The 2N3055 is a power bipolar transistor designed to handle high power loads in the range of 100 V, and 15 amps.
In this post I will comprehensively discuss the pinout function, electrical specification, and application designs for the power transistor 2N3055.
If you are an electronic hobbyist, you might have certainly used this very useful, and efficient power transistor at least once in your experiments. I have used 2N3055 transistor many number of times in many of my high current circuit applications without any issues.
Main Features
- DC Current Gain or hFE = 20 −70 @ IC = 4 Amps (Collector Current)
- Collector−Emitter Saturation Voltage − VCE(sat) = 1.1 Vdc (Max) @ IC = 4 Adc
- Outstanding Safe Operating Area
- Available with Pb−Free Packages
Pinout Diagram
How to Connect the Pinouts
Just like any other npn BJT, 2N3055 connections are also pretty straightforward. In the common emitter mode which is most frequently used configuration, the emitter pin is connected with the ground line or the negative supply line.
The base is connected across the input signal through which the transistor needs to be switched ON or OFF. This input switching signal can be anywhere between 1V and 12V ideally. A calculated resistor must be included in series with the base pinout of the transistor.
The base resistor value will depend on the load specifications attached at the collector pin of the transistor. The basic formula can be studied from this article.
The collector pin should be connected with one terminal of the load, while the other terminal connects with the positive supply line. The load current specs must be at any cost lower than 15 amps, in fact lower than 14 amps to avoid current reaching the breakdown limit.
MAXIMUM RATINGS AND SPECIFICATIONS OF 2N3055 TRANSISTOR
Maximum ratings are the highest tolerable values beyond which a permanent damage can occur to the device. These ratings specified to the device are stress limit values (not the standard operating criteria) for the particular device and are not valid simultaneously.
If these limits are exceeded, device may cease to function with its standard specifications, causing severe damage to the device and also affecting its reliability parameters.
- Collector to Emitter Voltage VCER = 70 Vdc
- Collector to Base Voltage VCB = 100 Vdc
- Emitter to Base Voltage VEB = 7 Vdc
- Continuous Collector-Current IC = 15 Adc
- Base Current IB = 7 Adc
- Total Power Dissipation @ TC = 25°C Derate Above 25°C PD = 115 W @ 0.657 W/°C
- Operating and Storage Junction Temperature Range TJ, Tstg = − 65 to +200 °C
THERMAL CHARACTERISTICS of 2N3055
Thermal Resistance from Junction−to−Case R0JC = 1.52 C/W
ELECTRICAL CHARACTERISTICS of 2N3055(TC = 25 C unless otherwise specified)
CHARACTERISTICS WHEN DEVICE IS OFF
- Collector−Emitter Sustaining Voltage at collector current IC = 200 mAdc, IB = 0) VCEO(sus) = 60 Vdc
- Collector−Emitter Sustaining Voltage at collector current IC = 200 mAdc, RBE = 100 fi) VCER(sus) = 70 Vdc
- Collector Cutoff Current (VCE = 30 Vdc, IB = 0) ICEO = 0.7 mA
- Collector Cutoff Current (VCE = 100 Vdc, VBE(off) = 1.5 Vdc) ICEX = 1.0 mA
- Emitter Cutoff Current (VBE = 7.0 Vdc, IC = 0) IEBO = 5.0 mA
CHARACTERISTICS WHEN DEVICE IS ON
- DC Current Gain (IC = 4.0 Adc, VCE = 4.0 Vdc) (IC = 10 Adc, VCE = 4.0 Vdc) hFE = 20 to 70
- Collector−Emitter Saturation Voltage (IC = 4.0 Adc, IB = 400 mAdc) (IC = 10 Adc, IB = 3.3 Adc) VCE(sat) = 1.1 to 3 Vdc
- Base−Emitter On Voltage (IC = 4.0 Adc, VCE = 4.0 Vdc) VBE(on) = 1.5 Vdc
DYNAMIC CHARACTERISTICS
- Current Gain − Bandwidth Product (IC = 0.5 Adc, VCE = 10 Vdc, f = 1.0 MHz) fT = 2.5 MHz
- *Small−Signal Current Gain (IC = 1.0 Adc, VCE = 4.0 Vdc, f = 1.0 kHz) hfe = 15 to 120
- *Small−Signal Current Gain Cutoff Frequency (VCE = 4.0 Vdc, IC = 1.0 Adc, f = 1.0 kHz) f hfe = 10 kHz
- *Indicates Within JEDEC Registration. (2N3055)
The transistor comes with a couple of limitations in terms power handling capability.
