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Please explain this circuit to me

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Hi, gang. I'm working my way through Simple Circuits Volume I from Jaycar in Australia. Here's a schematic. I've built the circuit. May I ask a few questions about this circuit:

1. What is the purpose of the 100k Ohm resistor?

2. Why is the transistor in this circuit at all? Why don't you put the probes between the flashing LED and its 470 Ohm resistor, for example?

3. What is the relationship between the Emitter->Base current and Emitter->Collector current? Is there some sort of graph that will document it?

Richard
 

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Smartie

Member
3.) The transistor is a semiconductor, made up of 3 layers of silicon. that have been doped. in your case the transistor is PNP. meaning the layers are doped P-Type, N-Type and then P-Type.
The relationship between the the Base of the transistor to the Collector of the transistor is Ic = Ib*HFe. (Current through the Collector is equal to the current through the base of the transistor times a constant value dependant on the transistor itself, typically 200 to 300).
I don't know everything about the transistor, but you can find more info about it here:
Bipolar junction transistor - Wikipedia, the free encyclopedia.
the current through the emitter is equal to Ic + Ib (collector current plus emitter current)

2.) Thre transistor is used as a electronic switch :p

1.) the 100kΩ resistor is used to bias the base of the transistor, bring up the voltage just enough to switch off the transistor (if the transistor was NPN, then it would be on)
 
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ash20

New Member
1.) I think 100k resistor is used to decrease the amount of current flow to the base. a transistor base is lightly doped and hence needs very
less current to drive it. hence we always use a high value resistor to decrease the current value for base... or else, transistor will not work
and may damage...

2.) This ciruit is a FIXED-BIAS PNP transistor circuit (part involving +ve termial of battery, 100k resistor, 47k resistor around the transistor).
In this configuration, the transistor is thermally unstable.
This is solved by putting the transistor under common emiter configuration.
The 470 ohm resistor is used to increase the stability of the fixed bias circuit making it a emitter-stabilized bias circuit.

I dont know the use of transistor in the circuit since we can omit it and simply use the 470 ohm resistor for the flashing led to work
properly. you can re-read the text for which the circuit is made... they may show some more details...

3.) One basic relation is --> IE = IB + IC + leakage current (may sometimes be omitted)
Rest relations depend upon the type of biasing circuit used... like for this emitter stabilized fixed-bias circuit, some relations are:

IB = (VCC – VBE) / (RB + (β + 1) RE)
VCE = VCC – IC (RC + RE)
ICSAT = VCC / (RC + RE)
 

audioguru

Well-Known Member
Most Helpful Member
1.) I think 100k resistor is used to decrease the amount of current flow to the base. a transistor base is lightly doped and hence needs very
less current to drive it. hence we always use a high value resistor to decrease the current value for base... or else, transistor will not work
and may damage...
No it is not. The 100k resistor turns off the transistor. The 47k resistor limits the base current.

2.) This ciruit is a FIXED-BIAS PNP transistor circuit (part involving +ve termial of battery, 100k resistor, 47k resistor around the transistor).
In this configuration, the transistor is thermally unstable.
This is solved by putting the transistor under common emiter configuration.
The 470 ohm resistor is used to increase the stability of the fixed bias circuit making it a emitter-stabilized bias circuit.
No, no and no.
The transistor is a stable on-off switch.
It is already a common-emitter configuration.
The 470 ohm resistor has nothing to do with the transistor, it limits the LED current.

I dont know the use of transistor in the circuit since we can omit it and simply use the 470 ohm resistor for the flashing led to work
properly.
The transistor allows a very weak current to turn on the transistor which applies a much higher current to the LED.
 
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ash20

New Member
No it is not. The 100k resistor turns off the transistor. The 47k resistor limits the base current.

if you make the circuit in a circuit analyser, then you will find that 100k is not turning the transistor off... or else, the author himself/herself would not have given it...

and the 470 ohm along with limiting the led current is making the transistor emitter stabilized under fixed bias mode... to stabilize the thermally unstable transistor.. it is performing dual function here....
 

audioguru

Well-Known Member
Most Helpful Member
if you make the circuit in a circuit analyser, then you will find that 100k is not turning the transistor off... or else, the author himself/herself would not have given it...
The 100k resistor turns off the transistor when the input signal becomes an open circuit.
But the 100k resistor really should be between the base and emitter of the transistor.

the 470 ohm along with limiting the led current is making the transistor emitter stabilized under fixed bias mode...
What is "emitter stabilized"? The transistor is simply an on-off switch.

