I also recall studying that a change in the voltage V(be) causes a change in Ic. And I believe in npn transistos the current can only flow from the collector to base due to the flow of electrons

 

Ib = IC/gain = 21/100 = .21 mA is required to turn the transistor on

Good job!

 

I also recall studying that a change in the voltage V(be) causes a change in Ic. And I believe in npn transistos the current can only flow from the collector to base due to the flow of electrons

True dat. Cant have current without voltage to push the electrons through. Speaking of which let's not get into an electron flow discussion. Since your getting the Cliff Notes here we can save that topic for another Oprah show.

 

OK, lets see what we know so far.

The collector current is 21 mA. This is the load consisting of one resistor and four LEDs.

We know that we need .21 mA to turn the transistor on.

Based on that I am going to recommend the use of a 2N3904 NPN transistor. Why did I pick that particular transistor? Because I am familiar with its properties from the data sheet. So now you will become familiar with the data sheet.

Your next assignment is to download the data sheet and report back on the following properties:

Absolute Maximum Ratings

Collector - Emitter Voltage (V)
Collector Current Continuous (mA)

Thermal Characteristics

Total device dissipation (mW)

On Characteristics

DC Current Gain (Ic = 10 mA)
DC Current Gain (Ic = 50 mA)
Base-Emitter Saturation Voltage (Ic=10 mA, Ib=1 mA)
Base-Emitter Saturation Voltage (Ic=50 mA, Ib=5 mA)

Good Luck!

 

Collector Emitter Voltage : 40V
Collector Current Continuos : 0.2000A
Total Device Dissipation : 625 mW
DC Current Gain: (Ic=10mA) = 50
DC Current Gain = 250
BE Saturation Voltage: V(t)log(Ic/Ib) = V(be)
Is V(t) the emitter voltage?

 

The main difference between and NPN and PNP transistors are NPNs will conduct when a bias voltage is applied to the base. The PNP transistor will not conduct when a bias voltage is applied to the base. One key thing to remember regarding semiconductors that use an arrow...the arrow points to the negative.

LEDs require a current limiting resistor, and you should try not exceeding 80% of the supply voltage when calculating LEDs in series for stability and battery pack longevity.

 

DC Current Gain

This was a trick question. I didn't give you enough information to answer it. You should have come back to me and asked about the base-emitter current. The data sheet will give you the gain based on a Vce of 1 volt and they also give you curves based on a Vce of 5 volts.

Base-Emitter Saturation Voltage

Forget the math. The info is right on the data sheet. Its in the table with ON characteristic and they also give you curves.

Study the data sheet and see what you come up with.

P.S. Use the Fairchild data sheet, it is by far more complete.

 

I will look at the datasheet right now
But I was just wondering this is the visual I have right now of where we are going
Is this correct?

 

Base-Emitter Saturation Voltage (Ic=10 mA, Ib=1 mA) 0.65 V(min) 0.85V(max)
Base-Emitter Saturation Voltage (Ic=50 mA, Ib=5 mA) 0.65 V(min) 0.95V(max)

 

The collector current is 21 mA. This is the load consisting of one resistor and four LEDs.

We know that we need .21 mA to turn the transistor on.

A better rule of thumb is that the base current is 10% of the collector-emitter current. Using this rule guarantees that virtually every transistor will be fully saturated.

Assuming a gain of 100 is dicey, and won't always do the job.

 

I will look at the datasheet right now
But I was just wondering this is the visual I have right now of where we are going
Is this correct?

Probably not what you want. As drawn, all the transistors and thus every set of LEDs is controlled by the single voltage applied to all the bases. Those switches are redundant; once any one of them is thrown, all the LEDs will light.

 

I will look at the datasheet right now
But I was just wondering this is the visual I have right now of where we are going
Is this correct?

Wont work as drawn, but we'll get to that a bit later after you are up to speed on transistors.

 

Base-Emitter Saturation Voltage (Ic=10 mA, Ib=1 mA) 0.65 V(min) 0.85V(max)
Base-Emitter Saturation Voltage (Ic=50 mA, Ib=5 mA) 0.65 V(min) 0.95V(max)

OK, that's better. What I want you to take note of is that there is a range - in this case between o.65V and 0.95V. It's not a hard and fast number and is a function of collector current and base-emitter current. For design purposes I typically use 0.7 Volts.

 

Let's go through one more exercise and then we'll finish designing your circuit. For this I want you to refer to the attached diagram.

Fig. 1a
What is the voltage on the collector? Explain.
What is the collector current? Explain
Fig. 1b
What is the voltage on the collector? Explain
What is the collector current? Explain
What is the base-emitter current? Explain
What is the gain of the transistor as configured in this circuit?
Is the gain for a specified transistor the same in all applications? Explain

 

A better rule of thumb is that the base current is 10% of the collector-emitter current. Using this rule guarantees that virtually every transistor will be fully saturated.

Assuming a gain of 100 is dicey, and won't always do the job.

We'll get to all that as we progress. The OP is woefully lacking in fundamentals so I am deliberately taking him through this process so that he will come away with some knowledge of why as well as what. We're zipping through course material that would normally take me several weeks to cover in class and in the lab with students who are well grounded in AC & DC circuit fundamentals.