2sb817 upgrade? Dual final 959 needs a better regulator.

[LeapFrog]

...
so, it would seem that the answer is no, changing to a higher current rated part in the same package will not produce a cooler running radio, all other things being equal.
LC

LC,
when I get some time on my hands I'd like to test this, I'll use a DX88HL with BJT's;
I will leave the radio configured the same way for both tests (and for equal duration/output) AM mod transistor being the only factor that changes.
I do not doubt your EE professor for one moment, but I'm stubborn and would like to test the theory in practice (so that I may post objective measurements here).

Unfortunately, people always want to stick bigger parts in and crank things up! If you need more power get a linear amplifier bjt, LDMOS or tube your choice.

Onelasttime,
Having observed a sizable difference in output (from my amplifying device) by simply increasing the output of the exciter, I see why some people would want modifications made to the radio.

Not all high-power modifications are performed to an equal degree of success,
In regards to spectral purity.

73

 

[unit_399]

I don't think upgrading to a higher current rated part will produce the desired effect (IE running cooler) than the other part. LC

LC is correct.

When a transistor conducts current between collector and emitter, it
also drops voltage between those two points. At any given time, the
power (heat) dissipated by a transistor is equal to the product of the
collector current and the collector-emitter voltage (neglecting b-e
current which is negligible).

In a given circuit, (regardless of what transistor is used) the amount of
power (i.e. HEAT) dissipated by the transistor will be the same since
the collector current and the voltage across the collector-emitter are
unchanged. Utilizing a transistor with a higher dissipation rating just
means that it will handle more heat ... it will NOT run cooler.

To protect the device obviously an external heat sink should be used.
On my dual final galaxy radios. I cut a 2 sq. in. hole in the chassis
behind the 2SB754. Then I fab a finned aluminum heatsink to fit the hole and bolt the transistor to it. The aluminum dissipates the heat much better than the steel chassis.

A word about transistor substitution. The DC current gain of the
2SB754, 2SB817, and TIP36 are not the same. The SB817 and TIP36
have higher current gain specs, and if they are used as a replacement
then their base drive should be reduced accordingly.

- 399

 

[LeapFrog]

LC is correct.

When a transistor conducts current between collector and emitter, it
also drops voltage between those two points. At any given time, the
power (heat) dissipated by a transistor is equal to the product of the
collector current and the collector-emitter voltage (neglecting b-e
current which is negligible).

In a given circuit, (regardless of what transistor is used) the amount of
power (i.e. HEAT) dissipated by the transistor will be the same since
the collector current and the voltage across the collector-emitter are
unchanged. Utilizing a transistor with a higher dissipation rating just
means that it will handle more heat ... it will NOT run cooler.

To protect the device obviously an external heat sink should be used.
On my dual final galaxy radios. I cut a 2 sq. in. hole in the chassis
behind the 2SB754. Then I fab a finned aluminum heatsink to fit the hole and bolt the transistor to it. The aluminum dissipates the heat much better than the steel chassis.

A word about transistor substitution. The DC current gain of the
2SB754, 2SB817, and TIP36 are not the same. The SB817 and TIP36
have higher current gain specs, and if they are used as a replacement
then their base drive should be reduced accordingly.

- 399

Thank You 399, your description {along with the community consensus} convinced me to not even bother testing this (it makes sense now).

I'm stubborn and would like to test the theory in practice (so that I may post objective measurements here)

No, now I see that would be a waste of time.
If I want/needed a beefier part I know what to do, but it will create the same amount of heat and offer no real thermal advantage

unless the mounting tab of the device is larger, then perhaps the device would be more suited to readily transfer any generated heat across the surface of the heatsink material; only because of a larger surface area.

That is not to say that in operation, a part with higher current limit will generate any less heat compared to the original (at normal operation)
Unless the other part was operating at its limit and attempting to dissipate excessive heat... no real advantage to be had other than a bigger surface area to transfer heat across.

Am I wrong on this one, or have I picked up on a clue?!

 

[StrangeBrew]

Seems to me that a part that dissipates the heat better while producing the same amount should itself stay cooler than a part that holds onto and accumulates the heat, am I missing something here?

 

[loosecannon]

Seems to me that a part that dissipates the heat better while producing the same amount should itself stay cooler than a part that holds onto and accumulates the heat, am I missing something here?

yes, the part you are missing is that just because a part has a higher current rating, does not mean that it dissipates the heat any better.

