Beginner NPN query

Thread Starter

outsidelimits

Joined Oct 28, 2020
4
I am slowly learning about transistors. In my head I thought I had the theory of an NPN - a small voltage ~ 0.7V via the base can "switch on" the connection between the collector and emitter. I am not sure I have this correct though?

I have a simple cicuit setup as attached. When there is light on the LDR the LED stays off. When the LDR is covered, the LED comes on. However, it appears that the full current isn't being passed from the collector to the emitter, as I expected. That is, the LED is lighting up, but not to full strength.

Can someone help me out? Once the base is activated, shouldn't the full current flow from the C --> E ? The LED takes 100mA @ 12v so the N3904 should be able for that. What I am seeing is a dimmer LED that it should be
 

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ericgibbs

Joined Jan 29, 2010
12,270
hi ol,
Welcome to AAC.
Firstly you must have a current limiting resistor in series with a LED.
The NPN transistor has two common operating conditions, Saturated and Active

The NPN will act a switch when sufficient Base current drives the transistor into saturation, your Base drive does not supply enough Base current for that to happen.
So the LED will be 'dim'
E

Update:
What is the light and dark resistance of your LDR.?
 
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MrChips

Joined Oct 2, 2009
23,090
So far you are almost correct with your assumptions. Yes, the base-emitter voltage will be greater than 0.7V when the NPN BJT is turned on.

The mistake you are making is to assume that the BJT is a voltage device whereas it is a current device. You need to have sufficient current through the base-emitter PN junction in order to reach 0.7V.

Let us do some basic calculations using Ohm's Law.
With a supply voltage of 12V and base supply resistor of 100kΩ, the maximum base current will be

IB = V / R = 12V / 100kΩ = 0.12mA

If you assume a current gain = Ic / IB = 100
Ic = 0.12mA x 100 = 12mA which is eight times lower than the 100mA needed to turn on the LED.

The general rule of thumb is to use current gain = 10 in order to put the BJT into saturation mode.
In other words, you do not have enough current gain in your circuit. You can either add an extra transistor stage or you can replace the BJT with a darlington transistor or a MOSFET.

Edit: As a test, remove the LDR and replace R1 firstly with 10kΩ and then with 1kΩ.
 

djsfantasi

Joined Apr 11, 2010
7,563
You are using the NPN transistor as a switch. The general guideline is that the current into the base must be 1/10th of the current needed by the switched load.

So, the current into the base must be 10mA. If this guideline isn’t followed, the load will only get a portion of the current required. I.e., the LED will only light dimly.

You need to calculate the resistance required to provide 10mA. Assume Vd is the voltage from the divider formed by R1 and the LDR. Id must be 10mA. Calculate the value of R1 and see what you get.
 

ericgibbs

Joined Jan 29, 2010
12,270
hi ol.
When calculating the Base current, you must allow for the fact the 100K in series with the 20k LDR.
The LDR will shunt current in parallel with the Base junction, so the Base current in you circuit will be only approx 60uA when the LDR is at 20K
, not 100uA.
Lets us know the Dark resistance of the LDR.

E
 

LvW

Joined Jun 13, 2013
1,220
The mistake you are making is to assume that the BJT is a voltage device whereas it is a current device. You need to have sufficient current through the base-emitter PN junction in order to reach 0.7V.
If you hadn't mentioned it (as a "mistake"), I wouldn't have mentioned it either, because I don't want to reopen this (endless) discussion we had already in the past.
But - clearly spoken - your remark is really wrong ! Therefore, the assumption of the thread starter is correct in this respect.
The bipolar transistor is a voltage controlled device: Ic=f(Vbe): Shockley`s famous equation, Ebers-Moll/Gummel-Poon -Model.

There are many proofs for this - there is not a single physical proof for current control.
The relation Ic=B*Ib is of course correct, because an unavoidable base current flows, but a correlation is no causality in the sense of cause and effect.
 
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MrChips

Joined Oct 2, 2009
23,090
If you hadn't mentioned it (as a "mistake"), I wouldn't have mentioned it either, because I don't want to reopen this (endless) discussion we had already in the past.
But - clearly spoken - your remark is really wrong ! Therefore, the assumption of the thread starter is correct in this respect.
The bipolar transistor is a voltage controlled device: Ic=f(Vbe): Shockley`s famous equation, Ebers-Moll/Gummel-Poon -Model.

