Achieving Transistor Saturation: Example with R, U, I Parameters Needed

How to put a transistor into saturation? I would like an example with parameter values (R,U,I). Thanks

Hello.

You use the general formula Ic = beta Ib. Knowing the collector current and assuming the beta parameter, you calculate the minimum base current of the transistor (at the saturation limit). In order to drive the transistor into saturation, the base current must be greater than Ibmin. you calculate the resistance value in the base circuit: Rb = (Ucc - Ube)/ 1.2 Ibmin. Now with the data: Ucc = 12V Ube = 0.7V Ic = 50mA beta I assume 50

Ib = Ic/ beta Ic = 50 mA/50 = 1 mA.
For saturation Ibn> 1.2 Ib (or equal) Ibn>1.2 mA

Rb = (12 - 0.7)/ 1.2 times 10 to the -3 power = 9.42k of the 9k1 series

At these values, the transistor will enter saturation and the voltage Uce will be equal to max 0.2V.

I built the system according to your instructions and started experimenting, and I got a little confused

why you can assume beta=50
I have BC547 base through 2k to LPT (Ib=1.31mA) catalog Ic=200mA
Uce=45.5mV Ube=0.76V

a relay with a diode connected in parallel to the collector for 12V

the calculations show that Ib = 4.8mA (it's probably a lot for lpt and why it shows less on the meter) Rb = 2.3K
So, to load the LPT to a lesser extent, I have to find a transistor with a smaller Ic ??
Why the transitor switched on at Rb=10k (b low base current consumption)

Why not connect the load between the emitter and ground, only between the collector and positive?
How does this transistor work?

Icmax is the maximum collector current in this system (so it depends on the load resistance, the voltage drop across the CE junctions and the supply voltage)
The load is connected between the collector and the supply for npn transistors because the transistor has a voltage gain and also amplifies the voltage differences between the base and the emitter and if you add a resistor between the emitter, the potential of the emitter will increase as the current increases and the difference between the base and the emitter will decrease (the transistor will disconnect). Take the gain of the transistor from the catalog data, and preferably measure yourself with a meter. And anyway, if you want to control eight relays from the printer port, use the ULN280x series system (just choose the right model) it contains eight identical transistor circuits with surge suppression diodes, controlled from TTL levels (built-in resistors) that can be loaded to 0.5A.

TO THE ELECTRICIAN.

As you write, do not make people brain water - for example, where in the transistor you have the CE connector. Coming back to the thread, it's not easier to go from the "back". The collector current can be calculated from the load resistance:
I=(VCC-UCEsat)/R
where VCC - supply voltage, UCEsat - saturation voltage 0.1-0.2V

transistor base current

Ib=(V-Ubep)/Rb

V - control voltage, Ubep - forward voltage ~0.7V

If Ib*beta>I

the transistor will be saturated

But a slip-up, good thing none of my lecturers knew about it. I also propose a "backwards" approach (maybe I didn't write it clearly, but you should start the calculations from the collector current at the saturated transistor, the voltage drop on the TRANSISTOR (not the junction ) )

Hello.

manfed

Beta is assumed, because in "serial" production no one will select transistors, even in a given group of reinforcements. The Worst Circumstances Method is always used in projects. Secondly, you measure beta "at a given current", which does not mean that it is the same for 1 mA and 10 mA. The catalogs give the measurement conditions with the hfe table.

Saturation conditions refer to the load current. Catalogs give average or max values.

Assuming you have full 5V at the output then
Ib = 5V - 0.7V / 2.3k = 1.87mA
for beta = 50, the maximum collector current can be 93mA. If it is larger, the transistor will not be saturated and the difference Uce > 0.2V.

If we want to reduce the load on the source (LPT in this case), we use darlington transistors. They have a very large beta, which is the product of the gains of the component transistors.

If you connect a load between the emitter and ground (for an NPN transistor), the load current will "raise" the potential of the emitter relative to ground, which will cause the transistor to clog and establish new equilibrium conditions (I assume constant base drive). The supply voltage will be divided into Uo and Uce.

Previously (for R = 2.3k), there was probably an overload of the port and the output voltage was lower, which consequently reduced the base current.

The working circuit of the transistor is a common emitter. Analyzing from the voltage side, it is an inverter.

greetings

Thanks for the explanation (good thing these transistors are cheap )
one more time:
I measured the relay consumption I=12mA at V=12V so this is my Ic
About 5V Uzb enters the base
Ib=(Ic/50)*1.2=(12mA/50)*1.2=0.288, which is about 0.3 mA (now it's probably very little)
Rb=(Uzb-0.7)/Ib=(5-0.7)/0.3=14.3k
The point is that everywhere I look the values differ significantly (Rb 1k-4.7k Ib is also larger
And one more thing, how to measure beta and what can be useful for me to measure hFE in the meter (whether it will not calculate beta)

Is it possible to isolate the LPT directly with optocouplers, as is done in practice (all outputs are isolated or what?)

off works, but I found other solutions in books, sometimes resistors are used between the emitter and the base (why?)
which gives me the use of a divider of 2 resistors on the base (less load on the signal source on the base? But the second resistor connected to Vcc, i.e. here about 12v, will change the drop depending on the load (is it a so-called pull-up resistor??))
i.e. that I should use the most extensive off with all the elk and at least 3 resistors because such an on is incorrect?
Is there any difference in a switch based on 1 tr. when using npn and pnp and if so, when to use

hFE=BETA
The resistive divider at the input of the amplifier is used to set the operating point so that the transistor DOES NOT enter the saturation state with an alternating signal and works in the linear region (alternating signal amplifiers)