... to stabilize the thermally unstable transistor.
The transistor is not thermally unstable. The input signal going low turns on the transistor through the 47k resistor (but the value is much too high, it should be 15k) and the 100k resistor turns off the transistor when the input signal becomes an open circuit.
 
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ash20

New Member
The transistor is not thermally unstable.

you need to re-read about transistor and the different biasing modes it is put into... your knowledge on transistors is limited to switches only.... please re-read transistor biasing techniques before commenting further...
 
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ash20

New Member
The transistor is not thermally unstable.

you need to re-read about transistor and the different biasing modes it is put into... your knowledge on transistors is limited to switches only.... please re-read transistor biasing techniques before commenting further...
 

Diver300

Well-Known Member
Most Helpful Member
Hi, gang. I'm working my way through Simple Circuits Volume I from Jaycar in Australia. Here's a schematic. I've built the circuit. May I ask a few questions about this circuit:

1. What is the purpose of the 100k Ohm resistor?

2. Why is the transistor in this circuit at all? Why don't you put the probes between the flashing LED and its 470 Ohm resistor, for example?

3. What is the relationship between the Emitter->Base current and Emitter->Collector current? Is there some sort of graph that will document it?

Richard

Audioguru is correct, as always. I think that Ash20 should be more careful.

To answer your questions in a different order:-

3: A guide to the collector current is that it is a fixed multiple (the gain, or hfe, or 200 to 300) times the base current. Also, the collector current can't exceed what the circuit will supply, in this case about 8 mA, limited by the 470 Ω resistor.

2: The transistor makes the circuit more sensitive, so far less current at the probes is needed to turn on the LED. The transistor would not be needed for simply testing continuity.

1: The 100 kΩ resistor is to stop the circuit being too sensitive. Without it, a little bit of moisture on the circuit would turn on the transistor and the LED.

The 47 kΩ resistor limits the base current so that nothing goes bang if the probes are shorted together.
 
How did you get the maximum current that the circuit will supply? Is it (voltage of battery / (resistance of transistor + resistance of resistor + resistance of LED) )? One should also add the additional pathways through the 47k and 100k resistors, but they'll be trivially small. I've put my ammeter into the circuit and it goes up to about 0.045A (when the LED is on).
 

Diver300

Well-Known Member
Most Helpful Member
If the transistor is turned on completely (saturated), its resistance can be ignored. The LED will not really have a resistance but will drop about 2 V whatever current is taken. So the calculation becomes (voltage of battery - 2)/resistance of resistor which is (6-2)/470 = 0.0085 A or 8.5 mA

I can't see how you can get 0.045 A. It's not possible with a 6 V battery and a 470 ohm resistor.

You might have:-
a faulty ammeter
a lower value resistor
an incorrectly connected ammeter
 

ericgibbs

Well-Known Member
Most Helpful Member
How did you get the maximum current that the circuit will supply? Is it (voltage of battery / (resistance of transistor + resistance of resistor + resistance of LED) )? One should also add the additional pathways through the 47k and 100k resistors, but they'll be trivially small. I've put my ammeter into the circuit and it goes up to about 0.045A (when the LED is on).

hi,
The only way you can get 45mA, is if you are using a 9V battery and a 47R in series with the LED.????

EDIT:
Image
 

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audioguru

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The 47k resistor does not supply enough base current for the BC558 transistor to saturate properly. It supplies a base current of only 113uA when the probes are shorted together. Then weak but passing transistors can saturate with a collector current only 20 times more which is only 2.2mA.
Its hFE is used only when it is a linear amplifier with a collector-to-emitter voltage of at least 5V so that it is not saturated.

About the thermal condition:
The datasheet shows that if the collector-base voltage is max at 30V and the junction is max at 150 degrees C then the leakiest collector-base junction conducts a max of 15uA when the base is not connected to the resistors. But the voltage and temperaturre are much less than max and the 100k resistor conducts away some base leakage current so normal temperatures have no effect. Thus the transistor is thermally stable.
 
Sorry, I put the decimal point in the wrong place. The circuit carries up to 0.0045 Amps. Note that the LED flashes faster than my DMM updates itself, so it's probably already an imprecise measurement. I'm using 4 AA batteries (MN1500 LR6) that together produce 6.42 Volts.
 

Diver300

Well-Known Member
Most Helpful Member
The 0.0045 A is 4.5 mA, or about half the 8.5 mA that I estimated. That is probably because the LED is only on half the time so that the current shown on the meter is about the average current, which would be about in line with what would be expected with a 470 ohm resistor, and the current only being there half the time.
 

