I have this exercise (not homework!) where I need to find at what point, using a pot, I will let enough current pass through a BJT base to let 70mA cross a relay's coil. I have trouble determining how to do this.

enter image description here

I know I first need to calculate the required BJT base current, which is 100 uA given that the transistor used in the exercise has a gain of 700.

Then, I am not sure how to use the variable voltage divider to determine the voltage at the BJT's base node, which will let me calculate the potentiometer's resistance at that point.

Can someone help me understand? Should I use KCL?


  • \$\begingroup\$ You need to decide how accurate the divider needs to be. The larger the current through the divider is compared to the 100uA that taps off to your BJT, the more accurate the divider is, but the more power it wastes. I usually arbitrarily go with a factor of 100x to 1000x. \$\endgroup\$
    – K H
    Feb 28, 2021 at 22:32
  • \$\begingroup\$ If I remember it right, you must apply certain voltage to the base (about 0.3-0.5 V) to open transistor. So, in effect you have 0.3 emf in parallel with 200 Ohm resistor and your goal is to drive 100uA into its branch. Now it's pretty simple to find necessary pot resistance. \$\endgroup\$
    – AlexVB
    Feb 28, 2021 at 22:58
  • \$\begingroup\$ One more thought. Transistor's gain doesn't matter. You use it as a switch. So you just need to fully open the transistor. If the relay is designed for 12V then it will draw exactly the current it needs. \$\endgroup\$
    – AlexVB
    Feb 28, 2021 at 23:02
  • \$\begingroup\$ @AlexVB the relay is theoretical and opens up at 70mA or more. The question exactly is at which potentiometer R value (starting at 5k and going down to 0ohm) will the the relay trigger? So It is not necessarily at the transistor's saturation level. It is simply when Ibe will be 100uA, generating a Ice of 70mA. This I am pretty sure! (and silicon's bias voltage in transistors is 0.7V) \$\endgroup\$
    – JCSB
    Feb 28, 2021 at 23:17
  • \$\begingroup\$ @JCSB Assuming small-signal BJT of \$I_\text{SAT}=10\:\text{fA}\$, I get from Ebers-Moll that \$V_\text{B}=769\:\text{mV}\$. You know the \$200\:\Omega\$ resistor current is \$3.845\:\text{mA}\$ and the current in the remaining two resistors must be \$3.845\:\text{mA}+\frac{70\:\text{mA}}{\beta=700}=3.945\:\text{mA}\$. From there, the \$1\:\text{k}\Omega\$ resistor drops \$3.945\:\text{V}\$. Therefore the variable resistor must drop \$8.055\:\text{V}-769\:\text{mV}=7.286\:\text{V}\$. This means \$\frac{7.286\:\text{V}}{3.945\:\text{mA}}\approx 1.847\:\text{k}\Omega\$ for the variable resistor. \$\endgroup\$
    – jonk
    Mar 1, 2021 at 7:37

1 Answer 1


If you estimate the Hfe of the transistor as 100 then 70mA Ice requires 700uA Ibe. Assuming that the Vbe drop will be about 700mV, the 200R resistor will draw 3.5mA and so the more positive part of the potential divider needs to supply 4.2mA. It has a voltage drop of 11.3V (12 - 0.7) and so the resistance should be 2690R. The Hfe of the transistor may be significantly different from this, but since the majority of the divider current flows through the 200R resistor, the circuit won't be particularly sensitive to the Hfe.

  • \$\begingroup\$ You want the transistor to be saturated with a low Vce. The hFE (beta or current gain) is used for an amplifier with a high Vce, not for a saturated transistor. The datasheet for every transistor shows that it saturates well when its base current is 1/10th its collector current (then its base current must be 7mA, not 100uA. \$\endgroup\$
    – Audioguru
    Mar 1, 2021 at 1:04
  • \$\begingroup\$ You do not need a voltage divider, the transistor base-emitter diode is shown in its datasheet as about 0.7V so simply calculate (12V - 0.7V)/7mA= 1614 ohms will saturate the transistor. Use 1.5k ohms. The 200 ohms resistor wastes 0.7/200 ohms= 3.5mA, use 10k ohms to turn off the transistor. \$\endgroup\$
    – Audioguru
    Mar 1, 2021 at 1:04
  • \$\begingroup\$ @Audioguru such a setup would be very sensitive to the Hfe of the transistor, which can change significantly with temperature (amongst other things). \$\endgroup\$
    – Frog
    Mar 1, 2021 at 1:18
  • \$\begingroup\$ Please remember this is an exercise, i cannot change the components! I am stuck with the values seen in the image \$\endgroup\$
    – JCSB
    Mar 1, 2021 at 3:30
  • 1
    \$\begingroup\$ OK then, looking back at the OP we have a required base current of 100uA and a 200R resistive load at 700mV, so we need to source 3.6mA from 11.3V, which requires a resistance of 3139R. You have a fixed 1K so if the variable resistor is much below 2139R the relay will be energised, and if it's much above 2139R then the relay will be de-energised. \$\endgroup\$
    – Frog
    Mar 1, 2021 at 5:00

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