I started learning Transistors with this tutorial from the Sparkfun website. The tutorial is good and pretty easy to understand. So I decided to put my learning to the test and assembled a simple NPN transistor as switch. Here is what the tutorial states as the condition to toggle it between Saturation and Cut-off states:

Saturation: Ve < Vb and Vc < Vb

Cut-off: Ve > Vb and Vc > Vb

Below is the circuit I assembled on a simple breadboard to test these conditions in each state.


simulate this circuit – Schematic created using CircuitLab

It works as expected. The LEDs turn on in the Saturation state and turn off in the Cut-off state. But when I measured the voltages to see if the conditions are satisfied as per the tutorial, I can see some voltage readings don't obey the above conditions. I have noted down my readings below:

Saturation state:

  • Vbe or Vb = 0.62
  • Vbc or Vc = -4.92
  • Ve = 0

In the saturation state, the conditions are satisfied Ve < Vb and Vc < Vb

However in the Cut-off state, the readings are:

  • Vbe or Vb = 0.02
  • Vbc or Vc = -7.24
  • Ve = 0

In this case the readings doesn't hold the conditions (Ve > Vb and Vc > Vb). Actual Vb is still greater than Ve and Vc. Why is this happening? Did I miss anything here? Why do the actual measured readings seem to not be obeying the above conditions?

Besides these, in the Cut-off state where there is no collector current, I can still observe voltage drops between those LEDs connected to the collector terminal of transistor. Why is this happening? When current is zero, voltage drop shouldn't exist, isn't that true?

  • 1
    \$\begingroup\$ How can Vc be negative with only one power supply. Think about what you are doing and measure all voltages relative to the emitter node. \$\endgroup\$
    – Andy aka
    Commented Oct 12, 2018 at 18:00
  • 1
    \$\begingroup\$ To achieve saturation, use a smaller resistor in series with the base. Maybe 4.7k or 2.2k instead of 14k. Although I am suprised that the collector was 4.92V. Suspecting a circuit or measurement error, in fact. In order to determine if you are in cutoff, measure the collector current indirectly. Put the voltmeter across the 120 Ohm resistor. If you put the voltmeter from collector to base, you provide an alternate path for current to flow (through the voltmeter). \$\endgroup\$
    – user57037
    Commented Oct 12, 2018 at 18:03
  • \$\begingroup\$ Thank you everyone for taking time to help me out in this. I realized I got the concept wrong and your answers helped me to understand it better. \$\endgroup\$ Commented Oct 16, 2018 at 13:27

3 Answers 3


First, you are not in saturation. Simply having VB higher than Ve is not enough. Saturation requires supplying enough current to exceed the current needs of at the collector, minimizing the drop across VCE, and taking it out of the linear region. A lower value base resistor is needed.

Second, you need to account for leakage currents. It will never be zero, as nothing is perfect. 20 mV may seem large but its not really. But again, a stiffer pull down resistor would help minimize it.


So, depending on your LEDs you're dropping between 4.5 and 7.5 (ish) volts. Assuming saturation, that leaves 4.5-7.5 volts across that 120-ohm resistor. That works out to 37 to 63mA through the LEDs -- that's a LOT for a typical LED, but let's put that aside.

You're supplying about 0.8mA to the base of the transistor. In order to develop 63mA of collector current you'd need an HFE of 76 at saturation -- and there's no way you're going to do that with an ordinary NPN. I think there may be some hifalutin' ones from Diodes, Inc., but I'd have to check.

To reliably saturate a 1970's-era transistor, like a 2N3904 or a 2N2222, you need to go for a base current of around 1/10th your target collector current. Note that the 1/10th figure is very much a rule of thumb -- study of the data sheet of the transistor you're working with is a Good Idea, but 1/10th is pretty safe for hobby work with hobby transistors. Unless you go to those new exotic types, that's what you're stuck with. So to drive that 63mA, you need to supply 6.3mA, which means you need (12V - 0.6V)/(6.3mA), or around 1800 ohms. So you are, as I would say to my boss, "A little bit off".

And -- you may want to check the ratings on your LEDs. 20mA is typical for not-so-special LEDs. If you go to DigiKey and get their super-bright LEDs (that are rated for 20mA), 20mA is painfully bright. 63mA wouldn't be good for them.

  • \$\begingroup\$ This is good, except for the 2nd paragraph. A P2n2222 can easily have a HFE of 76. Minimum 50 to Max 300 on it's data sheet. \$\endgroup\$
    – Passerby
    Commented Oct 13, 2018 at 1:56
  • 1
    \$\begingroup\$ I was (probably) misusing the HFE nomenclature. I just checked, the PN2222 data sheet (from ON at least) specifies saturation as having a 10:1 collector-base current ratio. There are parts, such as this one that have much higher collector-base current ratios at saturation, and much lower saturation voltages -- but you have to dig to find them. \$\endgroup\$
    – TimWescott
    Commented Oct 13, 2018 at 15:17

Partial answer in addition to the excellent ones already provided.

In the cut-off state with Vbe = 0, when you attempt to measure the collector potential Q1/Q2 the input resistance of your multi-meter biases your diode stack. If you have a "cheapie" meter with a Rin of 1 MOhm you can end up biasing your diodes with uA's of current causing the collector potential to decline. There is also a small static cut-off leakage through Q1/Q2. If you add a 10 kOhm pull-up resistor in parallel with your diode stack, you should observe near "text-book" readings.

If Q1/Q2 were inserted backwards (collector and emitter swapped) the transistor can still operate in the reverse-active region with very low current gain (\$\beta\$). You can double check Q1/Q2 are in the correct orientation.


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