# Why does this transistor limit the current so much?

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.

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

• You need 100mA into base to switch 10A load. Ib=Ic/B ,B=100 Try Rb = 20ohm.
– user208862
Commented Apr 2, 2021 at 12:23
• That means putting a negligible resistor value to feed the base, eg 1 ohm Commented Apr 2, 2021 at 12:25
• @MichalPodmanický B is the forward beta?
– Kris
Commented Apr 2, 2021 at 12:29
• @Kris Right. In real transistor you probably dont find the one with Beta=100 for such a high Ic current.
– user208862
Commented Apr 2, 2021 at 12:36
• 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? Commented Apr 2, 2021 at 12:38

## 1 Answer

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.

• 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"
– Kris
Commented Apr 2, 2021 at 12:36
• @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? Commented Apr 2, 2021 at 12:48
• @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. Commented Apr 2, 2021 at 13:55
• 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. Commented Apr 2, 2021 at 14:57
• 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. Commented Apr 2, 2021 at 14:59