Circuit with transistor - current in open state

Question

I am trying to understand how transistor works. Here is my example from electric circuit simulator (multisim). I cannot understand why PR4 (ampere meter) shows 113.42 nA current when circuit is open. Also I am confused about sign of current in PR5 (ampere meter). It shows -11.343 uA. Why is the sign negative. I am guessing the magnitude of current is so low because of transistor resistance., but why is negative is unclear.

Answer 1

Probably the simulated switch is not infinite resistance when open. That current indicates about 100M\$\Omega\$. There's probably an option to change it, but even a real switch will have some leakage (probably less leakage/higher resistance than that, though). There is a technical reason why simulated switches do not snap from zero ohms to infinity, but you needn't concern yourself with that at this point.

As you can see, this small amount of leakage is resulting in a fair bit of collector current flow, enough to light an LED visibly if it's a good one. In a real circuit you might put a resistor from the switch to ground to avoid that problem. Something like 10K would reduce the base current to almost zero, and below the point where a real transistor has much current gain.

If you look at your simulator's arrow indicating current flow, it is showing the current flow from the collector to the battery, so it will be negative. If that bothers you, there is probably a way to swap the ammeter "leads".

Note that your resistors are both too low and would destroy most small transistors (and some bigger ones too). Values in the k\$\Omega\$ are more plausible.

Answer 2

As a bonus for your question, here are the (suggested) base voltages, for some collector currents. Notice how the base_V is logarithm of the collector current.

• 10mA 0.66 volts'

• 1mA 0.60 volts

• 0.1mA 0.54 volts

• 0.01mA 0.48 volts

• 0.001mA 0.42 volts

• 10uA 0.36 volts

• 1uA 0.30 volts

• 100 nanoAmps 0.24 volts

• 10 nanoAmps 0.18 volts