Equivalent for 2N3055 Transistor HELP!!!!!!!!!!!!!!

Rob Henry

unread,

Jan 12, 1994, 10:11:34 PM1/12/94

to

I have a 14 volt 14 amp power supply with obviously can put out over 190 watts.
when it is at full output.
The transistorw,two of them are the 3055 which are rated 115 watts.
Now,I don't know why they are in there since they are obviously not big enough.
I need a 200 watt version that is the same as the 2n3055.

I keep burning the 2n3055 out because in my application since we use Almost all
of the rated current for long periods of time.
Please leave mail and tell me of anything you might know of an equivalent
transistor.
thANks

Jerry Ethridge

unread,

Jan 13, 1994, 1:14:54 PM1/13/94

to

If there are two of them, perhaps they are sharing the load.

Dean Youngquist

unread,

Jan 13, 1994, 3:27:07 PM1/13/94

to

Rob Henry (ah...@cleveland.Freenet.Edu) wrote:

: I have a 14 volt 14 amp power supply that can put out over 190 watts.

: when it is at full output.

: The transistors ,two of them are the 3055 which are rated 115 watts.

: Now,I don't know why they are in there since they are obviously not big enough.
: I need a 200 watt version that is the same as the 2n3055.

: I keep burning the 2n3055 out because in my application since we use Almost all
: of the rated current for long periods of time.
: Please leave mail and tell me of anything you might know of an equivalent
: transistor.
: thANks

---------------------------------------------------------------------
As far as I know 70 watts is the most power disapation you can get in a
TO-3 case. I have some 2n5301 that are rated for 30 Amps current, and
have a breakdown voltage over 170 volts. I have a couple hundred of them
in the TO-3 case style. They are actually rated only 40 volts but I have
tested them up to 200 volts. They are NPN and make a great heavy duty
replacement for an ailing 2n3055. I'm trying to sell them for $3 each if
anyone wants any just phone or email. I also have the data sheet on them.

Robert Casey

unread,

Jan 13, 1994, 4:29:44 PM1/13/94

to

In article <1994Jan13.1...@exu.ericsson.se> exu...@exu.ericsson.se writes:
>In article t...@usenet.INS.CWRU.Edu, ah...@cleveland.Freenet.Edu (Rob Henry) writes:
>>
>>
>>I have a 14 volt 14 amp power supply with obviously can put out over 190 watts.
>>when it is at full output.
>>The transistorw,two of them are the 3055 which are rated 115 watts.
>>Now,I don't know why they are in there since they are obviously not big enough.
>>I need a 200 watt version that is the same as the 2n3055.
>>
>>I keep burning the 2n3055 out because in my application since we use Almost all
>>of the rated current for long periods of time.
>

>If there are two of them, perhaps they are sharing the load.

Usually, when using a pair of transistors in parallel, one uses small ohmage
resistors connected to the emitters. Like:

q1 emitter---resistor--\
>-----output
q2 emitter---resistor--/

Idea being that the resistors will remove the effects of the transistors'
Base-emitter drops being slightly unequal. If there were no resistors, the
transistor with the smaller B-E drop will take most, if not all, the load.
And burn out. Then the other transistor tries to take over, and then
also burn out. (This assumes the bases are tied together, in an emitter
follower config.)

Resistors are selected to create something like 1/4 volt drop. Rig the
regulation feedback to compensate this out.

Dana Myers

unread,

Jan 13, 1994, 7:12:40 PM1/13/94

to

I think Rob ought to first understand how much power is being dissipated
by the devices first. A 190W power supply may only release a few watts
as heat if it is very efficient. A linear supply is especially easy to
evaluate; here's how:

1. Measure the unregulated voltage input to the pass transistors. This value
will change when the supply is loaded (i.e. go down). So, be sure to measure
this under the fuill rated load. Suppose you measure 17V.

2. Measure the total current delivered to your load. Suppose the rating is 14A.

3. Measure the output voltage at full load; ideally this shouldn't change from
no load, but don't expect this. Suppose you measure 13.8V.

You can now get a rough guess at pass transistor power by:
Ptotal = (Vin - Vout) * Iout, in our example:

Ptotal = (17V - 13.8V) * 14A = 44.8W

There's a little something here I skipped, and that is the base-emitter
current and voltage drop. This is harder to measure, but you can measure the
base-emitter drop with a voltmeter and guess at the current by dividing the
load current by transistor Hfe under load, i.e., total base current = 14A / 10
= 1.4A, this multiplies by the base-emitter voltage, use the highest one if
there are multiple pass transistors, say you measure .9V you get a total power
in the base-emitter junctions of (1.4A * .9V) of 1.26W. Add this to the 44.8W
total and you get a total of 46W. Assuming this is shared equally by the pass
transistors (optimum, not always the case) you get a power dissipation of 23W
per pass transistor, well within the rating for power dissipation as long as
the the emitter current is within spec (about 7A, which should be fine).

Maybe there's some other problem in Rob Henry's power supply?

---
* Dana H. Myers KK6JQ, DoD 466 | Views expressed here are *
* (310) 348-6043 | mine and do not necessarily *
* Dana....@West.Sun.Com | reflect those of my employer *
* This Extra supports the abolition of the 13 and 20 WPM tests *

Hoyt A. Stearns jr.

unread,

Jan 17, 1994, 10:39:21 PM1/17/94

to

>Rob Henry (ah...@cleveland.Freenet.Edu) wrote:
>
>
>: I have a 14 volt 14 amp power supply that can put out over 190 watts.
>: when it is at full output.
>: The transistors ,two of them are the 3055 which are rated 115 watts.
>: Now,I don't know why they are in there since they are obviously not big enough.
>: I need a 200 watt version that is the same as the 2n3055.
>
>: I keep burning the 2n3055 out because in my application since we use Almost all
>: of the rated current for long periods of time.
>: Please leave mail and tell me of anything you might know of an equivalent
>: transistor.
>: thANks
>---------------------------------------------------------------------

First of all, the dissipation in the transistors has little to do
with the power output, it's the voltage across them times the current.

Second, I've found that the brand of 2N3055 is very important. Use RCA.
RCA has thick wide bus bars welded to the large silicon wafer mounted on
a thick copper plate.

I opened a Motorola which used fine gold wires, and a much smaller wafer.