Inductors resist changes in current. To change the current in an inductor, you must apply a voltage. If you remove the voltage, the inductor will make its own voltage so that the current can keep flowing. Of course, making this voltage requires energy, and as that energy moves out of the inductor and into something else, the current will eventually reach zero, when there's no energy left in the inductor.

So, when the switch is closed for a while, there is a current in the inductor. When you open the switch, that current must keep going. Where can it go in this circuit, if its prevented from going through the battery by the open switch?

A diode allows current to flow only in one direction. It's no mistake that their schematic symbol looks like an arrow. If the switch is opened, there is only one place for the same current to keep flowing, and that's through LED1 and LED2. It will keep flowing here, even without help from the battery, until all the stored energy in the inductor has been transferred to the LEDs.

In your second circuit you have the Base-Emitter diodes of the lower two transistors in parallel. The division of the current between both will heavily depend on small variations between the two transistors. This might not show up in your simulation, but it is not a good situation.

The situation with the top two transistors is even worse. The right one will take almost all of the current, because the emitter of the left transistor is at a slightly higher voltage than the emitter of the right one.

That second circuit would work OK with MOSFET transistors, it is in fact the way ports were commonly made in the NMOS days (before CMOS became dominant). MOSFETS are voltage-controlled transistors, hence the problem of how the base current is split between the two transistors does not arise. Bipolar transistors are (base-) current-controlled, hence they each need at separate base resistor.

A potential problem with your first circuit can arise when you use its output to drive subsequent gates. When the output is not pulled down all the way below the Vbe of a transistor (which can happen in the NAND because two Vce's are in series) a subsequent transistor might still receive some base current. Two solutions:

• add (for each transistor) a resistor from base to ground of ~ 100k. this would also make the circuit faster.

• avoid the two-transistors-in-series circuit, use only the NOR gate as building block (you can still get an AND gate at the cost of two extra transistors by inverting the two inputs).