Hello, i have a PNP 2N5087 component which is connected as shown bellow.
for forward active PNP we need the Emitter-base forward biased and base-collector reversed biased.
(emitter)P-N(base) needs to be V_E>V_B to forward biased.
(base)N-P(collector) reversed biased V_base>V_collector.
The base is a voltage divider between V1 and VDD.
given the datasheet bellow Vbe=0.85 .
but its wrong it should be Veb=0.85 for active region.
i know that the difference between active and saturation is the Vce voltage drop.

how do i put in saturated region my PNP transistor?
Thanks.
https://pdf1.alldatasheet.com/datasheet-pdf/view/11478/ONSEMI/2N5087.html

 

How can it saturate without killing itself? It’s emitter is connected to a 5 volt voltage source and its collector is connected to ground. If it were to “try” to saturate it would short out the power supply.

What is it that you want this circuit to do?

 

Hello DickCappels,Could you please recommend me how to update the circuit so i could switch the PNP from active forward state to saturated?
Thanks.

 

You need some resistance between Vdd and Gnd in addition to the transistor. In no universe can you successfully short a power supply with impunity.

 

You need a load between the collector and ground. Saturated means the load is getting as much current as possible. With no load that current is infinite.

 

The whole point of having a transistor in saturation is so that it can switch power to a load which is connected between collector and the opposite power supply.
When a transistor is saturated is has <0.5V between collector and emitter (usually <0.05V), so you can't have collector connected to 0V and emitter to 5V.

 

Hello, two resistors were added to the collector and emmiter as shown bellow.
I know that Ic=(beta+1)I_b (in forward active state) beta=hfe
but i dont know if my transistor is in the active state.
its like what came first the egg or the chicken riddle.
I know its an itterative thing.hfe=beta is a range of numbers.
How do i calculate the biasing in this situations?(given that i am not sure what state i am in)
in order to know the Ve and Vc mathematically.
Thanks.

 

The text in the datasheet shows the maximum of 0.85V base-emitter On Voltage for some of them when IC is only 1mA and Vce is 5V. Figure 9 shows a typical On Voltage of only 0.63V for most of them for the same IC and Vce. The graph also shows a higher typical Vbe when the transistor is saturated with the same IC.

 

Hello Audio Guru,could you please guide me regarding the problem i have finding the bias mathematickly in post number 7?
Thanks.

 

You start making calculations from things that you know. For Example:

1. At node Vb, you know the sum of the currents Ib I(R1) and I(R2) must be zero.
2. You know tha (Vb - 1V)/R2 = I(R2)
3. You know that (Vdd - Vb)/R1 = I(R1)

You keep going until you have the right number of equations for the number of unknowns.

This is undergraduate stuff.

 

Now you added R3 which reduces the collector and emitter currents a lot since the base bias voltage is the same as before.
Since the collector and emitter currents are now very low then the transistor is active and is far from saturation.

 

Your problem is thinking that you need to generate a specific base-emitter voltage, but you don't.
You just generate the required base current to saturate the transistor (which from the data sheet is 1/10 of the collector current), and you don't want R3, as that is detrimental if you want the transistor to saturate as a switch.
The Vbe will then be whatever it needs to be for that base current (which is around 0.7V from the base-emitter junction).

You do not necessarily need R1, but it is sometimes included to shunt high temperate base leakage current.
For that it is usually a high value (e.g 10kΩ).

 

In figure 7, the base voltage is a little higher than 4.06V (the voltage divider of R1 and R2). You do not know the transistor hFE so you do not know its exact base current. The emitter voltage is close to +5V and the collector voltage is close to ground (use simple arithmatic) then the transistor is far from saturation.

 

Thank you for adding the 100 ohm collector resistor. That means that is you were to saturate the transistor the collector current would be 5V/100 ohms or 50 ma (milliamps) the most common recommendation is that the base current should be 10% of the collector current (a standard starting point) so your base current should be 5 ma (milliamps).

 

Hello,I have written the equation bellow.
but as you said correctly i dont have the HFE.
What could be done to overcome this obsticle?
Is there some thing i could use in the data sheet?
Thanks.

Vdd-R3*(hfe+1)*ib-0.65+R2*ib=V1
https://pdf1.alldatasheet.com/datasheet-pdf/view/11478/ONSEMI/2N5087.html

 

Obviously you didn't read my previous post or you wouldn't still be posting that incorrect circuit, so I will will not post any further comments.

 

Hello, two resistors were added to the collector and emmiter as shown bellow.
I know that Ic=(beta+1)I_b (in forward active state) beta=hfe
but i dont know if my transistor is in the active state.
its like what came first the egg or the chicken riddle.
I know its an itterative thing.hfe=beta is a range of numbers.
How do i calculate the biasing in this situations?(given that i am not sure what state i am in)
in order to know the Ve and Vc mathematically.
Thanks.
View attachment 291654

First, you have to decide what "saturation" means for YOUR purposes. Manufacturers need to specify test conditions for their datasheet parameters and long-standing tradition, for most non-power BJT transistors, is to measure saturation parameters under conditions in which the collector current is 10x the base current. But that may or may not be relevant for what is important to YOU with THIS circuit.

The first place to start your quest is a data sheet for the 2N5087 (that you linked to in your first post).

First, what is the limiting collector current under ideal conditions? That's been Vce = 0 V and Ib = 0 A. That gives you a transistor current of 25 mA.

Next, see what the TYPICAL value of Vcesat is at that current.



That's that bottom curve and, at 25 mA, you see that it's about 100 mV.

Next, take both Vcesat and Ib into account. What collector current is needed in order to result in Vce = 100 mV when 110% of that current is flowing in the emitter (due to the base current, which is 10% of the collector current)?

Now take 10% of that collector current, and that's the base current you need.

Keep in mind that all of these parameters have tolerances and only some of those ranges are documented in the data sheet. Plus, most of the data sheet values are at one temperature (25°C in this case), not across the range of operating temperatures that the device is rated for. How to adjust for this depends on what, exactly, is important for THIS particular application.

As a separate issue, you are specifying the value of R1 to five significant figures. Why?

If you really expect the circuit not to behave properly if R1 happens to be 230 Ω or 231 Ω, then you have much more important issues to deal with. More reasonable choices are 220 Ω or 240 Ω and you want the circuit to behave acceptably as long as that value is within 10% of that value. If needed, you can narrow that down some, but if you really need bias-circuit components that are much tighter than that, you are back into the bigger issues realm since your transistor parameters are going to swamp that.

 

Your base voltage is so high that some 2N5087 transistors will have a very small active voltage swing and others will rectify a very small output swing.

 

Hello WBahn,I am trying to understand the logic ,
you picked a point in figure 9 of the data sheet in which Ic/Ib=10 and Vce_sat=100mV
Ic=20mA Ib=2mA
this is my saturation conditions. In saturation i cannot assume Vbe=0.6 volts.
in the circuit bellow(please disregard the resistor value) i can controll the Vb voltage because its a voltage devider.
so Ic/ib Is the HFE (beta) value? correct?

there is also the Vbe sat plot .
So i take 20mA line up to the Vbe sat plot and get that given Ic=20mA my Vbe(sat) will be 0.8 V?
Correct?

 

You can choose the Vin as a 0V(Gnd) or 5V pulse and to calculates the Ic, Ic, Rb, etc...