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2N3055 heat sink

gary350

Well-Known Member
2N3055 data sheet says 20 watts. I would like to get this right the first time. I don't want to change heat sink 3 times to finally get it right. No cooling fan.

What physical size, shape, metal thickness, fins, etc. works for 20 watts for a 2N3055 transistor?

If I could generator 20 watts of heat some how I could figure out what size heat sink that will work.
 
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rjenkinsgb

Well-Known Member
Most Helpful Member
Look at the thermal resistance figure of whatever heatsink you are considering (or similar size ones in online catalogues).

A 1'C per watt heatsink would heat by 20' at 20W dissipation.
A 0.25'C per watt would heat by 80'C for 20W dissipation, etc.

Also note if it's in free air; a high power heatsink would often be on the outside of an enclosure to ensure adequate airflow.
 

unclejed613

Well-Known Member
Most Helpful Member
i don't know where you got 20W from, the [data sheet] says 115W. if, however you are going to run the device at 20W, i think the TO-3P package might have 20W dissipation in free air.
 

AnalogKid

Well-Known Member
Most Helpful Member
As always, it depends on the stuff you have not said. What is the environment? Is the transistor and its heatsink inside a shoebox, or on the outside of something, sticking out into ambient air? What is the maximum ambient air temperature? Indoors or outdoors? This matters because the range of relative humidity is much greater outdoors, and as the RH goes up, the ability of air to cool something goes down.

What is the maximum temperature you want the transistor to reach? Not the max allowed on the datasheet, but the max you want for your design, its application, its expected reliability, etc?

ak
 

crutschow

Well-Known Member
Most Helpful Member
think the TO-3P package might have 20W dissipation in free air.
I think you slipped a decimal point. 2W is more like it.
The TO-3 package has about a 39°C/W free air thermal resistance, so 20W would give a junction temperature rise of 780°C above ambient. :arghh:
 

gary350

Well-Known Member
Here is my siren circuit with two 555 timers, circuit with 2N3055 amplifier. I am going to test it with alligator clips connected to car battery.

While looking through boxes I accidently found the 2N3055 transistor I know I already had before ordering 12 more. Some of the heat sinks came with 2N3055 transistors. This large heat sink is made for 6 transistors all are 2N3055. Looks like I robber 4 transistors to build other stuff.

Some where I have a speak cabinet with 50w speaker and built in 50w amplifier powered by 120vac. I can use the cabinet & speaker with my circuit if I ever find it. Otherwise I build my own cabinet with speaker and my circuit.

I have several boxes of heat sinks. Circuit will not fun very long so heat sink maybe not need to be very large. I don't want the transistor to over heat in 2 minutes and burn out.

This is a goofy sounding siren with square wave or saw tooth it really needs sin wave to make it sound right.

I also have shelve full of other speakers.

118770

118771

118772

118773

118774
 

audioguru

Well-Known Member
Most Helpful Member
You are feeding DC pulses into a speaker that is designed for AC, not DC. The current in the speaker will be about 11.7V/8 ohms= 1.5A but for only about 1/11th the time it is sounding due to the ratio of R1k and R10k. Therefore the speaker power is about 1.6W and the heating in the transistor is only 0.04W which is nothing and it does not need a heatsink.

If the signal feeding the transistor is a triangle wave or a sinewave then it will get hot and it will need a heatsink.
 

gary350

Well-Known Member
You are feeding DC pulses into a speaker that is designed for AC, not DC. The current in the speaker will be about 11.7V/8 ohms= 1.5A but for only about 1/11th the time it is sounding due to the ratio of R1k and R10k. Therefore the speaker power is about 1.6W and the heating in the transistor is only 0.04W which is nothing and it does not need a heatsink.

If the signal feeding the transistor is a triangle wave or a sinewave then it will get hot and it will need a heatsink.
I copied this from TTL book, I did none of the math. OK I need to do the math to check output of the 555.118787
 

sagor1

Active Member
As for which heatsink, the gold colored on in the middle of the picture (with all the heatsinks) will provide the most cooling, IMHO. The black one to the right and the one to the bottom (1/2 of it would be about the same as the black one) should be enough for several dozens of watts...
Anything else may not be enough of a heatsink...
 

gary350

Well-Known Member
I am having trouble with duty cycle, I don't see in the book how to get 50/50 duty cycle. 1KHz should work best as 50/50 square wave. The other 555 #1 should work best saw tooth 2 seconds long, maybe 3 seconds, variable resistor might be nice. I need an online 555 calculator program to experiment with?

Wife had surgery so I have the grocery list and off to the grocery store in a few minutes. I have several stops, grocery, gas station, dollar tree.

