BASIC 
ELECTRONICS COURSE 
Page 43 INDEX

A CLEVER CIRCUIT
The very clever circuit below converts an analogue signal into a digital signal. It looks very simple, and it is. But it's very difficult to work-on and diagnose, (if you get a fault) as you will see in a minute. 

HOW IT WORKS
The way the circuit works is simple. The first transistor is biased ON and this means the second transistor is not conducting. The voltage across the LOAD resistor is zero, so that any device connected to the output (in other words the device will be the resistance of the LOAD) will not be operating. An output device can be a LED, globe, motor, relay, an input line of a gate or microcontroller, or any other device. 
The advantage of this circuit is the rapid response of the output. The output goes from one state to the other during the time when the signal is rising only a few millivolts and this will take a very short period of time. 

The circuit is not a Schmitt Trigger, however it has the advantage of the speed of a Schmitt Trigger with a lot fewer components.
We said the circuit is difficult to diagnose because you would expect the collector of the first transistor to change appreciably when it is detecting an input signal. But this is not the case. 
Since it is connected to the base of the output transistor, the collector of the first transistor cannot rise above 0.7v, even when it is not conducting, because the base-to-emitter voltage of the second transistor prevents any rise above 0.7v. 
When the first transistor is turned on, the collector will fall to about 0.2 to 0.4v and because this voltage is lower than that required by the second transistor to turn on, it is controlled. 
But this means the collector of the first transistor will only change between 0.2v and 0.7v during its two states of operation and this is a very small change. 
If you connect a voltmeter to the collector, you may be tricked into thinking the circuit is not working. It is not until you realise the voltage change is extremely small, that diagnosis can begin. 
You will need to set a voltmeter to a very low range to detect the difference between the two states. The ON state will be less than 0.6v and the OFF state will be above 0.6v     That's how close the two will be!
The animation below shows the to transistors in the ON and OFF states. When a transistor is ON, it is equivalent to a low-value resistor and when it is OFF, it is equivalent to a high-value resistor. 

The voltage across resistor R and the LOAD also change during the two states. The animation below shows the voltage across these two components. Don't go by the animation for swing on resistor "R" because it sees less than 0.3v change from one state to the other! On the other hand, the voltage across the LOAD will change by almost full rail voltage!


DESIGNING THE CIRCUIT
Designing the circuit consists of working out the value of the three resistors: 
1: The base-bias resistor for the first transistor,
2: The load/turn-on resistor
3: The LOAD resistor. 