- Average junction temperature
- Breakdown voltage
Safe operating area curves indicate the IC − VCE limits of the 2N3055 transistor that must be taken care of in order to ensure a stable and error free operation. Meaning the transistor must not be operated to increased dissipation levels than what is advised in the curves traces.
The data given in the figure below is plotted while TC = 25°C; TJ(pk) is variable in accordance with the power level.
Second breakdown pulse boundaries are legitimate for duty cycles upto 10% but must be derated for temperatures as indicated in the following figure:
Application Circuits using 2N3055
The 2N3055 is a versatile NPN power transistor which can be effectively applied for all medium power (current) delivering circuit. The few main of these applications are in the field of inverters and power amplifiers. Due to relatively high hFE range this device can be used in a wide range of circuits for handling high current efficiently.
It's metal TO3 case becomes ideally suited for attaching a fast cooling large heatsink quickly and easily allowing the device to work in its most favorable conditions.
I have plenty of 2N3055 based circuits in this website, glad to present a few of them here.
Amplifier Circuit using a single 2N3055
The circuit is the most basic form of power amplifier which can be build using a single 2N3055 BJT.
Although the amplifier above looks too simple to make, the low-tech design forces the 2N3055 to dissipate a lot power through heat.
For a more efficient and Hi-Fi amplifier design, I recommend the following mini crescendo, which is perhaps one of the most classic and efficient amplifier circuits using just a pair of 2N3055 transistors. For complete details you can read this article
Tiniest Inverter using 2N3055
I am sure you might have already come across this little inverter circuit. This circuit using only two 2N3055 and a transformer for creating a reasonably powered 60 to 100 watt 50 Hz power inverter. An ideal project for all new hobbyists and school students.
R1, R2= 100 OHMS./ 10 WATTS WIRE WOUND
R3, R4= 15 OHMS/ 10 WATTS WIRE WOUND
T1, T2 = 2N3055 POWER TRANSISTORS
Power Inverter 100 watts using 2N3055
If you are not satisfied with the power output from the above design, you can always upgrade it to a full-fledged, 100 to 500 watt power inverter, using a single or many 2N3055 transistors in parallel, as shown below:
Variable Power Supply Circuit using 2N3055
An awesome easy to build variable voltage and current work bench power supply can be built quickly using a single 2N3055 transistor and a few other complementing components, as shown below:
For more description and parts list you can visit this post
12V to 48V Battery Charger using 2N3055
This simple automatic 2N3055 based battery charger circuit can be used for charging any lead acid battery from 12V to 48V.
The high current handling capacity upto 7 amps of this device will allow an ideal, charging for any battery from 7 Ah to 150 Ah using the above circuit.
It has an automatic cut-off feature which will never allow the battery to get over charged.
Conclusion
From the above posting I have explained the main specifications and datasheet of the versatile workhorse transistor 2N3055.
This transistor is a universal power BJT that can be used in almost all higher power based applications where high current and efficient switching of current is expected.
The maximum voltage this device can handle is 70V which looks very impressive, and a continuous current for around 15 amp, when the device is mounted over a well ventilated heatsink.
We also studied a few cool application circuits using 2N3055, and how to connect it through its pinout diagram.
If you have any further doubts please use the comment box below for interacting.
Anton Coetzee says
What are the pinouts for 2N3055 transistor, eg in = base, out = emitter and adj = collector is this right?
Swagatam says
It is base, emitter and collector, there’s no ADJ in 2N3055 because it is a transistor not an IC.
A small voltage across base/emitter is amplified many times across collector and emitter.
Jack says
Dear Swagatam
Hello. I sincerely thank you for the extremely valuable information you gave me; as well the useful link. I think I should buy a 5 amp transformer as you have suggested since the motor consumes high current.
Best regards
Jack
Swagatam says
No problem Jack, wish you all the best.
Jack says
Dear Swagatam,
Hello
Hope you are doing well!
You are right Sir, but my aim was to increase the current since I thought that the output voltage should have been set on 6V while connecting to load, because of the existing of the IC7806 in the circuit.
I supposed that the voltage below 6v, is due to the low output current of the transistor 2N3055 and if I used another transistor in parallel with it, increasing the current would cause the voltage to be increased to 6V. I was wrong and I am grateful and thankful very much for everything you taught me.
I was doing your instructions during the past two days and Just to test things out, I added 8 Nos. of 1N4007 diodes to IC7806 which caused the output voltage to increase to 9V and after connecting the circuit to the load, (which is a 6V DC motor) it dropped to 3V. Adding the second 15 volt power supply to the input of the circuit caused the voltage to be increased to 4 volts, meaning a mere 1V increase occurred.
