I currently have a circuit, that consists of an LED connected between the collectors of a PNP and NPN transistor (ZTX550, ZTX450). the supply is +9v, connected to a 250 Ohm resistor, in turn connected to the emitter of the pnp while the emitter or the NPN is connected to ground.

The resistor drops the 9v, to around 4v, the led uses 2v. Leaving 2v for each transistor. In saturation each one would use less than 1v. I know the resistor values I should be using to achieve saturation on either transistor, but I cannot seem to achieve saturation in BOTH transistors in this configuration. While this isn't an issue, I haven't been able to find a straight answer to if I can saturate both transistors in this configuration. Your opinions would be very helpful , thanks in advance.

 

At first glance, you can drive enough current into each base to saturate both at once, but then almost all the supply voltage will have to be used up by the LED. LED's don't work that way. They will not increase their voltage drop to suit the current.

 

You need to post schematics to really make sense. Verbal and electronics don't mix well.

 

As requested, here is a schematic of the circuit in question.

 

For this circuit to be saturated,
1. Vce of each transistor will be less than 0.2V (somewhat arbitrary, but close enough).
2. Therefore, the cathode of the LED will be at 0.2V, the anode will be at 2.2V, and the emitter of the PNP will be at 2.4V.
3. This means the emitter current of the PNP will be (9-2.4)/250=26.4mA. LED current will be somewhat less, but still > 26mA.
4. The base of the PNP will be at (2.4-0.7)=1.7V.
5. This means the PNP base current will be ≈1.7V/47k ≈ 36uA.
6. This means the PNP beta would have to be ≈730 when saturated.
7. The NPN base current will be (9-0.7)/122k=68uA.
This means the NPN beta will have to be about 380 when saturated.

These are unusual transistors, in that their Vce(sat) is specified at Ic/Ib=100. Actually, at this low current level, only typical Vce(sat) is shown (see curves). Most BJTs are specified with Ic/Ib=10. Still , you are not providing enough base current to guarantee saturation of either transistor.

 

How much LED current do you want?

 

Hi Ron, I wished to deliver .02mA to the led. your observations helped me out a lot. I was able to get both transistors saturated thanks to you and DX3. I changed R1 to 330K dropping 6.6v at .02mA. R2 to a 10K and changed R3+R4 to 56K, this delivers 150uA - 160uA to each base, which should be overkill for saturation. Thank you all for your replies, you helped a lot. ^_^

 

R3 is still too large to guarantee saturation in the NPN. Ib should be: Ib>(20mA/100)>200uA.
R3<(8.3V/200uA)<41.5k.

 

I know. This project is more or less so I may better learn how to predict and correctly choose values. My multimeter has a hFe tester which measured both transistors to have 400+, So I based my minimum on that. This allowed me to use the resistor values in question.

 

I know. This project is more or less so I may better learn how to predict and correctly choose values. My multimeter has a hFe tester which measured both transistors to have 400+, So I based my minimum on that. This allowed me to use the resistor values in question.

Transistor Hfe (beta) testers probably do not measure beta at saturation. They probably measure it when Vce is several volts. Beta starts dropping significantly when Vce gets below a volt or so.

 

Hi Neoaikon,


In order for both transistors to saturate you must;

1. move R1 to be between the PNP collector and the LED.
2. change R3 from 56K to 10K.

The values of R2 and R3 can be much higher if you know what the minimum beta is over, time, Vce and temperature. If that beta value is 400 and the max Ic is 22mA then the minimum base resistor (Rb) value is (Vcc-Vbe)/(0.022/400)=145K.

Since knowing beta for sure of many devices, over temperature and Vce is difficult, many designers and transistor manufactures will assume the beta to garantee saturation will always be at least 10. Therefore Rb = (Vcc-Vbe)/(Ic/10) = 3.6K. IMO 10K is close enough.

Regards,
Ifixit

 

Hi Neoaikon,


In order for both transistors to saturate you must;

1. move R1 to be between the PNP collector and the LED.
2. change R3 from 56K to 10K.

The values of R2 and R3 can be much higher if you know what the minimum beta is over, time, Vce and temperature. If that beta value is 400 and the max Ic is 22mA then the minimum base resistor (Rb) value is (Vcc-Vbe)/(0.022/400)=145K.

Since knowing beta for sure of many devices, over temperature and Vce is difficult, many designers and transistor manufactures will assume the beta to garantee saturation will always be at least 10. Therefore Rb = (Vcc-Vbe)/(Ic/10) = 3.6K. IMO 10K is close enough.

Regards,
Ifixit

I don't think you read the whole thread.
The resistor does not have to be moved in order to saturate the transistors.
These transistors are guaranteed to saturate with a forced beta of 100, as opposed to most transistors, which are spec'ed for saturation with a forced beta of 10.

 

Hi Ron H,

Thanks Ron. I missed that part about the forced beta of 100. The PNP still won't saturate at a beta of 100, but would at 400, as Neoaikon has indicated.

Moving the resistor makes saturation independant of the LED voltage value. I.E. lower values could bring the PNP out of saturation. I like to make circuits independant of normal component value variations, but this is a non-issue if it is a "one of" circuit.

Regards,
Ifixit

 

With his new schematic (post #7), the PNP has more base drive than is needed to guarantee saturation, while the NPN needs more base current.
I agree that moving the LED current limiting resistor makes sense.

 

Thank you Ron and IFixIt for your replies. Yes this is a one off circuit, I'm aware I could move R1, but space constrictions make it's placement necessary .