troubleshooting
Why doesn't my transistor turn on?
Short answer: The base needs real current — a 1 MΩ resistor on 4.5 V trickles only 4 µA, barely nudging the collector. 10 kΩ gives 380 µA and the LED lights fully.
See it happen
This is the broken circuit, simulated live. The readout under the board tells you what the parts are doing.
simulating…
What's going on
An NPN transistor is a current amplifier. Push a small current into the base and it lets through a much larger current from collector to emitter — up to hFE times more. For a 2N2222 a typical hFE is around 200 at the currents these projects use, with a datasheet minimum of about 100 at low collector currents.
That small base current isn't free. The base pin behaves like a diode to the emitter: it drops a fixed V_BE ≈ 0.7 V once it starts conducting, and the resistor you put in front of it controls how much current flows:
- 1 MΩ base resistor: (4.5 V − 0.7 V) ÷ 1 000 000 Ω = 3.8 µA at the base. 200 × 3.8 µA ≈ 0.77 mA on the collector lane — a faint glow, far below the 11 mA the LED wants.
- 10 kΩ base resistor: (4.5 V − 0.7 V) ÷ 10 000 Ω = 380 µA at the base. hFE × I_B is now ≈ 76 mA of headroom, but the 220 Ω + LED lane only wants about 11 mA, so the transistor saturates (V_CE drops near zero) and the LED gets its full brightness.
With too much resistance on the base the transistor sits in its active region instead of saturating: it's conducting, but only a little, and the collector lane's current is capped by the base. Base current up → LED fully on.
The fix
Keep the switch and the collector lane the same; swap the base resistor to 10 kΩ — a tidy standard value that saturates this transistor with margin. Anything from about 3 kΩ to 20 kΩ also works; smaller R overdrives harder (more tolerant of hFE variation), larger R wastes less base current.
If you want the smallest R_B that still saturates for a different load, here's the design rule:
- Pick the collector current the load wants (I_C).
- Divide by hFE's worst case to get the minimum base current (I_B,min = I_C ÷ hFE).
- Overdrive that by 5–10× so parts variation doesn't matter (I_B = 10 × I_B,min).
- R_B = (V_in − 0.7) ÷ I_B, rounded down to the next E12 value.
The transistor base-resistor calculator does all four steps for you.
simulating…
Also check
- E, B, C pins swapped. A 2N2222 in a TO-92 case, flat side facing you, reads E · B · C left to right. Wire the LED lane to the middle pin by mistake and the "transistor" is just a diode drop. Fix: flat side forward, left pin to the battery − rail, right pin up to the resistor + LED.
- No shared ground. The emitter and the battery − need to be the same rail. If the signal source has its own battery, join their − wires too.
- A dead transistor. If an earlier experiment ran the collector or base past about 150 mA, the junction may be gone. Swap for a fresh one.
- Too much base current. Below ~100 Ω on the base you start burning the base-emitter junction. The design rule above keeps you safely between the two extremes.