Darlington transistor with LED

Design a circuit, using a darlington transistor, to monitor the voltage Vin which appears on an input line, and turn on an LED if Vin exceeds 4.8V. The circuit should run from +10V.

The current drawn from the line whose voltage is being monitored should not exceed 100uA.

Indicate component values.

(The correct answer is attached).

I am not sure about a couple of things:

*Shouldnt the 300k resistor be connected to the +10V positive rail (parallel to the LED), so that it is in line with the 100k (by voltage division)?
*Where does the input voltage (Vin) come from?
*Howcome the darlington transistor is connected as an earthed emitter, as opposed to emitter-follower?

The 300 and 100K are a simple voltage divider. The darlington will switch on when the base voltage is around 1.4V (2 silicon diode drops) - check the datasheet for a more accurate value. Vin is the voltage you are trying to compare to 4.8V

Markd77 said:

The 300 and 100K are a simple voltage divider. The darlington will switch on when the base voltage is around 1.4V (2 silicon diode drops) - check the datasheet for a more accurate value. Vin is the voltage you are trying to compare to 4.8V

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Would the current spilt at the point directly above the 100k resistor (one current going to 100k earth, the other going into base of darlington), which is what the simple voltage divider is doing?

I understand that vin is what you comparing to the 4.8V, but what voltage source would you use? The only source I can think of is the same +10V power supply. My datasheet says 1.2V.

How would you know that you must draw the darlington transistor connected as an earthed emitter, as opposed to emitter-follower?

An emitter-follower has no voltage gain so it does not switch. A common-emitter transistor has a high voltage gain so it can switch.
An emitter-follower has a DC voltage loss that varies with the current and temperature.

When the input is 4.8V then it is across both resistors in series. Then the lower resistor has 1.2V across it which turns on the darlington transistor that needs an extremely small base current to turn on.

Audioguru said:

An emitter-follower has no voltage gain so it does not switch. A common-emitter transistor has a high voltage gain so it can switch.
An emitter-follower has a DC voltage loss that varies with the current and temperature.

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How can an emitter-follower have no voltage gain if it is connected to the positive rail and what do you mean by switch?

Audioguru said:

When the input is 4.8V then it is across both resistors in series. Then the lower resistor has 1.2V across it which turns on the darlington transistor that needs an extremely small base current to turn on.

Click to expand...

How would you know to use the certain component values of the resistors, in order for the lower resistor to have 1.2V across it?

These should help:
http://en.wikipedia.org/wiki/Emitter_follower
http://en.wikipedia.org/wiki/Voltage_divider

An emitter-follower has no voltage gain. If its input voltage increases 0.2V then its output increases only 0.19V.

A common emitter transistor has lots of voltage gain so if its input voltage increases 0.2V then its output saturates and completely turns on its load.

The resistors are a simple voltage divider. Ask your teacher to explain how a simple voltage divider works, or use common sense and Ohm's Law.

Audioguru said:

An emitter-follower has no voltage gain. If its input voltage increases 0.2V then its output increases only 0.19V.

A common emitter transistor has lots of voltage gain so if its input voltage increases 0.2V then its output saturates and completely turns on its load.

The resistors are a simple voltage divider. Ask your teacher to explain how a simple voltage divider works, or use common sense and Ohm's Law.

Click to expand...

Oh so basically an emitter-follower will only produce an output that is equal to input, so there is no extra gain, while a common emitter's voltage gain is amplified?

I know how a voltage divider works (vout = V * R1 / R1 + R2), but it only makes sense to me in the new attached diagram (which I edited myself), as opposed to the actual answer (first attachment). Where your V is your positive rail, your vout is the same line as vin.

TsAmE said:

Oh so basically an emitter-follower will only produce an output that is equal to input, so there is no extra gain, while a common emitter's voltage gain is amplified?

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Yes, a common emitter transistor has voltage gain.
the emitter of a darlington transistor as an emitter-follower is the input voltage minus about 1.2V so will not be high enonough for most functions.

I know how a voltage divider works (vout = V * R1 / R1 + R2), but it only makes sense to me in the new attached diagram (which I edited myself

Click to expand...

You have the supply voltage turning on the darlington transistor.
I show the 4.8V signal divided then barely turning on the darlington transistor.

