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How does the ratio of R2 to R3 \begin{equation} Ratio=R_2/R_3 \end{equation} affect the transistor in the following circuit (note: V1 applies constant voltage of 4V): enter image description here

When the transistor switches ON the base node has a voltage of 0.7V and current flows through the lamp. It switches OFF when VBE reaches 0V, this is achieveable when R2 is low enough so that all current goes through R2 to the ground. How can we find the ratio of R2 to R3?

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  • \$\begingroup\$ Your question is unclear. The reason is because you are describing two different circuit operating points (transistor ON and transistor OFF) and yet the three resistors (R1, R2, and R3) all will define a single operating point depending upon their values. You really need to supply some additional information regarding what you are attempting to achieve AND the method by which the circuit will be switched between the ON and OFF state. The schematic you have shown gives no clue about how this state change is triggered. Also for the circuit to work you will have to show the 4V- connect to GND. \$\endgroup\$ Mar 24, 2018 at 17:15
  • \$\begingroup\$ @Arthur: If you use the schematic button on the editor toolbar your editable schematic gets saved inline without a grid and we can copy and edit it in our answers. A couple of minor points: (1) You didn't ground your 4 V supply so no current will flow in your circuit. (2) You have no load resistor. If your resistor combinations do turn on Q1 then it will short-circuit the supply. (3) You can de-clutter your schematic by adding GND symbols rather than 'wires' to the V1 negative, bottom of R2 and emitter of Q1. \$\endgroup\$
    – Transistor
    Mar 24, 2018 at 17:50
  • \$\begingroup\$ not a working circuit, but r1 and r2 must give >0.7v (out of 4). ve=0. vc = vb-0.7. that's almost enough to solve ;) \$\endgroup\$
    – dandavis
    Mar 24, 2018 at 17:51
  • \$\begingroup\$ Perhaps the OP misunderstood the question and is really asking about the ratio of R1:R2, not R2:R3. since the ratio of R1:R2 does matter directly. \$\endgroup\$
    – jonk
    Mar 24, 2018 at 17:52

2 Answers 2

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Follow the schematic simplifications below by following the direction of the blue arrows:

schematic

simulate this circuit – Schematic created using CircuitLab

The bottom schematic is the analysis equivalent. The Kirchoff's voltage law then provides:

$$4\frac{R_2}{R_1+R_2}-I_\text{B}\cdot R_X-V_\text{BE}=0\:\text{V}$$

Therefore:

$$I_\text{B}=\frac{4\frac{R_2}{R_1+R_2}-V_\text{BE}}{R_3+\frac{R_1\: R_2}{R_1+R_2}}$$

Typically, for a switch use, deep saturation is assumed and therefore \$\beta\approx 10\$ is typical. You can choose a different value. But this is a commonly accepted value.

If you know the intended lamp current when operating near \$4\:\text{V}\$ then you can work out that you need \$I_\text{B}=I_\text{LAMP}\frac{1}{\beta=10}\$.

So:

$$I_\text{LAMP}=\frac{4\frac{R_2}{R_1+R_2}-V_\text{BE}}{R_3+\frac{R_1\: R_2}{R_1+R_2}}\cdot\left(\beta=10\right)$$

This establishes all your relationships for the resistors. From there, you should be able to work out what you need.

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for calculating the current of the emitter which, IC=IE=B*Ib, you should first calculate the current of Base, and for calculating the current of the base, you should Obtain the ratio of R2 and R1:

Vx=4*R2/(R1+R2)

Ib=Vx/R3

IE=B*Ib

based on the how much current the lamp is sinking, you should design your circuit and the ratio of R1, R2, and R3

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