# Increasing the forward voltage required of the PNP transistor?

I am a newbie and was reading about the uses of a 555 TIMER. I came across a project to create 40 LED Bicycle Light in Led Circuits.

I can understand the diagram and most of the explanation except where it says

The high output level of the 555 timer is 1.7 volts less than the supply voltage. Adding the LED increases the forward voltage required for the PNP transistor to about 2.7 volts so that the 1.7 volt difference from supply to the output is insufficient to turn on the transistor.

I can not wrap my head around this. And I would appreciate if someone could help me understand why the 555 timer output is 1.7 volts less than the supply voltage and why adding a LED to the base would further increase the forward voltage causing it to turn off the NPN PNP transistor when the 555 Timer output is High.

• Minor note: It is the PNP they are trying to turn off when the output is high, not the NPN. Commented May 26, 2022 at 17:42
• Umesh, what they are saying is that if you didn't insert that weird LED then the 555 output voltage might never be able to turn the PNP off as its voltage would always be low enough to supply some current no matter its state. So the LED is added to increase the voltage difference required such that the 555 output state can, in fact turn off the PNP. It's still not all that good a circuit. But it may work well enough for DIY.
– jonk
Commented May 26, 2022 at 17:53
• Yes my bad...I incorrectly mentioned in the last paragraph as NPN instead of PNP. I fixed it now. Commented May 26, 2022 at 19:33

The PNP transistor, is common-emitter, switching on the high-side. That means the emitter is permanently tied to +6V. Given that the base-emitter potential difference must be less than 0.7V for this transistor to be switched off, it's clear that the base must have a potential of above +5.3V.

You are told that the 555 output, when high, is not guaranteed to rise closer than 1.7V from its positive supply potential, meaning that it's possible for that output to reach only 6.0V - 1.7V = +4.3V, instead of the full +6V.

This maximum of +4.3V is not high enough to switch off the PNP transistor, which we discovered requires above +5.3V at its base, to switch off. The 555 output is a full volt short of that!

The author's solution is to employ an LED, whose forward voltage is significantly more than 1V. When the 555 output is low, the part of the circuit that concerns us looks like this:

simulate this circuit – Schematic created using CircuitLab

We see that there will be about 24mA of base current, to switch on the transistor. This is possible because the LED D1 is forward biased, and permits current to flow.

Note that the base resistor R1 is smaller than the one used for the NPN transistor, because there's no diode in the NPN's base path. The diode used here, in the PNP base path, drops about 1.5V, and what remains across the resistor is therefore not as much as for the NPN base resistor. Presumably we want the same current in both bases, when the respective transistor is on, so the resistors must be scaled accordingly.

Now let's see what happens when the 555 output rises, say to +1V:

simulate this circuit

The thing to recognise here is that the voltage at the base, and the voltage at node A (the LED cathode) have not changed. That's because diodes and base-emitter junctions (which are technically also diodes) always clamp their voltages to some maximum. The voltage across the base-emitter junction cannot ever rise above 0.7V. The voltage across the LED cannot rise above 1.5V.

In other words, as long as current is flowing through them, the combined voltage across the base-emitter junction and the LED will be 0.7V + 1.5V = 2.2V. The only component here whose voltage is changing is resistor R1. Its voltage reduced by the same amount we raised the 555 output, because the voltage across the other two elements in the path, the LED and base-emitter junction, is clamped to a combined maximum of 2.2V.

If you raise the 555 output potential further still, eventually you will reach a point at which the voltage across R1 becomes zero, where Ohm's law tells you that no base current can be flowing, and the transistor will turn off. That potential is +3.8V. That's when the remaining 6V - 3.8V = 2.2V across the base-emitter junction and LED becomes insufficient to forward bias them.

The LED has effectively shifted the switching threshold down from +5.3V to +3.8V, and this new threshold is well within the output voltage range of the 555.

The reason the 555 can't output 6V is simply that the transistors inside it which are responsible for producing the output potential are not ideal, and in fact are quite weak. When "on", these transistors have significant collector-to-emitter resistance still, and drawing any amount of current through them (such as the 24mA base current this circuit requires) causes them to drop some voltage internally. Under no load, the output won't suffer nearly so badly, but as load current increases, output signal amplitude diminishes.

• Add 25 ohms to Rb to get a worst case answer Commented May 26, 2022 at 21:22
• Hi Simon, Thank you for the simple yet detailed and clear explanation! I truly learnt something new today! Thanks again!!! Commented May 26, 2022 at 21:58

You can always unwrap your head on the schematic and see each 0.7V diode drops in saturation and base bias voltages with Vce bulk resistance to understand the Voh and Vol specs

The public datasheet description is non-standard calling it a Output Voltage Drop rather than a limit voltage high Voh and low Vol. The operating equivalent output resistance, Ro is pretty close to 25 Ohms worst case at 100 mA for "Rol" and 22.5 Ohms for "Roh" the output driver as I have added below.

Thus to compute the base resistor for each transistor with Ic/Ib=10 at Vce(sat) reduce the compute value by 25 Ohms to get a good value for base current.

There are better drivers but you can understand by looking a low current PN junction drops then add IR drop including bulk resistance or choose a current and compute worst-case equiv. resistance.

The PNP LED on the base is not necessary but 0.1A * 4.2V is 420 mW demands a 1W resistor that will rise 42 'C above ambient. So it just reduces the cost of the resistor and shares the base drive loss.

FET switches are far more efficient. >99% vs 90% for a BJT switch if the RdsOn is < 1% of load Req.

• Hi Tony, I truly appreciate the detailed answer. Since I am a newbie, I am unable to understand the details/theory. I am a self taught amateur hobbyist (and a hard skull) who understands the basics of turning on and off a transistor, LED and 555 Timer. I have tried building basic circuits on a breadboard by following a book or an online tutorial. Commented May 26, 2022 at 20:04
• Learn Ohm's Law for drivers and resistors . Assume Vbe=0.7 for now Commented May 26, 2022 at 21:19