Thanks that makes a lot of sense now. So the transistor is connected as a common emitter as it has switching (2 states: on/off), as opposed to an emitter-follower which is continuously on? (as in question it asked to TURN ON an LED) Right?

A darlington transistor as an emitter-follower might not have enough output voltage to light a 3.6V blue or white LED when its base voltage is only 4.8V because its emitter voltage will be only 3.5V or 3.6V.

Audioguru said:

A darlington transistor as an emitter-follower might not have enough output voltage to light a 3.6V blue or white LED when its base voltage is only 4.8V because its emitter voltage will be only 3.5V or 3.6V.

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Why would the emitter voltage be 3.5V/3.6V, as by the time Vin (4.8V) goes through the voltage divider it becomes 1.2V, then vin - 1.2V (datasheet value of darlington) = 0?

TsAmE said:

Why would the emitter voltage be 3.5V/3.6V, as by the time Vin (4.8V) goes through the voltage divider it becomes 1.2V, then vin - 1.2V (datasheet value of darlington) = 0?

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You don't need a voltage divider to feed voltage to the base of an emitter-follower. The base of an emitter-follower also does not need a current-limiting resistor.

A transistor has a base-emitter diode that has a voltage drop of about 0.6V to 0.7V. If 4.8V is applied to its base then 4.1V to 4.2V will appear at its emitter.
But a darlington transistor has two transistors with their base-emitter diodes in series. Each base-emitter diode also has a voltage drop of 0.6V to 0.7V.
If 4.8V is applied to the base then about only 3.6V will appear at the emitter.

Thanks I understand I have another doubt. In the question when it states: "The current drawn from the line whose voltage is being monitored should not exceed 100uA."

Is this the Ic, as the collector is always the output of a transistor?

TsAmE said:

In the question when it states: "The current drawn from the line whose voltage is being monitored should not exceed 100uA."

Is this the Ic, as the collector is always the output of a transistor?

Click to expand...

No.
The resistor voltage divider and the base current of the darlington transistor both draw current from the 4.8V source being monitored.
The resistors draw 12uA (4.8V/400k) and the base of the darlington transistor (I looked at the datasheet of an MPSA14) is about 25uA max when its collector is saturated with 25mA. So the total drawn from the source is a max of 37uA but is higher when the input exceeds 4.8V.

Oh ok, but how would you work out the resistances in order to use them in the previous calculation of yours?

I got as far as (using voltage division):

vin = V * R2 / R1 + R2
4.8= 10 * R2 / R1 + R2

The 10V is for the switch and the LED, not for the voltage divider.
The 4.8V is the input that goes to the voltage divider.

The base of the darlington transistor begins to turn on with 1.2V. If the base to ground resistor of the voltage divider is 100k then its current is 1.2V/100k= 12uA.
The other resistor has 4.8V - 1.2V= 3.6V across it. It has the same 12uA of current as the other resistor so its value is 3.6V/12uA= 300k ohms.

Audioguru said:

The 10V is for the switch and the LED, not for the voltage divider.
The 4.8V is the input that goes to the voltage divider.

The base of the darlington transistor begins to turn on with 1.2V. If the base to ground resistor of the voltage divider is 100k then its current is 1.2V/100k= 12uA.
The other resistor has 4.8V - 1.2V= 3.6V across it. It has the same 12uA of current as the other resistor so its value is 3.6V/12uA= 300k ohms.

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I understand what you mean, but in the question you arent given the resistance values, which is why I dont get how you work out R1 and R2, it like too little info is given.

The LED must indicate when the input is more than 4.8V.
The circuit must not draw more than 100uA when the input is more than 4.8V.
You showed a voltage divider with 300k and 100k.

The statement "the input must not draw more than 100uA when the input is higher than 4.8V" does not make sense because of course the input current is higher when the input voltage is higher.

If the input is 4.8V then the input current is 4.8V/400k= 12uA.
If the input is 12V then the input current is (12V - 1.2V)/300k= 108uA.
If the input is 102V then the input current is 1.0mA.

The datasheet for the darlington transistor lists its minimum current gain at whatever LED current you want. Then calculate the base current and calculate the voltage divider resistor values. Simple.

I see. Thanks a lot for the patience haha .