# Why is the output voltage 5V when supply voltage is 18V?

I am building a headlight driver circuit for my solar car team at school. The circuit uses transistors to drive the headlights and an 18 V power supply. My HIGH and LOW signal is a 5 V signal and 0 V signal respectively. When I measure the output voltage it reads ~4.2 V, even though the power supply is 18 V. Why are the transistors giving me a voltage drop of nearly 15 V?

simulate this circuit – Schematic created using CircuitLab

(The old schematic below:)

• Please draw a proper schematic circuit diagram. Commented Mar 22, 2023 at 16:05
• empty-engineer - Hi, You asked the same question on Physics.SE and here - without deleting the one on Physics.SE. Then, regrettably, Physics.SE migrated their question to here, without noticing that it duplicated the one you had asked here - and both versions got answers here. It's a mess. The migrated question was kindly flagged as a dup by a site member here (and has been closed) but posting the same question to multiple parts of SE has caused a bunch of work. Please don't do that. Thanks. Commented Mar 22, 2023 at 16:56
• You have unconnected nodes on your schematic and you don't appear to have decouplers on your voltage regulator U4. U2 is not part-identified. Commented Mar 22, 2023 at 17:14
• @winny The 9V batteries were only used for the schematic. There is no way they could supply HEADLIGHTS. :D Commented Mar 23, 2023 at 10:13
• @EdinFifić I'm referring to OPs previous image here, with 9 V and breadboard: i.sstatic.net/OzFWt.png. But the first issue is the emitter follower. Commented Mar 23, 2023 at 13:23

Your bipolar transistor is driving the lamps in a voltage follower fashion, meaning the output voltage can't be higher than the controlling voltage. The basic rule to turn an NPN bipolar transistor ON (or just bias it) is to have about +0.6V on its base relative to its emitter, meaning the emitter can't go higher than the base potential minus 0.6V, in this case 5V-0.6V=4.4V.

simulate this circuit – Schematic created using CircuitLab

In order to apply the "full" supply voltage, you need to connect the bulb(s) on the opposite end of the transistor, that is, to the collector pin. In that case, the emitter is connected to ground, and even a 1.0V logic or control voltage can turn the transistor on fully for whatever output/supply voltage (up to the transistor's withstand voltage limit, of course). This type of a circuit is also used in logic level translators due to this fact, as the output can swing from 0V to the full supply voltage.
The collector can now reach almost 0V (almost at the ground level), so if the bulb is connected to the positive end of the battery on one side and to the collector of the transistor on the other side, it will have almost the full 12V across it (let's ignore the voltage drop across the transistor for the sake of simplicity). Here is the way it's done:

simulate this circuit

The R1 resistor serves to limit the current through the base which acts like a diode (base-emitter diode) and would burn without the resistor. The base-emitter junction stays at about 0.6V or slightly more while the transistor is on.
The collector-emitter junction now acts like a switch, open with no current through the base, and closed (ON) with a sufficient current running through the base.

Because a bipolar transistor has significant voltage drop across it while conducting a significant amount of current, it will dissipate a lot of wasted energy as heat and possibly burn out. At 0.5V across it and a 10A current, that is 5W of wasted energy and heat that needs to be removed from it.
You are also wasting energy to keep the transistor ON via the current running through the base. If you're running 10A through bulbs, you may need 1A through the base, which is not only additional 0.6-1W of heat added to the transistor, but a full 5W of energy needed from the 5V supply.

A much better and more modern solution would be a MOSFET transistor as a switch. It needs practically no current to keep it ON, while having a very small resistance and voltage drop across it. Here is an example schematic:

simulate this circuit
As you can see, there is a very small voltage drop (a few times smaller than for a bipolar transistor) across a MOSFET when it's turned on. This means there is a few times lower loss and a few times less heat dissipated in a MOSFET.
Additionally, almost no current at the controlling leg (gate) means that basically no energy is needed to keep a MOSFET on. For example, while you may need from at least about 50mA and up to about 2A of base current for 10A at the output with a bipolar transistor, you would only have a gate leakage current of 0.0002mA (0.2µA or 200nA) in a typical MOSFET at the most. That is less than the self-discharge current of a typical CR2032 3V coin cell!
To put it another way, a 3V coin cell could keep a MOSFET turned on for years, while with a bipolar transistor the same amount of energy would barely last a few hours at best (this could be improved with a "darlington" arrangement, but it would cause higher output losses, and it would still not come close to the MOSFET efficiency).

The included MOSFET model is just an example, you could use one with a lower resistance while on (called RDS(on)) to have an even lower voltage drop across it, or use a logic-level MOSFET which works better when you don't have 7-15V typically used to turn it on. The only drawback to using a MOSFET is the voltage required to turn it ON needs to be significantly higher than for a bipolar transistor, at least 3-4V though typically 7-10V vs. 0.5-1.0V
You could also use an 8V to 10V regulator (LM7808 to LM7810) to supply the NE555 (which can run at up to 16V at most, but uses more quiescent current with higher supply voltage, unless it's a CMOS version like TLC555 or ICM7555), which would let you use any MOSFET that has the lowest RDS(on).

For the sake of simplicity, I have kept the answer to the most often used and practical examples, but you can also use a PNP bipolar transistor or P-channel MOSFET (the above was an N-channel).

• Thank you. This helps a lot. Most literature I found online didn’t explain very well Commented Mar 22, 2023 at 18:47
• You're welcome. Feel free to ask me to clarify anything more in my answer here regarding this topic. Commented Mar 23, 2023 at 4:24
• I have edited, clarified and added more, which could help you make an even better solution. Feel free to re-read and give me feedback. Commented Mar 23, 2023 at 7:07
• Thank you so much, is it okay if I use this to explain better to my students? Commented Mar 23, 2023 at 12:18
• Of course it's OK. I could even help you with more details, as I have done, and I'm still doing a couple solar projects. Commented Mar 23, 2023 at 17:38

Hard to say for sure without a proper schematic, but it appears as though you are using the transistors as emitter followers (common collector). If this is the case the voltage at the emitter will be the base voltage minus around 0.7 V, so if you have 5 V at the base, 4.22 V at the emitter is reasonable.

One solution to this is to use the transistors in a common emitter configuration with the load between the positive voltage and collector. In this case you will need a base resistor to limit the base current to a safe level.

You will also need to use a transistor capable of driving the loads. Your diagram shows a small signal transistor driving multiple incandescent lamps which it seems unlikely to be able to do. Incandescent lamps have a low cold resistance which will draw appreciably more current at turn on then they do when hot and this needs to be taken into account. The schematic you added is incomplete and lacks information on the type of transistors and lamps, so if you can add that it would be helpful.