Attending: Tony, Elizabeth, Kieran, Joe, Mike, and welcome to a new visitor, Stuart, who has just moved to Blackpool.
Tony installed a LAMP stack into virtualbox with the intention of doing some web work on the likes of Drupal, Wordpress, Mambo, and/or Joomla.
Elizabeth has been dual booting Suse Linux and Linux Mint for a while now, and has come to the conclusion that Mint is 'The One'
Kieran was working on a game using Python, and Joe was working on a game server.
Stuart described his introduction to computing:
"Started in 86 after college doing old dos support with things like wordperfect and lotus123 and have been doing it for so long that I guess I am caught now.First computer a ZX81 in 79 Oooof those where the days :)"
The breadboard version of the logic probe working.
It looks a bit scary with all the jumper wires, but there are only a few components involved.
The next step is to assemble it into the barrel of a pen.
A logic probe is a useful piece of equipment to have when fault finding on digital circuits.
Here is a description of a logic probe on Wikipedia http://en.wikipedia.org/wiki/Logic_probe
Some very basic theory using this diagram.
The two jagged lines labeled R1 and R2 are resistors.
R1 will be connected to the positive supply, and R2 will be connected to the negative supply,
which will provide Point A (R1) with the voltage of the circuit under test, which can be up to about 15Volts for the CMOS chip we are going to use later. Here we will use 12volts, and Point B (R2) will be 0volts.
With point C unconnected, current (electricity) will flow between point A and B through the resistors R1 and R2.
If point C is used as a probe and connected to 12volts, no current will flow through R1 between points A and C, because they are both at 12volts. This is described as there being no potential difference (PD) between point A and Point C.
However, there is a 12volt PD between point C and point B, and current will flow through R2.
If the connection is now reversed, and the probe (point C) is connected to 0volts, the opposite happens, and current flows through R1
Electricity always takes the path of least resistance
When point C is connected either way around, it cuts out one of the resistors from the circuit.
Add two Light Emitting Diodes (LED) to the circuit, back to back as shown.
Now when the probe, point C is used, one LED will light up when connected to 12volts, and the other will light up when connected to 0volts.
LEDs are diodes, and will only pass current in one direction, which is why both LEDs do not light up at the same time. If the probe was connected to an output that pulsed high and low, then both LEDs would flash on and off, but in most cases, that would happen too fast for the eye to see.
This circuit can not be called a logic probe for a couple of reasons, but it does supply a basic understanding.
A third LED is needed to indicate pulsing outputs. This LED needs to be connected to additional circuitry to be able capture and latch a single transient pulse, or to stretch out high speed pulse trains long enough for the eye to see that pulses are happening. A CMOS 4001 integrated circuit will be used to achieve this.
Also, attaching the probe should have as little effect as possible on the circuit under test, or in other words, attaching the probe should not cause the circuit under test to behave differently. This can be a very complex topic. Here is an article which covers oscilloscope probes called "understanding the probing problem" http://teledynelecroy.com/doc/probes-probing
Here is a full circuit for construction, provided by http://www.zen22142.zen.co.uk/Circuits/Testgear/lprobe.html
The two resistors and two LEDs previously described can be seen in this circuit, near the probe. The LEDs are marked HI and LO.
IC1, the CMOS 4001, costs a few pence.
The logic probe can display four output states, High, Low, Pulsing and tri-state (or high impedance). The tri-state output is a high impedance state in which the output pin has no value, it is not at logic 0 and not at logic high.
IC1a (Pinout pins 1,2,3) is wired as an inverter with a 2M2 feedback resistor. With the probe not connected to a logic circuit, the output of gate 1a, pin3 is fed back to the input, pins 1 and 2 via the 2M2 resistor. This gate will oscillate at a very high rate and resultant output voltage at pin 3 will be approximately half the supply voltage. The Hi and Lo logic indicator LED's are also connected to a potential divider consisting of the two 1k resistors. The voltage at the resistor junctions is also half supply voltage so with no input, no output LED's light representing the tri-state.
A Hi or Lo logic condition at the probe input, will cause IC1a to rest in a permanent state indicated by either the Hi or Lo LED illuminating.
With a fast oscillator or clock signal input both Hi and Lo LED's will light but appear quite dim.
To increase brightness, the input signal is slowed down. This is achieved by gates IC1b and IC1c which are wired as a monostable.
The time constant for the monostable is determined by the 100n capacitor and 4M7 resistor. A fast input pulse now continually triggers and re-triggers, the monostable, effectively slowing the input signal. The output of the monostable is inverted by IC1d, wired as an inverter, and increasing output current to the pulsing LED.
http://electronicsclub.info/cmos.htm supplies this info about CMOS chips
- Supply: 3 to 15V, small fluctuations are tolerated.
- Inputs have very high impedance (resistance), this is good because it means they will not affect the part of the circuit where they are connected. However, it also means that unconnected inputs can easily pick up electrical noise and rapidly change between high and low states in an unpredictable way. This is likely to make the IC behave erratically and it will significantly increase the supply current. To prevent problems all unused inputs MUST be connected to the supply (either +Vs or 0V), this applies even if that part of the IC is not being used in the circuit!
- Outputs can sink and source only about 1mA if you wish to maintain the correct output voltage to drive CMOS inputs. If there is no need to drive any inputs the maximum current is about 5mA with a 6V supply, or 10mA with a 9V supply (just enough to light an LED). To switch larger currents you can connect a transistor.
- Fan-out: one output can drive up to 50 inputs.
- Gate propagation time: typically 30ns for a signal to travel through a gate with a 9V supply, it takes a longer time at lower supply voltages.
- Frequency: up to 1MHz, above that the 74 series is a better choice.
- Power consumption (of the IC itself) is very low, a few µW. It is much greater at high frequencies, a few mW at 1MHz for example.