There are many proofs for this - there is not a single physical proof for current control.
The relation Ic=B*Ib is of course correct, because an unavoidable base current flows, but a correlation is no causality in the sense of cause and effect.
I understand your objection but in my defense nowhere in my post did I use the word "control" or "current control". I understand that Vbe is the independent variable and Ib the dependent variable.

As a low impedance device, it is difficult to determine I-V device characteristics by controlling the applied voltage. Instead, we usually adjust the current and then measure the junction voltage.
 

Thread Starter

outsidelimits

Joined Oct 28, 2020
4
I am more than a little confused now! I had thought the base was voltage driver, maybe back to the text for me!

The LDR 1M ohm dark and 10-20K in light (according to the spec of it)
 

MrChips

Joined Oct 2, 2009
23,090
I am more than a little confused now! I had thought the base was voltage driver, maybe back to the text for me!

The LDR 1M ohm dark and 10-20K in light (according to the spec of it)
Yes, the two resistors constitute a voltage divider. But as with all voltage dividers you have to determine the available current. There is not enough available current to the base-emitter junction of the BJT to put the transistor into saturation mode.
 

LvW

Joined Jun 13, 2013
1,220
I understand your objection but in my defense nowhere in my post did I use the word "control" or "current control". I understand that Vbe is the independent variable and Ib the dependent variable.
OK - agreed. But I was afraid that the questioner could misunderstand your formulation "You need to have sufficient current .....in order to reach 0.7V. ". This sounds as if the current would produce the voltage Vbe.
As a low impedance device, it is difficult to determine I-V device characteristics by controlling the applied voltage. Instead, we usually adjust the current and then measure the junction voltage.
Yes - in some cases. In particular, for switching applications - that`s correct.
However, while "adjusting" (calculating) the current, we are using - of course - a voltage of app. Vbe=0.7V.
 

ericgibbs

Joined Jan 29, 2010
12,270
Hi guys,
Remember the idea is to help the TS solve his simple LDR Topic problem, not to confuse him with arguments about V versus I in Base control.

Please start a new Thread if you feel the need to discuss V/I in Base operation.

E
 

ericgibbs

Joined Jan 29, 2010
12,270
hi outside,
What current is required for the LED type you are using.?
Also colour of the LED.
We can then calculate a series resistor value.
E
 

MrChips

Joined Oct 2, 2009
23,090
OK - agreed. But I was afraid that the questioner could misunderstand your formulation "You need to have sufficient current .....in order to reach 0.7V. ". This sounds as if the current would produce the voltage Vbe.

Yes - in some cases. In particular, for switching applications - that`s correct.
However, while "adjusting" (calculating) the current, we are using - of course - a voltage of app. Vbe=0.7V.
We both understand each other. However the TS does not understand that he is using a voltage divider with high resistances in an attempt to drive a load with low resistance. Vbe will never reach 0.7V because there is not enough available current.

Impedance matters.
 

ericgibbs

Joined Jan 29, 2010
12,270
hi,
Lets assume a White LED, say 3.2Vfwd nominal
So 12V-3V, 9V/0.1 =90 Ohms, lets use a 100R series resistor with the LED

Rwatt= 9V*0.1A ~ 1Watt , use a 2Watt

E
 

LvW

Joined Jun 13, 2013
1,220
We both understand each other. However the TS does not understand that he is using a voltage divider with high resistances in an attempt to drive a load with low resistance. Vbe will never reach 0.7V because there is not enough available current.
Impedance matters.
Yes, again I agree.
In this context - I think the primary reason for misunderstandings and misinterpretations is as follows:
* Undoubtly, the BJT is a voltage-driven device Ic=f(Vbe)
* However, in some cases (during design and analysis of circuits) it is sufficient (and easier) to treat the BJT as if it would current-driven Ic=B*Ib.
 

Thread Starter

outsidelimits

Joined Oct 28, 2020
4
hi,
Lets assume a White LED, say 3.2Vfwd nominal
So 12V-3V, 9V/0.1 =90 Ohms, lets use a 100R series resistor with the LED

Rwatt= 9V*0.1A ~ 1Watt , use a 2Watt

E
Thanks, for simplicity I've dropped the LED back to the white LED, as above, and the circuit works as expected. The problem now is that the actual LED is @12v ~ 100mA. I think, as mentioned above, I just can't get enough current into the base in this configuration.
 

ericgibbs

Joined Jan 29, 2010
12,270
hi ol,
With that basic circuit you will not be able to control the 100mA LED current, using a single transistor.
Do you have any more transistors in your stock.?
We could improve the circuit by adding another transistor.
E
 
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