I have been trying to understand how a transistor works. I bought a simulator to be able to understand it better.

Below is a small circuit I designed. The load is a lamp. The characteristics of the lamp are 5 V, 50 W, so it should draw 10 A.

When I design the schematic without the transistor etc, it does draw 10 A. However when I use the transistor weather I put on the lamp 50 W or 20 W etc it will not draw a "big" current.

Of course this is a "generic" transistor from the sim.

The sim gives me the following settings to configure the transistor (please check the bottom left of the picture)

I do not understand how to configure this transistor since those settings do not match with the datasheet information at all.

enter image description here

EDIT: Without Resistor at all on the base Ib is too high but still Ic is "too low" enter image description here

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    \$\begingroup\$ You need 100mA into base to switch 10A load. Ib=Ic/B ,B=100 Try Rb = 20ohm. \$\endgroup\$
    – user208862
    Apr 2, 2021 at 12:23
  • \$\begingroup\$ That means putting a negligible resistor value to feed the base, eg 1 ohm \$\endgroup\$
    – andrea
    Apr 2, 2021 at 12:25
  • \$\begingroup\$ @MichalPodmanický B is the forward beta? \$\endgroup\$
    – Kris
    Apr 2, 2021 at 12:29
  • \$\begingroup\$ @Kris Right. In real transistor you probably dont find the one with Beta=100 for such a high Ic current. \$\endgroup\$
    – user208862
    Apr 2, 2021 at 12:36
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    \$\begingroup\$ You're using a generic transistor from the simulator, and you say "... do not match with the datasheet information at all.". What datasheet are you expecting the generic transistor to conform to? \$\endgroup\$
    – marcelm
    Apr 2, 2021 at 12:38

1 Answer 1


Your lamp needs 10 A to be fully on.

So your transistor then needs to be able to conduct at least 10 A. If the transistor wants to conduct more, that's OK, only 10 A will flow as the lamp will prevent more current from flowing.

So the collector current of the NPN transistor needs to be 10 A. All bipolar transistors have a certain amount of current amplification which is called \$h_{FE}\$ or \$\beta\$. This \$\beta\$ is the ratio between collector and base current:

\$\beta = \frac{I_C}{I_B}\$

When you want a collector current of 10 A then your base current will be about \$\beta\$ times smaller. Transistors for small currents (much less than 1 A) often have a \$\beta\$ of around 100. Transistors for large currents usually have a smaller \$\beta\$ of maybe 30 (it depends on the actual transistor).

But you're in a simulator so we can do anything we like. So let's assume that \$\beta\$ = 100 that means that your base current would need to be: 10 A / 100 = 100 mA.

I see that your base current is only 516 uA so it needs to be about 200x higher!

If you want to be able to switch on/off a 10 A lamp with a small current (smaller than 1 mA) then I would recommend using an N-channel MOSFET. MOSFETs do not need a high current at their input. Do realize that MOSFETs do need a high voltage at their gate, it depends on the MOSFET how much that needs to be. Some MOSFETs can work with the 3.3 V you're using in your schematic.

  • \$\begingroup\$ I just edited the post. I tried to even remove the base resistor and it draws a lot of current now.. however Ic seems to be still "low" \$\endgroup\$
    – Kris
    Apr 2, 2021 at 12:36
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    \$\begingroup\$ @Kris You're now hitting the limitations of the models in your simulator. Applying 3.3 V directly to the base-emitter like that will destroy such a transistor immediately. I suspect that this transistor model is also not suited for switching 10 A. So I suggest that you "scale the currents down", I mean, change the lamp to 5 V, 100 mA and add a base resistor such that around 1 mA is flowing into the base. Then what happens? \$\endgroup\$ Apr 2, 2021 at 12:48
  • \$\begingroup\$ @Kris though I would also use a MOSFET gor this, if you want to stick to BJTs you can look for "Darlington Pairs". It's double BJT configuration with much higher DC gain than a single BJT. \$\endgroup\$
    – Sim Son
    Apr 2, 2021 at 13:55
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    \$\begingroup\$ In the maximum ratings there's a pulsed drain current which has a maximum of 10 A. The fact that they measured up to 12 A means they're "on the edge" which can work for short pulses but there are no guarantees. You CANNOT use this MOSFET to switch a 10 A lightbulb. I would select a MOSFET that is rated for at least 20 A even for a 10 A bulb. Why? Because, when cold, a lightbulb has a lower resistance than when it is hot. So when you switch it on, much more than 10 A will flow. That will destroy your MOSFET unless it is properly rated for that. \$\endgroup\$ Apr 2, 2021 at 14:57
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    \$\begingroup\$ The 10 A pulse current is a really short pulse, think in the order of less than 100 us. The "cold" current of a lightbulb will last longer than that so you cannot use the "pulsed" rating. \$\endgroup\$ Apr 2, 2021 at 14:59

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