for various circumstances. This is a completely different story, it is the linear operation of the transistor in the active region, not the bi-state one.

ety · Answer

I mean, they strive for the same thing (deep saturation), but probably in a slightly different way?

Xantix · Answer

No - then the transistor is not entered into deep saturation, because it takes the transistor a relatively long time to get out of this state (compared to the frequency of the signal), which would disturb the switching of the transistor. Therefore, in such a case, one should try to keep the saturation state as shallow as possible. What exactly? And it depends on many factors...

mahdh · Answer

The Vce should be taken into account when designing circuits. It is representing the saturation voltage when the transistor is in its active region, conducting current. In most cases, this voltage drop is not negligible, and it can have a significant impact on the performance of the circuit.

Understanding Vce Voltage in BC337 Transistor Circuit Design: Analysis and Interpretation

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