Discussion in 'General Electronics Chat' started by curry87, Jun 24, 2010.

1. ### curry87 Thread Starter Member

May 30, 2010
101
0

http://www.rapidonline.com/netalogue/specs/81-0048.pdf

Where is the maximum base-emitter current ?

Where is the current required to fully turn on the transistor ?

When it says vcb,vce,vbe how are those measurements taken with a multimeter (ie where to place the probes) ?

What formular is used for total power consumed in a transistor ?

How do you recalculate the current Ic when a change of Ib has occured ?

If Rb = 0.2 x RL x hfe(dc) then how is Rb calculated with a led + limiting resistor to the collector ?

Last edited: Jun 24, 2010
2. ### zxsa Member

Jun 11, 2010
31
2
Hi Curry,

One of the very first things that I teach junior engineers under my supervision, is that you only get the datasheet from the actual manufacturer. You don't get info from a company selling the component, and you don't use datasheet archive websites.

Sure, these places serves a purpose when you're dealing with old components, but the best and most accurate info is always from the manufacturer.

According to your link, the actual transistor part number is BC238B. This part has most recently been manufactured by ON Semi, but is now obsolete. They replaced it with the BC237B.

Here's a link to that part:

I would assume that the two transistors will be similar enough for you to use this new part's datasheet.

I hope this helps you ...

3. ### Potato Pudding Well-Known Member

Jun 11, 2010
684
92
You don't really worry about the base currents normally except to know they should never need to be very high.

The Collector current is max of .35 amps and the current gain is between 200 to 460. Base Current of 2 mA max should be about right. That should turn the transistor fully on.

VCE is voltage collector to emitter
VBE is Voltage Base to Emitter
VCB is voltage Collector to Base

Looking at the flat side of the transistor body the pins are left side collector, middle base, right side Emitter.
Edit:For BC238B this is true according to the datasheet you provided but other transistors can have different configurations

The multimeter will read plus or minus depending on which way you connect the leads to measure these, unless you used an old analog needle meter that only measures forward voltage. For NPN you want the red lead on the positive = collector side.

Total power is normally integrated as Ice x Vce. So switched fully on with max current and near zero voltage and fully off with almost no current and max voltage are both low power because of the multiplication by a near zero value. Your max power situation is simplest to estimate as somewheres in the middle.

So say you have 20 volts V+ and a 100 ohm resistor on the collector (larger would be recommended). The mid point is 20volts shared equally between the transistor and the resistor so 10 volts each and the resistor would put the current at 10V/100ohms = 100mA. That would be 10 volts x 100mA = 1 watt. You might want at least 1kiloOhm resistor on that collector because the max power for that transistor is only 100mW. If you were using the transistor as a switch then you can use 100 Ohms, because the Power in the transistor will go between 20 volts x a few microamps leakage current and 200mA with .2 volts for Vce saturation. So microwatts of leakage in the off state and 40mW in the on state are all the transistor will feel. At least that should be true, until something goes wrong and you get thermal runaway anyways. If it is an oscillator for example or any use that requires switching at high frequency for long cycle counts then it is spending enough time in crossing the high power value middle range that you have to go with the safe collector load of 1kiloOhm or more for a V+ of 20V.

The current gain calculation will multiply the Base current by Hfe to give you Collector current, but the collector current is limited by the collector *and or emittor resistors/load.

An LED has a Voltage drop Vf. To get an RL equivalent when combined with any/all LEDs in series this should work...

RLequivalent = (RLEDseries x VCC) /(VCC - VfLEDseries)

So if you have
VCC of 20 volts
sum of 4 LEDs @ 3.5V each for VfLEDseries= 14 volts
RLEDseries of 200 Ohms (about 30 mA)

RLequivalent = 200Ohms x 20V / 6V = about 667 Ohms

I think you could use that in your equation as RL to find Rb but I am not 100%. If I correctly recognize your equation then that 0.2 is a power factor for a bias set point. That is normally around .5 but .2 works here because you need to give the load enough of the voltage to get above Vf of your LEDs before you even turn the LEDs on.

*Note that any emitter resistor load will affect the transistors operation, because it pushes the emitter away from ground with every increase in current. Most basic transistor stats are for a common or grounded emitter circuit.

Last edited: Jun 25, 2010