Transistor hfe

[sureshot]

Recently i've built a couple of voltage regulator transistor emitter follower circuits, using a single TIP2955, and two TIP2955 transistors. Curiosity has led me to a more powerful transistor.

At this stage i am unsure as its gain minimum is quoted at 1K or 1000mA, its power rating is a lot more than the TIP2955 which i have had really good results with building these two 12 volt power supplys.

The transistor i am looking at, or hoping i can use is the following: Type MJ11015 PNP, VCB VMAX 120, IC mA MAX 30, PTOT mA MAX 200W, HFE MIN IC mA MAX 1K@20A, pin out is quoted as 23A epitaxial TO3 package. I was hoping some one could help me through pitfalls if any of using this transistor, i have posted the schematic below. Where TIP2955 transistors are i wanted to use the following transistor above. i would be very grateful if anyone can help me with this. Many thanks.

 

[Bluejets]

Gain is just a number not 1000mA as you say.
Basically the possible amplification ratio.
Reason it is high is the transistor is a darlington arrangement where the gain of the first is multiplied by the gain of the second to get the final result.
It means that relative small signals can feed the transistor with little or no signal degradation.
Others may like to comment on it's use in your series pass arrangement.
Far as I can see if you still maintain the ballast resistors it should be ok but then I've never done it.

 

[Colin Mitchell]

The original would never work. The base current needed to deliver 30 amps would need to be about 1 amp. 1 amp flowing through 100 ohm resistor would drop 100 volts. So the circuit is one of those junk designs that has never been tested.
You transistor is a Darlington and will need less base current. You will have to fiddle with the value of R7 yourself. If R6 is 0.1R, 3v will be dropped across it and the base emitter voltage has to be 1.4v so you can see the mess you are in.

 

[(*steve*)]

ummm... Colin. The base current does not flow through the 100 ohm resistor.

It flows through the emitter resistors and the base-emitter junctions.

The purpose of the 100 ohm resistor is to allow the regulator's quiescent current to flow without turning on the pass transistors.

 

[sureshot]

Thank you for your replys, yes I don't know if this more powerful transistor would work for a higher current gain ! Thanks for putting me strait on the HFE gain mA description. I can confirm the circuit diagram above using those components works.. Well with two transistors I've been driving a 8.2 Amp load at 12 volts, and its really not to bad at all.
My voltage drops to 11.70 under full load, but I've found from using a single trasnsistor , the more transistors added the less the voltage drop. I'm looking to see if the transistor I've quoted above would be a viable option, given the circuits layout above, and other components in it. Thank you for your replys ! Hope some one knows this question, thanks again.

 

[(*steve*)]

The main problem with the transistor you have chosen is that the minimum voltage between collector and emitter is significantly higher.

This may necessitate a higher input voltage ant thus higher power dissipation.

You will have to replace all your transistors with this type, but you can probably leave everything else as it is.

 

[(*steve*)]

Oh, your input voltage looks easily high enough not to cause problems (other than heat),

 

[sureshot]

So kind of to sumerise, I need to find out if this more powerful transistor is a viable option ! And really could use some help understanding the math involved in calculating a base resistor, and ballast resistors, if this more powerful transistor is an option. I would say my maths is a bit slow to be honest, I'm not the sharpest knife in the draw when it comes to complex equations. But thank you for looking in ! I was really made up this circuit works with the TIP2955 transistors. And really from here found this more powerful transistor from a supplier here in the UK and its power specifactions looked interesting as a PNP transistor for this circuit. But as I said I'm unsure on a few levels, like resistor values, current gain in each transistor, and really the viability of the MJ11015 PNP transistor. Any additional help would be appreciated, kind of like a walk through the math involved, thanks again for your help.

 

[sureshot]

Thanks for helpingvme out Steve ! So my rectified input voltage is to low do you think ? Or is it that the 7812 voltage regulators 12 volts is to low here for this transistor ? These are things I'm unsure of. Or could I close the input to output voltage to say 15 volts transformer secondary input, again variables I'm unsure about with the MJ11015 transistor. Thanks for helping me with this.

 

[Colin Mitchell]

I made a mistake with my reasoning above as I normally use 2N3055 transistors for the pass transistor.
But when you are getting up into the high currents, you are much better off buying a switch-mode design.
The losses in a linear circuit are 20% and everything has to have enormous heat-sinks.
To deliver 30 amps would need 5amp from each transistor and the regulator will only deliver 1 amp total. You may get a gain of 30 out of each transistor and you may get 30 amps. The collector-emitter voltage is at least 4v when the current is 5amps so that 30 x 4 = 120 watts dissipation (to start with). But it will be much more as explained below. And could be as high as 500 watts.
You could use the power supply as a soldering iron and to heat the bath water.
The whole idea is to spread the losses between as many transistors as possible.
I don't think you are going to gain anything by using a set of Darlington transistors.