118801
 

AnalogKid

Well-Known Member
Most Helpful Member
It looks like you are using one 555 to switch another 555 between two frequencies. If so, add reference designators to *ALL* components, and I can tell you what to change.

You are correct, the output 555 wants to be 50/50 duty cycle at both frequencies. What duty cycle do your want the left 555 to be?

ak
 

audioguru

Well-Known Member
Most Helpful Member
Before you had the 555 pin 2 capacitor charging with 1K+10k= 11k but discharging with 1k so the duty-cycle was 1/11.
The 555 with the diodes allows charging and discharging to be the same (50/50 duty cycle) but the 47k resistor that changes the frequency messes up the duty-cycle.

When the duty cycle is 50/50 then the speaker and transistor current will be 1.47A for half the time so the power in the speaker will be 8.6W and the heating of the transistor will be 0.2W (it will be slightly warm with no heatsink).
Actually, the speaker is 8 ohms at about 400Hz but since it has inductance it will be maybe 12 ohms at 1kHz so the powers above will be less.
 

gary350

Well-Known Member
Before you had the 555 pin 2 capacitor charging with 1K+10k= 11k but discharging with 1k so the duty-cycle was 1/11.
The 555 with the diodes allows charging and discharging to be the same (50/50 duty cycle) but the 47k resistor that changes the frequency messes up the duty-cycle.

When the duty cycle is 50/50 then the speaker and transistor current will be 1.47A for half the time so the power in the speaker will be 8.6W and the heating of the transistor will be 0.2W (it will be slightly warm with no heatsink).
Actually, the speaker is 8 ohms at about 400Hz but since it has inductance it will be maybe 12 ohms at 1kHz so the powers above will be less.
I found 2 online 555 calculators if I find R1 & R2 with one calculator the other calculator gives me a different answer and doing math in book gives me a 3rd answer. I only trust the book. So far R1 = 1000K which is minumum ohms according to the book. R2 = 90K this is 50% duty cycle according to online calculator but book math says NO. Online calculators must be wrong.????????? Once somewhere I read 555 will give saw tooth output but I am not seeing that in book or online. All I see is square wave. Math is a lot of trial and error. R1 + R2 has to be 2 times R2 to be 50% duty cycle according to the book. I need to build prototype with variable resistors so I can see output on scope. I also need to connect output to small speaker to hear what it sounds like maybe if its not 50% duty cycle it still sounds good enough.

Look at this circuit 555 #1 only, R1 =1K and R2 = 90K online calculator says 50% duty with 1uf cap. Do math by book 1k + 90k/ 90K = 1.111% duty cycle. I only believe the book and I don't think 555 will do saw tooth output.

118807
 
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alec_t

Well-Known Member
Most Helpful Member
I need to build prototype with variable resistors so I can see output on scope. I also need to connect output to small speaker to hear what it sounds like
LTspice is freely downloadable software (from Analog Devices) which enables you to simulate the effects of resistor changes, see waveforms and create a .wav file output to hear the results. Highly recommended and used by many members on this site.
 

gary350

Well-Known Member
OK according to book math this should be right. Just have to build it and see. It I play with R1 & R2 I might get it up to 1.3 or 1.4 or 1.5 seconds with duty cycle not exactly 50% they need to add up R1 = R2 = 1000 ohms

118808
 

audioguru

Well-Known Member
Most Helpful Member
You have the capacitor charging with 420+580= 1000 ohms but it is discharging with only 580 ohms so its duty cycle is about 63/37.

Before in the siren circuit you showed a low value for R1 and a much higher value for R2, producing a duty cycle of 91/90 which is almost 50/50.
The datasheet shows that the 555 on the left side has such low value resistors that its frequency is 911Hz, not 0.4Hz.

In the 555 on the right side the diodes separate the charging and discharging resistors R5 and R6 so for a 50/50 duty cycle their resistances should be the same. The 47k resistor R4 from the 555 on the left side messes up the duty-cycle.
 

gary350

Well-Known Member
You have the capacitor charging with 420+580= 1000 ohms but it is discharging with only 580 ohms so its duty cycle is about 63/37.

Before in the siren circuit you showed a low value for R1 and a much higher value for R2, producing a duty cycle of 91/90 which is almost 50/50.
The datasheet shows that the 555 on the left side has such low value resistors that its frequency is 911Hz, not 0.4Hz.

In the 555 on the right side the diodes separate the charging and discharging resistors R5 and R6 so for a 50/50 duty cycle their resistances should be the same. The 47k resistor R4 from the 555 on the left side messes up the duty-cycle.
I used online 555 calculator it gives me stupid answers.

I did math on R1 = 1K and R2 = 67K the frequency formula answer = .000214 Hz
 

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