# Role of a resistor in a specific, simple circuit

i am currently reading "make:electronics" from Charles Platt. On page 129 in Figure 3-87 i found a circuit, which i have reconstructed in the following image: The circuit is supposed to do the following: If the switch SW1 is open, transistor T1 should close and the LED should light up. If the switch closes, the transistor should cut off, the led should turn off.

My question is, why did the author add R3? In a simulation I removed R3 and changed R1 to 11k ohm. The circuit behaved exactly the same way, at least as far as I can tell. 11k ohm is a nonstandard value, is that why the author had to construct it by putting 10k and 1k in series? If yes, why did he add it after the switch, and not before?

In the circuit, R1 and R3 are acting as a voltage divider, but what purpose does this serve?

Overall my question is, why include R3?

In the circuit, R1 and R3 are acting as a voltage divider, but what purpose does this serve? Overall my question is, why include R3?

R3 is indeed redundant. Why is it included?...perhaps to illustrate that to turn off the LED you only need reduce the transistor's base voltage (with respect to GND) below +2.6V DC, since the LED refuses to conduct current should its anode fall below about 2V.
As you've pointed out, shorting base directly to GND via SW1 certainly turns off the LED.
When switched on, SW1 sends 1.091 mA toward GND in the original circuit. In your modified circuit eliminating R3, 1.2 mA is shunted to GND via SW1. Transistor base current is zero. With SW1 open, about 0.94 mA current flows into transistor base.

Also, be aware that transistor collector and its associated resistor R2 is only active once the transistor turns on...turning on-or-off is controlled by base-emitter voltage. Transistor here is acting mostly as a switch (on or off).

Should you slowly raise base voltage, LED will begin to glow only when base voltage rises about 0.6V above the LED's threshold voltage (about 2V?). As you raise base voltage slightly higher, LED brightness increases quickly. If you raise base voltage above about 2.6V, transistor base-emitter junction passes a lot of current to the LED...too much base current destroys transistor base-emitter junction (or the LED) from over-current.

• To turn off the transistor (reference designators - !), the base voltage must be below the sum of the transistor base-emitter voltage (Vbe) and the LED forward voltage (Vf). For many common components, that total is around 2.3 V to 2.7 V, not 1.2 V. Commented May 4 at 16:08
• @AnalogKid Corrected...to get LED off, reduce base-to-GND voltage below about 2.6V, not 1.2V Commented May 4 at 16:32

If I'm honest with you, the arrangement of D1 and R2 is very odd, and really doesn't help to understand much. There are better circuits to illustrate the transistor's and resistors' roles, but I'll try to explain what's going on as best I can, and I'll try to do it in stages.

Before considering anything else, you must first understand these:

1. The LED is a diode.

2. Between the transistor's base and emitter terminals is another diode.

You must also understand that:

1. When current is flowing the "correct" way through a diode, the diode will try very hard to prevent the voltage across it from exceeding some limit.

For a red LED the limit is about 1.5V. For the base-emitter junction (and any regular silicon diode like the popular 1N4148 and 1N4001) that limit is 0.7V or so. To see this behaviour, let's simulate a test circuit, in which we try to raise the voltage across an LED, a transistor B-E junction and a regular diode:

simulate this circuit – Schematic created using CircuitLab

If I slowly increase supply voltage V1 from 0V to +3V, we will see that initially the voltages across each element, $$\V_A\$$, $$\V_B\$$ and $$\V_C\$$ will follow nicely, but eventually, when they reach their rated "forward voltage" they start conducting very heavily, opposing further increase:

The LED is a little "weaker" in its ability to oppose increases beyond it's 1.5V limit, but the other two are really trying very hard to prevent their voltage from rising beyond 0.7V. In any case, diode behaviour is clear enough. A diode's effective resistance drops dramatically as the voltage across it approaches some threshold, which tends to prevent the voltage across it from rising any further.

In your circuit, we have R1, the transistor B-E junction, and D1, all in series:

simulate this circuit

You need to be familiar with Kirchhoff's Voltage Law (KVL) to understand my next claim: If V1 is large enough, due to the upper voltage limits imposed by the two "diodes" here, potential $$\V_A\$$ will be:

$$V_A = 0V + (+1.5V) = +1.5V$$

Potential $$\V_B\$$ will be:

$$V_B = V_A + (+0.7V) = (+1.5V) + (+0.7V) = +2.2V$$

The remaining voltage across R1 will be the difference $$\V_1 - V_B = 7.8V\$$, permitting us to calculate current $$\I_B\$$ flowing into the base, out of the emitter, and ultimately through the LED, using Ohm's law:

$$I_B = \frac{V_{R1}}{R_1} = \frac{7.8V}{10k\Omega} = 780\mu A$$

That's easily enough to light the LED.

