# Transistor switching

I am using transistor as a switch, But am having a slight confusion. I am using the diagram below where I am using a 10ohm resistor. As I start switching the transistor instead of going into saturation goes into active. I know the different regions and their working, but why isn't the transistor going into the saturation region in this case? As I change the load to 1k, it works fine then. I am confused in this a bit.

• It's active mode because the 10 Ohm resistor won't drop much voltage and the collector voltage will be higher than the base voltage. The 1k resistor will drop a lot, 100 times more, so it goes into saturation then because the collector voltage will be lower than the base voltage. (I don't see your complete circuit, so all I can do is project/guess. But I think I'm projecting/guessing correctly.) Why don't you post your exact circuit with values provided? Jul 3, 2023 at 10:17
• It is good that you are asking questions like this. But you should provide more detail about the schematic and values you are using. We can help you see much better when we see what you are seeing. Specific quantities do matter. Jul 3, 2023 at 10:25
• @periblepsis I have editted the question but why isnt the 10 ohm not dropping a large voltage. I mean If a large current passes it can cover that.
– user340506
Jul 3, 2023 at 10:40

With the 10Ω collector resistor, the maximum current that can pass through it (and down into the collector) is $$\I_C = \frac{15V}{10\Omega}=1.5A\$$.

When that resistor is 1kΩ, the maximum current would be $$\I_C = \frac{15V}{1k\Omega}=0.015A\$$.

Given that the transistor is only able to multiply base current by $$\\beta = 100\$$ or so (varies with transistor model), with a collector resistor of 10Ω this requires base current to be $$\I_B=\frac{1.5A}{100}=0.015A=15mA\$$, whereas with 1kΩ at the collector it's only $$\I_B=\frac{0.015A}{100}=0.00015A=150\mu A\$$.

Now, assuming you're connecting the input to +15V, to switch the transistor on, base current will be $$\I_B=\frac{15V-V_{BE}}{10k\Omega}=\frac{14.3V}{10k\Omega}=1.43mA\$$. So the question is, can that amount of base current produce either of the two required collector currents?

Clearly that's less than the 15mA base current needed to saturate the transistor with 10Ω at the collector, and the transistor will be somewhere between cut-off and saturation, in the active region.

It is however significantly more than the 150μA needed to fully switch on the transistor with 1kΩ at the collector.

• Sir why not using the beta forced in case of operation in saturation mode ie 10 as stated in the previous answer.
– user340506
Jul 3, 2023 at 13:19
• My goal was to find what exact base current was required to just saturate the transistor in each case, which requires that I not assume the transistor to be well into saturation. That is, $\beta$ is not some forced value, just the regular "active" value, applied to determine the cusp of saturation. Above this prescribed base current, well into saturation, then $\beta$ will plummet, but that's another story. Jul 3, 2023 at 13:33
• Simon, a transistor with a beta of 100 is when it is far from saturation with a collector to emitter voltage that is frequently 10V. The forced beta of 10 is almost always used for a saturated transistor. Jul 4, 2023 at 1:58

For a transistor to be guaranteed to go deep into saturation you need to feed the base a substantial portion of the collector current (perhaps 1/10, which is what the datasheet generally specifies, or 1/20). The transistor also has to be capable of the collector current. We can't see what transistor model is being used, but most likely neither condition is being satisfied in your circuit.

In this case, the collector current when the transistor is in saturation would be almost 1.5A. A common TO-92 transistor such as the 2N3904 or BC547 behaves well at up to 100mA or so, but not so well above that.

For example, below (from the BC547 datasheet) shows the behavior with Ic/Ib = 10:

So even with Ib = Ic/10 (corresponding to a base resistor of less than 100Ω for a collector resistor of 10Ω and Vh = 15V) this particular transistor can't handle more than about 200mA (corresponding to a collector resistor of about 75Ω with a 15V supply) and it works much better at 100mA or 20mA.

Assuming Vh is 15V, the base current with a 10kΩ base resistor will be about 1.4mA, so a BC547 will be good switching 15mA or so, corresponding to a collector resistor of 1kΩ. You can play with the simulation and see where it actually (typically) comes out of saturation. Usually for a reliable design we prefer to give it enough base current compared to collector current that the variations in individual transistor characteristics (from part to part, or between manufacturers of the same part number), temperature etc. don't have significant effect on the performance. You just build one piece or 100,000 pieces and every one should work.

Ic/Ib (in saturation) is sometimes called "forced beta", because the ratio is higher than the gain of the transistor if it had a few volts from collector to emitter. In other words, the gain is less because the transistor is saturated. If you allow a higher voltage then the behavior of the transistor can be seen in a different light (again the BC547):

As you can see, with 5V Vce, the gain of the transistor is almost constant at (typical) 200 from 1mA up to 50mA or 60mA, and above 100mA or so it rapidly deteriorates.

There are beefier transistors that can switch higher currents, even in TO-92, but to switch 1.5A at 15V generally we'd prefer to use a MOSFET in 2023.

• So in case of the transistor in saturation we have to consider beta forced. One thing more to clarify, if I change the base current then will the circuit give me a behaviour according to this Ic=betaforceIb as it gives in active with Ic=betaIb.
– user340506
Jul 3, 2023 at 11:42
• You cannot assume that in all cases because the transistor you have may not be capable of saturation even with a forced beta of (say) 10 if the collector current would be too high for the part. Both conditions must be satisfied. Jul 3, 2023 at 11:52

# Learning to dream

In addition to specific knowledge and tinkering circuit tips, it is good to make efforts to reveal the most general ideas behind circuit phenomena. What is more, it is useful to even consider unrealistic circuit ideas... to dream. In this way, we learn not only to do but also to think and deeply understand circuit solutions.

