Inverter question

Thread Starter


Joined May 18, 2008
Hey guys,

In one of my labs, I had the task of constructing this inverter with a transistor.

I constructed it, and applied a 5V square wave of 50kHz to it, and then looked at it on the oscilloscope. We could the see the output was indeed inverting the signal, but the output wasn't square like the input.

Is it because of capacitance inside the transistor? Which ones and what are they called? is it Cje or Cjc??

When we turned down the variable capacitor in the diagram, all the way down, we noticed the output was almost square like the input. Very nice.. But I have literally no idea how this 'speed-up' capacitor works.

I've been staring at it for ages!! But still I can't see how that var. capacitor on the base of the transistor works.. Can anybody help?



Joined Feb 12, 2009
The capacitance between the base and the collector is the main reason why the rising and falling edges are slow. This is not helped by the high value of the base resistor

I take it that points 5 and 6 are joined together. Adding capacitance here will bypass the base resistor at high frequencies.

Best to drive the base from a low impedance source. This could be a potential divider made of fairly low value resistors. Less than 1k. Also don't drive the transistor fully "on" or "saturated". This means not letting the collector volts going lower than about 0.6V. On old TTL logic circuits this was done by connecting a reversed shottkey diode between base and collector.


Joined Feb 4, 2008
The circuit arrangement is not very good because if you set the 100K pot at 0 ohms then the transistor will blow. Also, 100K is very high to drive the transistor base. Use a 1K fixed resistor and a 5K variable resistor if you want to vary the base resistance. When you turned the capacitor all the way down you have actually set it to maximum capacitance and thus the input square wave was transferred more effectively because of the lower impedance of the capacitor. The purpose of this capacitor is to reduce the turn on time of the transistor by supplying more current at the rising edge of the input square wave. Also, it reduces the turn off time (its main purpose) by removing the stored charge in the base due to the diffusion capacitance faster (it applies an instantaneously negative voltage at the base when the square wave is 0V).