The values can be worked out by mathematics or simply determined by experimentation. You can spend lots of time working out the values by computation, but what values of current-flow are you going put into the equations? It's much better to generate the values by experimentation because there are lots of hidden tricks in driving different devices and the only way you will find out is to actually drive them in reality. 
For instance, relays, motors and solenoids have back EMF (a high voltage produced by the device) that can destroy the output transistor if it is not protected. 
The easiest way of protecting the transistor is to provide a reverse-biased diode across the device. Other devices take a high start-up current and unless this is factored into the design, the circuit will not work properly.
We start by deciding on the load we wish to drive. If it is a LED, we only need about 20mA. If it is a relay we need about 50 - 150mA. If the load is a globe or motor, we need to be aware of a very important fact. A globe requires about 6 times more current to get it to turn on. This is because the cold resistance of the filament is much lower than its operating resistance. Once it is turned on, the current requirement is as stated on the globe. A motor requires at least 4 times the running current to get it to start. In both cases this is called start-up current and is especially noticeable when you are trying to drive larger globes and motors. 
Due to the design of the circuit, you can see current flows through resistor "R" whenever power is applied to the circuit and this makes it a very wasteful design if you are using batteries. The circuit is only really suitable for low-current loads.
But if you are using a plug-pack, consumption will not be a factor.  
Firstly we need to decided on the current-value of the "LOAD." 
For a LED it will be 20mA, for a globe it can be from 100mA to 1,000mA (1A) and for other devices it can range from a few mA to 1amp or more. 
Once we know the current through the LOAD, we can work out the value of "R" in the diagram above. 
The way the circuit works is this: Current flows through R and through the base-emitter circuit of the output transistor, to turn the transistor ON.  The current through R is multiplied by the "gain of the transistor" and this current will pass through the LOAD (and collector-emitter circuit). If this current is exactly what is needed by the device, the circuit will work perfectly. This is called the DEVICE CURRENT. If the current is not enough, the globe will not turn on fully or the motor will not reach full-speed. 
This is a very dangerous situation as the transistor will get much hotter than expected and may overheat. This is a topic for further discussion, so we will concentrate on driving the transistor FULLY in this discussion. 
Now, here's the unusual part. Suppose the current through the collector-emitter of the transistor is mathematically MORE than required by the device! The result is the device will only allow the DEVICE CURRENT to flow. The additional current capability of the transistor is "up-your-sleeve" (available - but not used at the moment). 
Why do we need additional current-capability?
If we are driving a globe or motor, the starting current is 4 - 6 times the operating current and the value of R must be worked out using these values. 
During start-up, high current will flow through the LOAD and as soon as the globe or motor starts-up, the current will drop. But the high base-current will remain ALL THE TIME. 
We will give an example to show what we mean:
If a globe requires 100mA for its operation, the start-up current will be as high as 500mA. The gain of a transistor under high-current conditions will be about 50, so the base current will be 500/50 = 5mA. 
Suppose the rail voltage is 10v. The value of can be determined from Ohm's Law.  The value of resistance = 2k. You can use 1k8 or 2k2. 
The current through "R" flows all the time. When the first transistor is not turned on, the current will flow into the base of the second transistor and activate the LOAD. 
When the first transistor is turned ON, the current will flow through the collector-base leads of the first transistor and since the voltage between base-emitter of the second transistor will be less than 0.6v, the output stage will NOT be activated. 
We can now work out the value of the base-bias resistor. The collector current for the first transistor is 5mA. Suppose the gain of the first transistor is 100, the base current will be 5/100mA or 50 micro-amps. 
This current can be obtained from the base-bias resistor or from an external source that delivers the current to the base. The value of base-bias resistor = 200k. You can use 180k or 220k. 
The current to activate the circuit (the input current) must be a minimum of 50uA (50 microamps) and must be higher than 0.7v, to turn on the first transistor. 
All values of current depend on the voltage of the rail as well as the gain of the transistor(s). 
  
QUESTIONS
Question 151: What is the base-emitter turn-on voltage for a transistor?


Question 152: What is the voltage across the collector-emitter junction when a transistor is turned ON?


Question 153: In the circuit above, titled: "CONVERTS ANALOGUE SIGNALS TO DIGITAL," the two transistors are directly coupled. Explain how the first transistor turns the second transistor ON and OFF.

Question 154: If the collector-emitter voltage of a turned-on transistor was 0.8v, how would the circuit above work?

Question 155: In the circuit above, if the first transistor turned ON? What is the state of the second transistor?

Question 156: Describe how to change the state of the circuit:

Question 157: Why is the circuit a "wasteful" design?

Question 158: Why is the circuit only suitable for low-current loads?


AN IMPROVED DESIGN

In the next circuit we have an improved design. The circuit turns ON a lamp (or other device) when a very small current is delivered to the input. In this case the current is provided by a finger, touching two TOUCH-PADS. The circuit also has another feature. It will "latch-on" and the globe will remain illuminated. A second touch pad is provided to turn the circuit off. 

Mouse-
over:


to see
circuit
work


The TOUCH-PADS deliver current from the power rail to the input of the circuit, via a moist finger. The finger acts as a very high resistance but it will allow enough current to flow to turn the circuit ON. 
As the circuit turns ON, voltage is developed across the globe and this is passed to the "front-end" via a 4M7 FEEDBACK resistor. This resistor effectively takes the place of your finger so that when the finger is removed, the circuit will stay active. 
To turn the circuit OFF, Q1 is made active by the OFF touch-pad and this transistor removes the "turn-on" voltage to transistors Q2 and Q3. This turns off the output transistor Q4 and the voltage across the globe reduces. When the finger is removed from the OFF pads, the circuit has completely turned off and absolutely no current will be drawn.
The only current taken when the circuit is off is called LEAKAGE CURRENT and this will be LESS than 1uA and will flow through the junctions of the BC 557 when the transistor is OFF. 
This circuit is a vast improvement over the previous design and it can be used for all types of applications where an electronic "latch" is required. 
There are a number of technical features in this simple circuit and these will be discussed on the next page.


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