I don’t know What other question I should ask or how do I want you to help me. May be this could be useful to say that my two 15 volt power supplies without their boxes are totally 1700 grams.
Thanks a lot for all your help!
Best regards,
Jack Smith
Swagatam says
Hello Jack,
In your circuit the transistor is configured as an emitter follower, which means the voltage at the emitter of the transistor will be always 0.7V less that the input supply voltage. In your case since the input is 6V, so the emitter voltage of the 2N3055 will be 6 – 0.7 = 5.3V.
Instead of using two power supplies, try using a single 5 amp power supply and check the results.
If the voltage is dropping due to the motor it means either the motor current rating is too high or the power supply is not correctly rated to satisfy the motor current requirement.
Jack Smith says
Dear Swagatam,
Hope you are doing well!
Just to test things out, I added another 2N3055 as parallel ( to the already existing 2N3055 of the current circuit ); over a separate aluminum heat-sink. However, the output voltage with load increased to 3,5V only; meaning a mere 0.3V increase occurred. I added another 15V, 3A power supply (same as the current power supply), as a parallel to the input of the circuit to see if the output voltage increases to 6V. This time, the output voltage increased to 3.8V, instead of 6V that I expected. I did the installation of the second transistor over another aluminum heat-sink to see if the voltage would increase, yet it did not happen.
I would be very glad if you would kindly tell me if you think that mounting the second 2N3055 over the common, single aluminum heat-sink box and close to the present 2N3055 would increase the output voltage to 6V, or if I should do as your instructions and use TIP35 ones?
Thanks a lot for all your help!
Best regards,
Jack Smith
Swagatam says
Hello Jack,
Parallel transistors are used to increase current not the voltage. In your initial question you said you wanted to increase the output current, right?
If you want to increase the output voltage then you must use 1N4007 diodes in series with the 7806 ground pin.
In that case you can remove the extra 2N3055 transistor.
Jack Smith says
Dear Swagatam
Do you mean that I can use TIP35 instead of the second 2N3055 or both of them should be TIP35
Wish you healthy
Jack
Swagatam says
Dear Jack, you can use TIP35 for both the transistors, which can greatly enhance the current handling capacity of the design.
You also consider using only one TIP35 in place of the existing 2N3055 and check if that helps you to get the required full current output.
Jack Smith says
Dear Swagatam
Hello
Thank you so much for your precious response.
All the best
Jack
Swagatam says
No problem, Jack!
Jack smith says
Dear Swagatam, hello
Thank you very much for so soon response and valuable suggestion, as well.
All the best
Jack
Swagatam says
You are most welcome Jack!
Jack says
Dear Swagatam,
I must thank you for your very interesting and educational website, as well as your comprehensive responses to clients. I had a quick question. I need to increase the output current of my 6V, 3A power supply which consists of one IC7806 and one 2N3055 Tr. I have installed them on top of a big aluminum box, so as to function as a heat-sink. Moreover, there is a fan installed on top, which cools the box.
I am planning to add another 2N3055 as a parallel transistor, alongside the already present 2N3055; and I am not sure how to go about this.
The following is the description of the whole circuit diagram:
1. Base of the 2N3055 is connected to the output of the IC7806.
2. Collector of the 2N3055 goes to the input of the IC7806.
3. I get 6v from the emitter of the Tr., which also connects to the ground through 10uf and 100nf capacitors.
4. Common pin of 7806 goes to the ground through a 1N4148 diode.
5. Both the input and output pins of IC7806 connect to the ground through a10uf and a 100nf capacitors on each side.
Would it be enough if I connected the B, C, and E of the two aforementioned transistors together with pieces of wire?
Thank you very much in advance for your attention!
All the best,
Jack
Swagatam says
Thank you Jack, yes definitely you can add another 2N3055 in parallel with your existing 2N3055 by directly joining their EBC pins together. Just make sure to mount them close to each other and over a common single aluminum heatsink. If your box is made up of aluminum then you can mount the new transistor just beside your existing transistor.
No other modifications would be required.
You can also consider using TIP35 instead of 2N3055 which is more powerful than the 2N3055
Wanjohi says
Thanks again sir! For good job you do to enlighten us,in the field of electronic..Am proud of you sir..God bless you mightly.
Swagatam says
Thank you Wanjohi!
Emmanuel says
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Swagatam says
No Problem, you can solve all your queries here!