We have established that base potential $$\V_B\$$ will effectively be "clamped" to a maximum near +2.2V, and this will be true regardless of anything else. This is the nature of diodes and base-emitter junctions, and I encourage you at this point to verify for yourself, in your own simulator, that this clamping to a combined maximum of $$\1.5V + 0.7V = 2.2V\$$ is occurring, regardless of the state of SW1.

You correctly identified R1 and R2 as forming a potential divider, where (ignoring any other influences) base potential $$\V_B\$$ will be:

$$V_B = 12V \times \frac{R_1}{R_1 + R_2} = 12 \times \frac{1}{11} = +1.1V$$

So, when SW1 is closed, base potential $$\V_B\$$ should fall to about +1V. Remember that the diodes (B-E junction and LED in series) will not oppose a combined voltage of less than +2.2V, and will therefore not prevent the condition $$\V_B = +1V\$$ when SW1 is closed:

simulate this circuit

Note: Because current is being diverted around R3, via the base and LED which "load" the divider, $$\V_B\$$ is slightly under the predicted +1.1V.

With SW1 closed base potential $$\V_B\$$ falls, and so does LED voltage $$\V_A\$$, to well below the value necessary to illuminate it, and so the LED goes dark. The answer to your question "why include R3?", is that without R3, base potential would not fall below the level required to extinguish the LED, when SW1 is closed.

Up until now we have not connected the collector, and yet with SW1 open the LED will light, and with SW1 closed, the LED will go dark. In other words, with or without collector connected, SW1 controls LED state. That's why I claimed that this circuit is a bad example, and only serves to confuse; SW1 would (poorly) control LED state with or without the transistor!

However, there can be some redemption, in that the circuit will still permit us to see how the transistor acts as a current amplifier. If we now connect the collector and R2, and measure currents, we will witness the transistor doing what it was born to do, amplify:

simulate this circuit

The ammeters show that base current $$\I_B\$$ stays around 700μA, but the transistor is drawing 160 times that amount into its collector, as current $$\I_C\$$. The sum of base and collector currents emerge at the emitter, and that total of 12mA is lighting the LED much brighter.

I'll say it again, this circuit is not a good example of anything, but at least you can see the transistor's amplification role. Other than that it doesn't really make any sense to put the LED at the emitter, in this application where a switch is being used to either completely illuminate the LED or completely extinguish it. There are much better circuits, much easier to understand.

When a "load" (an LED in this case) is connected at the emitter like this, the transistor is in a configuration called "common-collector" (also known as "emitter follower"), as opposed to "common-emitter". I encourage you to study them, because their behaviours are very, very different. Unfortunately the author of that circuit has combined elements of both, which will be difficult for a newcomer to transistors to understand. Start with one or the other, then examine what happens when you place elements both at the collector and the emitter, as is the case here.

• Thank you very much for your very detailed answer, i've learned a lot from it. I have switched to "The Art of Electronics" by Horowitz and Hill, which left me with less questions. When i asked why R3 was included i meant replacing R3 with 0 ohms, which would pull the Transistor base to 0V when the SW1 is closed. This would practically achieve the same functionality. Regarding your comment about the educational value of this circuit, the author expands this circuit to become a "home security system", SW1 would be replaced by reed switches in series.
– balu
Commented May 18 at 18:58
• @balu I see, with R3=0Ω, or even just some low value like 100Ω to represent an imperfect switch, that would certainly make more sense, but it's still strange to put the LED in the emitter path. There could be a very good reason for doing that, I don't know what the author is trying to achieve. I love that book, AoE, it was invaluable for me in the early years, and I still refer to it from time to time. Commented May 19 at 2:50

EE engg. So the transistor is a voltage controlled current source, but we know due to KCL law the current flowing into the base also affects the source and drain current. So you need to create a voltage divider with R1 and R2 that drives some current then the transistor should pull atleast 10 times less current into the base so ass not to affect the bias circuit(when more current is pulled through a resistor the voltage also goes up at the base too changing operating point of the transistor.) So this voltage divider creates a more stable bias during operations. You can follow up with more questions if you need a better explanation on any part of it. I hope this helps.