a 22 ohm resistor in series with 5 volts applied would pass 190 ma or so to drop 4.3 volts. a little high? use a 220 to 1000 ohm resistor.

212 Ohms for 20mA

hmmm as stated above

oh well

 

220Ω - 470Ω works fine in most cases.

 

So that can have the pin I/O down to 10mA which I presume is ok?

I mean is there a min current on the b e?

if that makes sense..... fingers crossed

 

In your circuit collector current is in range ? What is the load that you gonna to connect at collector.

 

You need to factor in the transistor current gain and the expected load to work out what the base current should be.

 

Ok yes I see now.

I looked back and that stuff from that forum makes sense now.

Thanks for help and for patience

 

Would this:
http://www.farnell.com/datasheets/78775.pdf
current gain be the 20mA dc

giving me 200mA or so?

 

As a conservative rule-of-thumb, use a current gain of 10 when using the transistor as an ON/OFF switch.

So if you need to switch 200mA, use a base current of 20mA.

4V/20mA calls for a base resistance of 200Ω.

220Ω will do.

 

Thanks

I think I am looking at current gain incorrectly on the datasheet.

How is it worked out?

Is this the info I need?

DC Current Gain hFE
(IC =0.1 mAdc,VCE =10Vdc) 35 --
(IC =1.0 mAdc,VCE =10Vdc) 50 --
(IC =10mAdc, VCE =10Vdc) 75 --
(IC =10mAdc, VCE =10Vdc,TA =--55°C) 35 --
(IC = 150 mAdc, VCE = 10 Vdc) (Note 1) 100 300
(IC = 150 mAdc, VCE =1.0 Vdc) (Note1) 50 --
(IC = 500 mAdc, VCE = 10 Vdc) (Note 1) 40 --​

--​

​

 

If transistor in our circuit will work as a switch, you don't need Hfe from datasheet.
Most BJTs vendors define saturation region when Ic/Ib = 10 (called forced beta) .
And this is why we assume Hfe_sat = 10
http://www.onsemi.com/pub_link/Collateral/P2N2222A-D.PDF (figure 11).
And the most datasheet show Vce_sat for Ic/Ib = 10
So to be one hundred percent sure that your BJT will be in saturation you must use this so-called forced beta technique.
For european BJT Ib/Ib = 20

But if we design an amplifier, we use a minimum hfe value from the data sheet.

 

With a Microcontroller driving the base of a transistor the Microcontroller will normally stall out at 20mA or less but it might also die and it is never a good practice to connect a loadof more than rated current on the pin.

The resistor acts as a voltage dropping resistor. It will limit the current through the pin and the base of the transistor.

If you have an MCU pin output of 5 Volts and a transistor that has 0.7 volt for NPN, or 1.4 for a Darlington NPN+NPN Base to Emitter, then you have 4.3V or 3.6V that will be dropped as current flow through the resistor.

Without the resistor the only current control is what the MCU has for impedance or limiting and the transistors intrinsic base emitter resistance. You won't be able to pull the MCU pin high which can break the chip, or you will put 5 Volts across the Base Emitter Junction, which many transistors are surprisingly willing to survive, leading to very bad habits.

As a habit know what the pin output current rating is, make a note of the resistance that will limit current of full on voltage to less than that current and try to have one of those resistors or larger on all pins.

5 Volts output at max of 20mA would use about a 270 ohm resistor.

3.3 Volts at max 5mA would use about a 680 ohm resistor.

For inputs it is often a good idea to go much larger with the dropping resistors because the MCU imputs are already high impedance meaning it won't change readings, and if an overvoltage or undervoltage is ever accidentally applied, it will hopefully drop through the pin protection diodes safely.

 

Hi I am still unsure as to how to find this 0.7 approx voltage from base to emmitter in a datasheet


Where is it? and do I use it alike an LED forward voltage for my resistor calculations?

Looking here:

http://www.farnell.com/datasheets/78775.pdf

Is it this

Base--Emitter Saturation Voltage (Note 1)
(I​

​

​

​

C = 150 mAdc, IB = 15 mAdc)
(IC = 500 mAdc, IB = 50 mAdc)
VBE(sat)​

0.6

 

I measure Vbe(base-emitter voltage) vs Ib (base current) for BC337-40
The power supply voltage was Vcc = 10V

RB = 680kΩ....Vbe = 0.614V....Ib = 13.8µA

RB = 470kΩ....Vbe = 0.616V....Ib = 20µA

RB = 220kΩ....Vbe = 0.624V....Ib = 42.61µA

RB = 100kΩ....Vbe = 0.639V....Ib = 93.61µA

RB = 50kΩ......Vbe = 0.659V....Ib = 187µA

RB = 10kΩ......Vbe = 0.719V....Ib = 928µA

RB = 5kΩ........Vbe = 0.748V....Ib = 1.85mA

RB = 2kΩ........Vbe = 0.787V....Ib = 4.6mA

RB = 1kΩ........Vbe = 0.819V....Ib = 9.18mA

RB = 500Ω......Vbe = 0.856V....Ib = 18.29mA

RB = 200Ω......Vbe = 0.989V....Ib = 45mA

And as you can see the base current change 45mA/13.8μA = 3260 times. But Vbe change only by 375mV. This very small change in Vbe compared with the huge changes in the base current can be ignored in some cases. And this is why we use Vbe= 0.6...0.7V in hand calculations.

Also look at figure 2
http://users.ece.gatech.edu/mleach/ece3050/notes/bjt/thebjt.pdf
Figure 2 shows a typical plot of Ic versus Vbe for Vce constant. The plot is called the transfer characteristics. There is a threshold voltage above which the current appears to increase rapidly. In the forward active region, the base-to-emitter voltage is typically 0.6 to 0.7V.

 

I may be new to this forum but I am not new to electronic design. I would like to express the issue in a different point of view that may make more sense.

Any I/O pin has enough current to turn almost any transistor fully on (saturated). What the I/O pins do not have is a guaranteed logic low voltage the insures the transistor will be turned off. If the I/O pin will not drop to a low enough voltage, it doesn't matter what resistor you put in series with the base, the transistor will never turn off.

A PIC16F690 (representative of the PIC family) has a max low volt output (Vlo) of .6 volts. This is a high enough voltage to keep a transistor at least partially on.

A better strategy is to set the transistor to remain fully off until the I/O pin rises to over 1.5 volts. All that is required is to add a resistor, of equal value to the base resistor, between the base and emitter.
The two resistors will act as a simple voltage divider and will keep the transistor off until the output pin voltage rises above 1.5V. When the I/O pin voltage rises to its full high value (>2.4V) the base current will be determined by the base resistor. The base-emiter resistor will have little impact.

In other words, add one resistor to your I/O transistor and avoid a lot of problems.