First – A little ramble-on: -

There were so many new posts appearing on this blog – and then, all of a sudden, they stopped. – What happened?

The answer is that I haven’t written any for a while as I’m going through a bit of a personal crisis. – BUT, having said that, this is, in itself, a post; so I can’t really be accused of not posting.

I bought a new second-hand scanner yesterday. – I don’t even know if it’ll work with Windows 7 64-bit (the only operating system I use now) yet as I haven’t even managed  to get it set up. That’s something I’ll need to get round to this weekend. The reason I wanted to use the scanner – for the purposes of this post – is to illustrate what I was saying in the form of scanned-in hand-drawn circuit diagrams.

Update: As I thought; it’s ages old and has no available driver for Windows 7 64-bit. – That was a waste of £5!

Right at the moment I can’t do that, so I’m going to try this without illustrations; just to find out how easy/difficult it is. Hopefully I’ll have a working scanner next time I post, but for now we’ll make do with words, and hope that words are enough. I hope that you can follow this. Just a note to say that some working knowledge of electronics is required when reading this article.

Down to business: -

Okay; so I want you to visualise, if you will, an NPN bipolar transistor – maybe a 2N1111, connected in a circuit as a very simple DC Amplifier: That being with a potential divider connected to the base which is also the input terminal. The collector is connected to the positive rail via a collector-load-resistor, and the emitter is connected to 0v via an emitter-load-resistor; therefore essentially placing the device in the centre of another potential divider.

Values for the components? Well let’s say that, with Vs at 9 volts, we’ll give the resistor, R1, connected between the base and + rail, a value of 10 kilohms, and the resistor, R2, connected between the base and 0v, a value of 1.1 kilohms; thus biasing the base at near exactly 0.7 volts, the transconductance threshold.

The collector load resistor (R3) can be given a value of 1 kilohm, and the emitter load resistor (R4) a value of 110 ohms. – That’ll set the transistor’s working beta at a factor of almost dead on 10. – Those 4 resistors and the transistor give us a very basic DC amplifier.

If you were to connect the circuit as an emitter-follower and take the output from the top of R4 at its connection to the transistor’s emitter then the device acts as a non-inverting amplifier stage. If you take the output from the transistor’s collector at the bottom of R3, however, the device acts an inverting amplifier (of sorts). This is because, when the transistor is conducting, when a positively-biased DC input signal is present at the base, and Vb>0V7, the value of Vc drops as the value of Ve increases due to the potential divider effect across the transistor’s substrate; thus causing the potential difference, Vce, to be equal to the device’s transconductance threshold. When the base voltage swings negative beyond 0V7, however, the device stops conducting and the value of Vc returns to rail voltage while Ve returns to 0 volts; thus increasing Vce to Vs.

We can do a lot with that potential difference, especially if we connect the inputs of a 741 op-amp across the device: The 741 will output the actual potential difference of 0V7 or 9V, rather than the input’s characteristic waveform, and if we were to take it a stage further still and connect the output of the 741 op-amp to a low-current 5V regulator device – then we’d have a crudely digitised representation if the input signal; a square-waveform at either 0V or 5V.

– Ladies and gentlemen we have just reinvented the world’s simplest analogue to digital converter. Electronics is such fun!

Cue the adverts: -