# Transistor in cutoff mode but still allows current to flow Hello everyone, I am a freshman in computer science, and as we are dealing with a lot of hardware, we stumbled upon transistors. We learned that if a transistor is in cutoff mode, it basically acts as a switch that is not closed.

In the picture I provided, we can see that the transistor is in cutoff mode, as the voltage supplied to the base is not enough (it should be around 0.6V). But, if I had an open switch, the voltage between the collector of the NPN transistor and the ground should be 5 Volts. In this case, however, it says 4.881V. I have no idea where the voltage is lost along the way, because as the current does not pass through the transistor, no current should flow through the entire branch, and therefore there should be no loss of source voltage. What am I getting wrong?

• Look up the leakage current specification (usually Iceo) in that transistor's datasheet. – user_1818839 Nov 30 '20 at 23:23
• What is the thing that looks like a gray circle between +5V and the 2.2k resistor? – Bruce Abbott Dec 1 '20 at 3:56
• BTW I suggest using a proper simulator such as LTSpice. – Bruce Abbott Dec 1 '20 at 3:58
• @BruceAbbott I've used the simulator here before a lot; that's an LED. It's not the standard symbol because it will "light up" when current flows through it. – Hearth Dec 1 '20 at 4:39

Transistor in active mode. Calculate base current is 0.2-base emmiter drop voltage. For small current it is less then 0.7 V. Multiply that current on the hfe (it may be different, depends on current) you will get some collector current and voltage drop on collector resistor. Check parameters of transistor model. It is close to real.

This answer is just demonstrating that even with quite modest base-emitter voltages, you will generate enough collector current to drop around 100 mV across a 2k2 resistor.

When you apply a small forward bias voltage to the base with respect to the emitter (basically a forward biased diode), there will be a small base current flow. But, you have to dig around a bit to estimate its value and, I normally turn to the forward characteristic of the 1N4148/1N914 diode because it has decent documentation from ON semiconductors. For instance, consider this graph: - It shows how much forward current is taken for a range of applied voltages. At an applied voltage of 0.275 volts, the current taken is 1μA and, if you extended the straight line down to 0.2 volts, the current would be approximately 350 nA: - So, 350 nA is going to be "about the right" sort of forward current into a 1N4148 diode when 200 mV is applied between anode (base) and cathode (emitter) and it will be roughly the same for a typical NPN transistor.

If the transistor has a current gain (β) of 100, you can expect a collector current of 35 μA and 35 μA through a collector resistor of 2.2 Ω will drop 77 mV thus, the voltage at the collector will be 4.923 volts.

In your simulation of an unspecified BJT you see a volt drop of 119 mV so it's not a million miles from the 77 mV I calculated above. You also have "something" in series with the 2k2 resistor (the circle) that may increase the volt drop.

• The 'something' is an LED, so 119 mV at a very low current is quite believable. – Bruce Abbott Dec 1 '20 at 20:44

Every semiconductor device has some leakage current, even when the device is "off".

You are getting a bit more that leakage current, because even at a base bias of 0.2v, some base current will flow, and this will cause collector current to flow.

• Like @Math said, the transistor is not OFF in the ideal sense. The 0.6V that's usually used as the Vbe value is typical of a transistor in it's a linear region. But, depending on the device, it could start conducting at a much lower voltage. Look the Ice leakage spec for some typical devices to see what we mean. – SteveSh Nov 30 '20 at 23:08