Tips and advice on transistor biasing

Hi,

Can any one help me the proper way of computing the gain of a bipolar transistor. Let's say the 2N2222 or BC547. I use them to control the Common Cathode 7-segments. I saw lots of circuit examples on the net multiplexing 7-segments in digital clock circuits. Im a hobbyist of PIC Microcontroller and I have a little background on the engineering perspective of electronics. I'm using a regulated power source (+5vDC) generated by L7805CT regulator. The collector is connected to the Common Cathode of a 7-segment. The Emitter is connected to ground. And the base is filtered with 1K resistor going to the I/O pin of PIC.

My question is, what is the difference of changing the base transistor to 22K, some circuits, they use 470Ohms? What is the effect with respect to gain of the transistor if the resistor in the base of the transistor is small or big.

I also read statements like CONSTANT CURRENT source especially driving LEDs and 7-segments. What is it:?:

Re: Transistor Biasing

The way that you are using the transistor in your design is as a simple switch. As such the gain of the transistor is based entirely on the hfe of the device. By changing the base transistor value you change the value of collector current a lower value gives you a higher collector current and more load drive (segment may appear brighter) and conversely a larger value will give you less load current. The important thing to make sure if you change the value of base transistor is that you are still saturating the device. So look a the the current that your LED segment requires then check the minimum hfe value of the transistor that you are using then you can work out the value of base resistor. to keep thing simple lets say that the minimum hfe of the device is 100 and the load needs 100mA. So the minimum base current required is 1mA (100/100) to ensure saturation add 10-20% to this so lets say 1.2mA now to work out your base resistor value it is just (supply voltage/ (max current/min hfe *1.2) = (5/(.1/120) = 6kohms. (this is just an example your LED will not need 100mA of current)

A constant current circuit is a circuit that is designed to supply a constant current to a load even when its impedance changes value. THere are many examples to be found of these if you search on the web. You could make a simple transistor one to drive your LED segments.

Re: Transistor Biasing

emavil,
In addition to the good advice from kevpat, here a couple of additional considerations:
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Look at the spec of the output driver that is used to drive the transistor. If the driver has a passive pullup resistor, then there is no prblem with using as low a value resistor as you want. Keep in mind, however, that the passive pullup resistor will limit the amount of current that the driver can supply. For example if the pull-up resistor is 10K, then the maximum current that it can supply will be 5/10 = 0.5 mA, even with the driver connected directly to the base.
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If the driver has an active pullup, do not exceed the current at which the "1" state is specified, otherwise, you risk forcing the driver into its linear operating region. This condition can result in exdessive power dissipation in the driver.
.
Regards,
Kral

Re: Transistor Biasing

Thanks a lot for the advice. very comprehensive.

My followup question is can you explain to me further (layman) what is meant by SATURATION. I download datasheets of the mentioned transistors and I saw this SATURATION in a chart form.

Could you please explain to me the reasons why we need this so called PULL up resistor. What is the role of this thing? As I tried, the circuit will still work without this pull up resistor.

What is the effect if we configure the transistor to deliver more current than its maximum current gain?

I also saw some circuits using DARLINGTON transistor such us TIP120 NPN, TIP127 PNP. These transistors can be interfaced to microcontrollers directly with having to compute the base current. And it is also a good choice especially driving loads the require supply voltage higher than +5vDC, such us relays. What can you say about this?

Re: Transistor Biasing

Emavil,

Saturation simply is the term used to imply that a bipolar transistor is fully conducting i.e you will no longer see an increase in collector current by increasing the base current.

Pull-ups can be used for several functions for example limiting output current in a driver or with open collector outputs you can interface a 5v output with a 12v input device by using a pull-up to 12v on the open collector ouput. The reason your circuit works without one is that your PIC will most likely have an internal pull-up.

In your application using a very low value of base resistor to give a high collector current (high current gain) will saturate the transistor but the LED segment that you are driving will probably have an internal series resistor so the collector current will be limited to , (supply voltage- (Vcesat+voltage drop of LED)/ series resistor value.)

In other applications you would normally use feedback techniques to set a gain level much less than the minimum hfe of the transistor so that your circuit is more stable and reliable.

A Darlington transistor is a package that contains two transistors, they give you a very high gain but have higher saturation voltages and hence higher power loss. For your application there is absolutely no reason to use a Darlington pair in fact I would go as far as to say there drawbacks out weigh there good points and would advise against using them at any time.

Re: Transistor Biasing

emavil,
By definition, saturation is the condition in which both the base-emitter diodes and the collector -base diodes are forward biased. This means that the collector voltage is at or below the base voltage.
.
Darlingtons can be useful in situations where the driving device can not supply enough base current to saturate a conventional transistor. One big disadvantage of a Darlington is that when its output transistor is saturated, its collector voltage is equal to the sum of the saturation voltage of the output transistor and the base-emitter voltage of the input transistor. In other words, a darlington can not supply a "zero-state" voltage as low as that of a saturated single transistor.
Regards,
Kral

Re: Transistor Biasing

Transistor can be thought like a water tap.
saturation is like turn on the tap knob to it's maximum.
You might be able to turn the knob more,
but the water that is coming out of the tap is the maximum,
or the tap is in saturation condition.

The transistor you are using seems to act like a switch.
Using transistor to driven the LED display.
It is easy to use the transistor as a switch (driving it to saturation),
instead of worrying about the details about the gain.

You might want to adjust the brightness of your LED,
by adjusting the gain.
I find it easiler to use a PWM control circuit to adjust the brightness.
Less computation to wok out.


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