It is still the same amount of current, flowing through the same size part, on to the same heatsink.

check my post above where we asked an electrical engineering professor about this.
LC

 

[StrangeBrew]

.... the larger rated device will get the heat from the silicon to the metal on the back of the transistor more efficiently, which will keep the junctions under the volatile temperature.

....

This part here seems to imply that while the total heat in the system may be the same the important parts should be cooler, with the back of the transistor being the part that's in contact with the heat sink wouldn't getting the heat to that part of the transistor more efficiently be a good thing?

 

[loosecannon]

that only matters while the part/heatsink is actually heating up, and at some point during that process, the heat created on the heatsink begins to heat the part back up.

again, im not the expert on this, but the lesson we got from the expert was that we would not notice the slight decrease in the heat on the part itself because of this.
LC

 

[StrangeBrew]

OK, I think I got it now. The new part doesn't eliminate the bottleneck, it just moves it. Eventually the whole shebang is going to reach the same temp anyway.

Thanks.

 

[unit_399]

that only matters while the part/heatsink is actually heating up, and at some point during that process, the heat created on the heatsink begins to heat the part back up.LC

LC -
I hope you didn't mean this the way I read it.

The heat sink does not create heat. It absorbs heat produced by the transistor. If the heat sink is of adequate size, and correctly fastened to the device, it will heat up until it reaches an equilibrium point. This is the temperature where the heat absorbed by the sink is equal to the heat radiated by it. If the device produces more heat than the sink can radiate, the device will eventually fail. At this point, there are three choices: 1. Install a more efficient heat sink, 2. Replace the device with one that has a greater dissipation spec, or 3. Both.

This is why I install an finned aluminum sink on all of my export rigs. Using a sink like this will actually allow the transistor to run cooler. The same amount of heat is being produced, but the more efficient sink dissipates it faster so the equilibrium point (heat in = heat out) is reached at a lower temp.

The Galaxy 959 converted to dual mosfet finals (as described by the original poster) also needs to have a heat sink installed on the back of the radio to protect them. I use the sink from the RCI 2950 because it is the right size and is readily available. I cut holes in the steel chassis so that the transistors can be bolted directly to the sink. This allows the heat to be dissipated more efficiently. 73s.

- 399

 

[loosecannon]

LC -
I hope you didn't mean this the way I read it.

The heat sink does not create heat. It absorbs heat produced by the transistor. If the heat sink is of adequate size, and correctly fastened to the device, it will heat up until it reaches an equilibrium point. This is the temperature where the heat absorbed by the sink is equal to the heat radiated by it. If the device produces more heat than the sink can radiate, the device will eventually fail. At this point, there are three choices: 1. Install a more efficient heat sink, 2. Replace the device with one that has a greater dissipation spec, or 3. Both.

- 399

I see how you could interpret it that way once i read it again.

I meant what you better explained above.
That they reach a point of equilibrium and equalize in temperature.
LC

 

[Ranch55]

I see how you could interpret it that way once i read it again.

I meant what you better explained above.
That they reach a point of equilibrium and equalize in temperature.
LC

They can only reach a point of equilibrium if the heat sink also generates/creates its' own heat, which it cannot.
Whenever the output power ceases (unkeying), the heat will also cease rising. Shortly after that point, the temperature will stabilize for a time, then start decreasing as the heat sink dissipates it heat into the surrounding air.

 

[StrangeBrew]

So I guess for all practical purposes the regulator and it's heatsink work together and for best results they should both be upgraded together.

 

[Shadetree Mechanic]

I am learning alot here. Let me see if i have this straight. The main idea of having a larger transistor package is to be able to conduct more heat away from the junction and into the heat sink. A bigger conductor can carry more heat. Heat seeks cold. Cold is absence of heat. Heat will move through a conductor to where it is colder. The spot that is colder is the heat sink. The junction stays cooler for a given load because the heat is removed more efficiently because of the bigger conductor.

 

[Shadetree Mechanic]

They can only reach a point of equilibrium if the heat sink also generates/creates its' own heat, which it cannot.
Whenever the output power ceases (unkeying), the heat will also cease rising. Shortly after that point, the temperature will stabilize for a time, then start decreasing as the heat sink dissipates it heat into the surrounding air.

Equaliberiam meaning heat in equals heat out.

 

[Chevboy0167]

To me, sounds like the only way to remove more of the heat created by this "upgrade part", is to mount it on the rear of chassis, next to the finals, and bolt it to the already factory supplied heatsink. Then run some wires to the board where it came out of originally.

At this point, if you hotrod the radio or are long winded, you may need to replace the factory heatsink with a lager unit like they use on the Stryker/Magnum radios which may require some drilling.

Thoughts???