 

[sureshot]

With the TIP2955 transistors I've seen the heating effect as wasted energy, they are bolted to a large fan assisted heatsink. My transistor tempreture with two TIP2955 transistors is about 70°C and voltage regulator 7812 around 62°C although I could increase heatsink size. Currently I've used PC CPU heatsinks with fan cooling via a 50°C thermal switch contact normally open, closing on tempreture rise to keep things cool. But I could get a larger heatsink if the more powerful transistors are a viable option. So my question now looks like, what input voltage would be sufficient to get the 12 volts high current regulated output.

 

[sureshot]

Its the simplicity of this circuit that interests me, I know there are losses using this method, but if I can keep my input voltage margin as low as possible, with respect to the voltage regulators input to output level maybe I can build a modest viable high current low voltage power supply. My thinking with the MJ11015 transistor in place of the TIP2955 transistor is its greater power handling in the specifactions. I won't need anywhere near 30 Amps at 12 Volts, my use will not exceed 15 Amps really, at 12 Volts, to power radio gear. This for me is more a curiosity of the MJ11015 transistor being more efficient on heat and current carrying ability as its higher powered. Kind of hope my reasoning makes sense. Although I do know its far from the most efficient linear power supply circuit.So I'm still wandering is the MJ11015 transistor a more efficient power handling device, over the TIP2955 ? I'd really like to build it if this higher power transistor is a viable option.

 

[Colin Mitchell]

It depends on what we call the REGULATION OF THE TRANSFORMER.
Normally, a 24v , 30 amp transformer will produce 27v AC on NO LOAD and 24v AC when fully loaded.
24v AC will convert to 33v6 , minus about 4v after it passes through the bridge.
This means you could have 30v before the transistors and 12v after.
This is 18v across the transistors x 30 amps = 500 watts.
The minimum voltage across the transistors is 4v when 5 amps flows and if you can get a transformer that produces a lower voltage than 24v AC, you will have less heat to dissipate.

 

[Colin Mitchell]

At the end of the day, both transistors will dissipate EXACTLY the same wattage.

The only difference will be the regulator will dissipate 12 watts or more and about 1 watt.

 

[sureshot]

Ok here again I'm unsure, not about the transformers secondary, as I was planning on a secondary voltage of 15 Volts before rectification and filtering. The bit I'm unsure about know is where and under what conditions does the regulator dissipate 12 watts, and 1 watt, is this 12 watts the regulator dissipates with the TIP2955 transistor, and 1 watt with the MJ11015 transistor ? Or have I got that reasoning completely wrong there.

 

[sureshot]

So would the regulator behave the same with the MJ11015 transistor, as it does with the TIP2955 transistor ? And as its a more powerful transistor could it handle current more efficiently over the TIP2955 transistors specifications ? Again questions using the MJ11015 transistor I'm not certain about. And thank you for all the help given so far, its much appreciated.

 

[Colin Mitchell]

You would be pushing the regulator to its maximum with the TIP and 30 amps.
"And as its a more powerful transistor could it handle current more efficiently over the TIP2955 transistors specifications " No - the same. However the drive current for the TIP would be much higher.

 

[sureshot]

Ok, I know the voltage regulator passes about 800 mA of the total current in the 30 Amp schematic using the 6 x TIP2955 transistors. I know heat is generated carrying the current via this emitter follower circuit, and its a trade off between input voltage to output voltage in terms of losses. But what I'm looking to find out is will the MJ11015 transistor run cooler with its higher power ratings. Given the TIP2955 under the same conditions. So if both types of transistors where subjected to the same conditions, would the MJ11015 transistor run cooler, there by having better power current carrying capability, over that of the TIP2955 transistor. I've asked these questions as the specifactions for the HFE of the TIP2955 transistor over the MJ11015 transistor is very different where HFE is stated in the data sheets. Could the same resistors in the TIP2955 voltage regulator circuit above be used, if the MJ11015 transistor was used in place of the TIP2955 transistor. The benefits if my thinking is anywhere close, being that with a more powerful transistor would it run cooler for the same current carrying as the TIP2955 transistor. Again thanks for helping me here.

 

[Colin Mitchell]

As far as you are concerned, the losses produced by the transistors will be ABSOLUTELY IDENTICAL.
The only difference is the base current for the MJ will be 1% of the TIP.

 

[(*steve*)]

The only advantage of using higher gain transistors is that you may not see the same degree of voltage "sag" at higher currents.

note that you should not draw a dc current more than 71% of the rated current of the transformer.