# Transistor Analysis

Discussion in 'Electronics Homework Help' started by Akshatha Venkatesh, Mar 27, 2017.

Not open for further replies.
1. ### Akshatha Venkatesh

145
0
Jan 14, 2017
Hi, can anyone help me with the circuit given below. what's the "voltage at this point" when the transistor is in saturation ? Is it Vce(sat) from the data sheet of the respective transistor ? and I also know the gain (hfe), so how do I calculate the collector current of the transistor when it is fully on (saturation) ?

I'm trying to calculate the collector current , so that I can calculate the base current since I already know the gain.
Thank You.

File size:
26.4 KB
Views:
136
2. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
Assuming the transistor is in saturation, the voltae from collector to emitter is Vcesat. Since the emitter is at 0 V (common potential), this is also the "voltage at this point".
Assuming further that the "ADC pin of MCU" has an infinite input resistance (At least much much higher than the resistances involved in the circuit shown), you can then calculate the currents through R0 and R1 from Kirchhoff'S and Ohm's laws.

Akshatha Venkatesh likes this.
3. ### Akshatha Venkatesh

145
0
Jan 14, 2017
So if the voltage is 0 V , how will the current flow there ?

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,496
2,837
Jan 21, 2010
Why are you concerned that the voltage at the emitter is zero volts?

Do you understand that the voltage at the negative end of the battery would also be labeled as 0V, and the emitter is at this potential because it is connected to it?

The voltage at the collector is the voltage at the emitter plus Vce(sat). You'll have to look this up from the datasheet. This won't give you the exact base current because the relationship ib = ic / hfe only applies when Vce >> Vbe (and even then it is only approximate). In saturation Vbe > Vce. Ib will be significantly larger than the equation predicts and there is nothing much preventing it from increasing without bound.

In addition, if you are actually using a different device (say a biased transistor) then the calculation simply doesn't hold. If you look at the datasheet there may be something else.

Akshatha Venkatesh likes this.
5. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
Battery-plus -> R0 -> transistor-C-E -> Battery-minus
This holds for Vce > 0 V, too.
A simple application of Kirchhoff's voltage law.

6. ### Akshatha Venkatesh

145
0
Jan 14, 2017
What should I look for in the datasheet to find out the base current to fully turn on the transistor ?Please help

File size:
91.5 KB
Views:
111
7. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
An NPN transistor is in saturation whenever Vbe < Vce (see here, 'regions of operation'). Therefore the threshold for entering saturation is Vbe=Vce. Any higher Vbe will only saturate the transistor more.
The datasheet for this particular transistor only gives you Vcesat and VI(on). You'll have to use one of these values. Once you know the collector current in the on state you get at the required base current by using the DC current gain parameter hfe: Ib = Ic/hfe.

8. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,496
2,837
Jan 21, 2010
I think you meant Vbe > Vce.

You may need to have 3 to 10 times that base current for saturation.

The datasheet may have enough information specified for its quoted Vce(sat) to allow you to compare the ib in this case with what you would calculate above. It's also worth noting that hfe varies between device and is not constant across large changes of ic.

Do you have a link to the datasheet?

9. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,496
2,837
Jan 21, 2010
Vce(sat) is given an 0.15V when Ic is 10mA. 0.5mA of base current (Ib) is required. The hfe is only listed as a minimum value (30). The rest of the datasheet might give more information, but clearly for saturation ib > Ic / hfe.

10. ### Audioguru

3,309
703
Sep 24, 2016
This must be a little BCxxx European transistor since its base current is 1/20th the collector current for saturation. Most little 2Nxxxx American transistors specify a base current that is 1/10th the collector current for saturation. hFE is not used for saturating a transistor. Power transistors (2N3055) sometimes use a base current that is 1/3rd the collector current and even with that much base current the saturation voltage is fairly high with a max of 3V.

11. ### Akshatha Venkatesh

145
0
Jan 14, 2017
http://assets.nexperia.com/documents/data-sheet/PDTC114E_SER.pdf

12. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,496
2,837
Jan 21, 2010
That's not just a transistor. Much of what has been started in this threat does not apply to that device.

If you can change it to a normal transistor, or go back to the thread where we were discussing this circuit we may be able to help you with your homework.

13. ### Audioguru

3,309
703
Sep 24, 2016
That transistor has a 10k resistor in series with its base. It saturates not very well (0.3V) when its input voltage is 1.8V or 2.5V. If the input voltage is higher then it will saturate better.

14. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
Right, a type, sorry.
The equation gives the minimum requirement. In practise it is common to use a higher base current as you say. However, note that if you use the minimum hfe from the datasheet, the typical value will be higher wich reduces the minimum base curent required. Therefore thebase current calculated from min. hfe will always be more than the typically required base current.
Using Ib>=3*Ibmin is a useful rule of thumb.
Right. The charm of this device is that it can be purely voltage controlled without the need for an additional base resistor as this is integrated.

That having been said, I think it is not too useful for @Akshatha Venkatesh. A back of the envelope calculation using currents as requested could go like this:
What is the max. collector current in the original circuit?
Assume Vce=0V and no current flowing out of the MCU input (right side). Then Ic = Vdc0/R0. (note: a better value for Vcesat can be taken from the datasheet, figure 7.)
Assume hfe = hfe(min) = 30 then Ib = Ic/hfe(min)=Vdc0/R0 * 1/hfe(min).
You now have to make a further assumption about Vbe in saturation. Assume Vbe=0.7 V (this is an arbitrary but typical value. The datasheet does not state that value.)
The current hrough R1 (internal) is I(R1) = Ib+I(R2). With above Vbe=0.7V: I(R2) = Vbe/R2.
The required input voltage to drive this input current is Vin >= Vbe+I(R1)*R1.
Do the math yourself.​

Alternatively: from figure 8 of the datasheet take minimum input voltage to fully turn on the transistor at the collector current required.

15. ### Akshatha Venkatesh

145
0
Jan 14, 2017
Which resistor is this R2 that you have mentioned above ?
Thank You

(mod edited for more specific quoting)

Last edited by a moderator: Mar 28, 2017
16. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,496
2,837
Jan 21, 2010
They are the ones inside the biased transistor packages.

17. ### Akshatha Venkatesh

145
0
Jan 14, 2017
R1 is referring to the biased transistor package too ? or the one in the above schematic which i have named as R0 and R1

18. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,496
2,837
Jan 21, 2010
They are the ones inside the biased transistor packages.

19. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
Right.

It is unfortunate that the resistor in the transistor package has the same name as the one in the circuit.
I should have made that much clearer. I beg to excuse that lapse.

20. ### Audioguru

3,309
703
Sep 24, 2016
You need to allow for the awful +/- 30% (!) tolerance of the resistor values inside that transistor.