Output current of collector

I am confused about the switching of this circuit. I can control the V1 input line. Output line is R4 load line. My Vbase (V1) is 0V, and Vbe is 0V. Transistor is off and collector should see 0V. That is the graph which I added. But, the 5V still supplies voltage and current to the R4 load.

How can i make this switching circuit? In this even if its off I see 5V on R4 and current.

Some other scenarios I tried: remove R3 100ohm. In that case there is 5V and 5mA on R4 which makes sense. Another case: remove R4 and have R3. In that case I see 5V across Q1 and no current. But there is still a potential difference that my device input will be able to read correct (despite no current)?

  • 3
    \$\begingroup\$ WIth V1=0, Q1 is off and therefore the collector current is zero. With the collector current zero, it's not pulling down the voltage at the collector, so all you have left over is a resistive divider made up by R2 and R4. It's like Q1 isn't even there. \$\endgroup\$
    – jonk
    May 16, 2019 at 22:39
  • 2
    \$\begingroup\$ RE "Transistor is off and collector should see 0V"...this is your misunderstanding. When the transistor is on, it pulls the collector down towards ground. When the transistor is off, R2 is able to pull the collector up to 5 V. \$\endgroup\$
    – The Photon
    May 16, 2019 at 22:41
  • \$\begingroup\$ You can replace Q1 with an ideal switch and work out the usecases of your circuit. From there, try to understand how to use a BJT as a switching element. \$\endgroup\$
    – sstobbe
    May 16, 2019 at 22:52

2 Answers 2


Transistor is off ...

That means there is an open-circuit between its collector and emitter.

... and collector should see 0 V.

That's your error.

  • You have R2 at 100 Ω giving a strong pull-up to V2.
  • You have R4 at 1 kΩ giving a weak pull-down to 0 V.

The combination gives \$ V_C = \frac {1000}{1000 + 100} V_2 = \frac {10}{11} V_2 \$.

With Q1 off it can be removed from the circuit. All you've got is a potential divider.


simulate this circuit – Schematic created using CircuitLab

Figure 1. An equivalent circuit using a relay. When the relay is off the contact is open and VC is almost 5 V.


That's a misunderstanding, in switching circuit the Load is connected into main current path, so in your case the Load is R2. R4 is no needed.

R1 is optional (no needed in your case) for high speed designs or can act as voltage devider with R3.


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