# Conceptual transistor switch

The problem that puzzles the OP arises from the fact that a transistor switch is not a true digital device with two states, but an analog device with a proportional relation between the input and output quantity. This is the simple truth about digital circuitry - there are no digital circuits; there are only analog circuits made to act like digital ones. But under certain circumstances (like the ones here) they go analog again, and that can be surprising.

## Conceptual transistor

Then a transistor can be conceptually represented as a voltage-controlled current source (VCCS). Fortunately for us, there is such a transconductance device in the CircuitLab library. We just need to modify it a bit by shunting it with a backward-biased "ideal" diode D to bring it closer to a real transistor (at the end of the answer it will become more clear why).

simulate this circuit – Schematic created using CircuitLab

## Exploring the transistor switch at various RL

To turn the transistor into a switch we only need to add a load in the collector circuit. Unfortunately it will be "floating" (ungrounded) and if this is a problem, we will connect the load in parallel with the transistor.

The OP's question is about the role of the load (collector) resistance on the transistor mode. Then let's examine our conceptual transistor switch at some typical collector (load) resistance values. To observe the response of the "transistor", we have to connect measuring devices.

RL = 0 Ω: We start with the extreme case when the load has zero resistance (short circuit). We apply a constant 1 V voltage to the VCCS input and, as a result, a 1 A current begins flowing through the output loop (the VCCS ratio is 1 S or 1 Ω-1).

As we can see from the DC live simulation, the "transistor" manages to push the desired 1 A current through the "load". But how? At what cost? How does it do it? We want to know and not just take it for granted.

simulate this circuit

The explanation is very simple - the output "collector-emitter" part behaves like a resistor with a resistance of 15 Ω. Thus, a voltage of 15 V is applied across it and a current of 1 A flows through it (Ohm's law); the voltage across the zero-resistance load is zero. The transistor is able to do this because it has some resistance; it is "active" as the textbooks say. It is active in the sense that it can change its resistance and thus regulate the current.

RL = 10 Ω: CircuitLab has a very handy option to set some internal resistance to be non-zero (for ammeters) and non-infinity (for voltmeters). Let's take advantage of it by replacing the load resistance and the "ideal" ammeter with zero internal resistance with only one real ammeter with 10 Ω resistance.

simulate this circuit

The transistor is still "active" and successfully doing its job of being a current source - to maintain the same 1 A current, it reduces its resistance to 5 Ω so that the total resistance remains 15 Ω. Now the total voltage of 15 V is distributed between the load (10 V) and transistor (5 V).

RL = 15 Ω: In an effort to maintain 1 A current, the "transistor" reduces its resistance to zero so that the total resistance remains 15 Ω. It succeeds but is at the limit of its capabilities. As they say in textbooks, the transistor is "saturated". The meaning of this verbal cliché for us is that the transistor has exhausted its resistance and has nothing more to reduce. All supply voltage is applied to the load.

simulate this circuit

Up to this point, the "conceptual transistor" behaves like a "passive" real transistor - it sinks current and its collector voltage is positive.

RL = 20 Ω: But if we keep increasing the load resistance more than 15 Ω, the "transistor" begins producing current and its collector voltage tries to become negative. That is why we have connected the diode D - it turns on and closes the current path through itself. As a result, the voltage across the "transistor" remains zero and the current begins to decrease.

Thus, the current of the saturated transistor is determined entirely by the load and not by the transistor (in contrast, the current of the transistor in active mode is determined entirely by the transistor and not by the load).

simulate this circuit

Let's now see this graphically using the DC sweep simulation as the load resistance changes from zero to 20 Ω. As you can see, there is a horizontal part (saturation) after 15 Ω.

# Exploring a fancy transistor stage

You probably understand that my goal is not only the transistor switch, but the functioning of the transistor in general. In the name of this great goal, let's jump over the limitations of reality and dream a little... because life is too short to allow ourselves to waste our time with boring and meaningless formal "explanations".

Removing the shunting diode. So let's imagine that inside our "transistor" there is a built-in voltage source. This is actually the original CircuitLab VCCS, but we just have to remove the diode that we put in at the beginning. As a result, our "transistor" begins behaving very interestingly - in order to maintain the same 1 A current, it adds a 5 V voltage to the 15 V supply voltage! Now it really lives up to its name "active device".

simulate this circuit

Let's now see this graphically using the DC sweep simulation as the load resistance changes from zero to a large value of 1kΩ (as the OP has done). As you can see, there is no horizontal part (saturation) in the curve; the "transistor" is always "active". It manages to maintain the same 1 A current even at the cost of 1 kV voltage. Well, it is a bit difficult to implement this real experiment in practice, but that is why CircuitLab helps us. Everything is possible here and dreams do come true :-)

Removing the power supply. But why do we even need a power supply now that our fancy transistor has it? Let's remove it and see if it makes a difference. It works! The "active transistor" fully powers the load by increasing its voltage to 20 V.

simulate this circuit

The graph is almost the same (the only difference is that the curve begins from zero voltage while above it begins from Vcc).

# Final conclusions

From the above experiments, we come to the conclusion that inside a current source there is a voltage source. Its voltage (also known as a "compliance voltage") is the maximum voltage that the current source can create across the load with maximum resistance by passing its current through it; the rest of the voltage across the regulated element is zero.

If we increase the load resistance further, the current source needs more voltage to maintain the current but has nowhere to get it from (unless it is made with a "fancy transistor" :-) As a result, the current source becomes a "transistor switch". So, our final conclusion is:

The transistor switch is a current source that has reached its compliance voltage.