JCLrd

New Member
So with all this said and done, are you understanding that this circuit is being used as a current amplifier, where a small signal current is introduced into the base emitter loop, and a much larger current is outputted from the transistor into the load (LED).
 
Yes, I understand that. What confuses me is that the E is common for both C and B. I understand relays much more intuitively. For a relay, the two circuits can be entirely separate.
 

JCLrd

New Member
The power supply is 6v.
Looking at the data sheet for that transistor, shows a VCE typ.~=90mV. @ IC of 10mA.
So lets use that data to try to analyse this circuit.

Since were using 10mA. for the collector current, then the voltage drop across the 470 ohm resistor will be 4.7V. Also from the Data sheet for the transistor, we can assume a value of 90mV. for the drop across this transistor, when it is saturated.
That makes a toal of ~=4.8v.
Subtract that from the power supply voltage and we have ~=1.2V. left over which is the drop across the flashing LED.

Now to make this transistor produce this current, we have to introduce a current that is around 20 times less than it's collector current. This is the base current.
The data sheet shows a value of IB to be 0.5mA. when IC is 10mA.

Now what value of resistance would be needed across the probes to introduce 0.5mA. of current into the base.

The data sheet shows 700mV. for the VBE, under saturated conditions. (transistor is used as a switch).

So to calculate the resistance needed across the probes would be like this.

(VCC - VBE) / IB = Rprobe. (6V. - 0.7V.) / 0.5mA. = 10600 ohms.

Now follow carefully what this is showing.

If you put a resistance of around 10 thousand ohms across the probes, it will allow a very small current of 0.5mA. to flow through these probes. BUT this small current through these probes, is also flowing into the base of the transistor to turn it on. When this transistor is turned on by this amount of current into its input base, then it opens up a vast amount of current iup to 10mA. to flow from its collector to emitter, since this collector to emitter of the transistor is in series with the LED, and the LED must have at least 10mA. for it to work, than this 10mA. that the transistor is supplying is able to work the LED.

WHILE at the same time it only took a current of 0.5mA to make all this happen.

Remember this 0.5mA. can NOT turn on the LED, it is too small of a current.

So why the transistor in the circuit, it acts like a relay, its base circuit is in effect a seperate circuit like a relay coil, ant the collector to emitter is like the relay contacts to the load being driven.

So it is like a relay, only the terminals are all connected with the same voltage, however the base loop is seperate from the collector emitter loop, as it is a branch off of the collector emitter loop.

The emitter always has the full amount of current flowing through it, while a small portion flows into the base, and the rest into the collector.

So see if that helps better understand the purpose of the transistor in this circuit.

PS,:
as adioguru said the 47K won't allow saturation.

I forgot to add the 47K resistor into the equation.


That should be 4.7K ohms so that a 5K resistance across the probes would make this circuit work.

But you get the idea of how thios could work with the proper resistor values.
 
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MrAl

Well-Known Member
Most Helpful Member
Hi, gang. I'm working my way through Simple Circuits Volume I from Jaycar in Australia. Here's a schematic. I've built the circuit. May I ask a few questions about this circuit:

1. What is the purpose of the 100k Ohm resistor?

2. Why is the transistor in this circuit at all? Why don't you put the probes between the flashing LED and its 470 Ohm resistor, for example?

3. What is the relationship between the Emitter->Base current and Emitter->Collector current? Is there some sort of graph that will document it?

Richard

Hello there,

This looks to me like a relatively simple continuity tester where the LED lights when the probes have a relatively low resistance between them and does not light otherwise.

Because of the 47k resistor and the 470 ohm resistor it looks like the designer was counting on a DC current gain of around 100, so it may not be designed for large temperature swings but that probably wasnt the intent anyway. The 100k resistor is there to help turn off the transistor when the probes are open circuited. It is connected to one of the input probes rather than across the base emitter junction because it works slightly better connected to the probe as the input dynamic is somewhat better when the entire probe voltage controls the transistor base rather than just part of it (if the 100k was across the base emitter it would form a voltage divider which would steal some base current all the time rather than just when we really want it to do so). When the probes are shorted for example, the base current becomes high enough to cause the transistor collector emitter to conduct more heavily thus turning the LED on.
The advantage of having a transistor is of course that it has current gain and that current gain can be used to raise the apparent impedance of the LED. If we drive the LED directly from the probes we would find that the resistance between the probes would have to be relatively small to turn the LED on, but with the transistor in the circuit that resistance can be much higher and thus causes less current to flow in the test circuit connected to the probes. This also means the circuit under test is less affected by the test itself by having the transistor there.
 
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