16 Replies Latest reply on Jun 7, 2019 6:48 AM by andrewj

# Question regarding thermal resistance of a UA7805 regulator

From the datasheet, the TO-220 package has the following thermal characteristics:

Rja = 19C/W

Rjc(top) = 17C/W

Rjp(top) = 3C/W  (jp in this case means Junction-to-exposed pad)

Am I right in thinking that for a TO-220, the Rjp value is a reference to the tab when used with a heatsink?  Typically, Rjc would be used, but if this was the case, then adding a heatsink of any thermal resistance > 2C/W would make it worse!  An older version of this datasheet just referred to:

Rja = 19C/W

Rjc = 3C/W

This would tie in with my interpretation.

Thanks,

Andrew

• ###### Re: Question regarding thermal resistance of a UA7805 regulator

It sounds like you are thinking the R between case and air is 2 C/W.

I would have expected Rca can be greater than Rjc.

• ###### Re: Question regarding thermal resistance of a UA7805 regulator

Not sure I understand Doug.  When I've been doing a thermal calculation it is along the lines of:

Pdiss = V*I

Tj = (Pdiss * Rja) + ambient  (no heatsink)

Tj = (Pdiss * (Rjc+Rha)) + ambient (heatsink with thermal resistance of Rha)

Typically, Rjc is a lot less for an IC than Rja because it's a value to use with a heatsink:  Junction-to-Case-to-heatsink-to-air (ignoring thermal paste or thermal pad).  In the datasheet, Rjc is only 2C/W less than Rja so adding a heatsink would make it worse.

Example

Ambient: 25C

Pdiss: 2W

Tj = 2w*19c/w + 25C = 63C

Tj = (2W * (17c/w+10c/w)) + 25 = 79C (with heatsink Rha of 10c/w)

If the Rjp value was used:

Tj = (2W * (3c/w + 10c/w)) + 25 = 51C

Hence my question.  In the slightly older version of the datasheet, TI only gave a Rja = 19c/w and Rjc = 3c/w.

2 of 2 people found this helpful
• ###### Re: Question regarding thermal resistance of a UA7805 regulator

Hi Andrew,

I think Rjc(top) value can be ignored, it is most likely the thermal resistance for the junction to the top of the case (i.e. the plastic bit) which won't be in contact with the heatsink.

This heatsink calculator website is helpful (useful for getting approximate values for popular semi packages, which should be in the correct ballpark). That website suggests a Rjc value which fits your Rjp value, so they are referring to Rjp. Also, although the TI doc says Rjp(top), that may be a typo, or may refer to interfacing on the top side of the metal bit, either way your heatsink will be contacting metal, so you can use the Rjp value.

4 of 4 people found this helpful
• ###### Re: Question regarding thermal resistance of a UA7805 regulator

Thanks Shabaz.  I think they could have been a little more confusing with their datasheet - clearly not trying hard enough!  I need to rethink some things now.

• ###### Re: Question regarding thermal resistance of a UA7805 regulator

I think we are saying the same thing. The case-to-air resistance should be the main component in the junction-to-air resistance number (Rja = Rjc + Rca).

Heatsinks increase the surface area which reduces the metal-to-air resistance which decreases the case-to-air resistance because case-to-heatsink resistance (metal-to-metal) resistance is small.

• ###### Re: Question regarding thermal resistance of a UA7805 regulator

I'm not sure - it reads like you are saying: 17C/W from junction to case plus a further 2C/W from case into air = 19C/W?  That on the basis Rjc(top) = 17C/W and Rja = 19C/W.

Perhaps you are saying that the formula to use with a heatsink is  Tj = (Pdiss * ((Rja - Rjctop) + Rha)) + ambient  rather than Tj = (Pdiss * (Rjp + Rha)) + ambient?

I feel thick.

• ###### Re: Question regarding thermal resistance of a UA7805 regulator

Hi Andrew,

I think you're doing fine, since you spotted this in the datasheet. I personally don't know it in detail either, I don't calculate heatsinks often enough, and it wasn't taught at uni either, I guess it's more "on the job".. I just knew where to look for the ballpark, because I've used that website before, and found it useful.

It's a confusing datasheet, where their Rjc(top) doesn't seem to mean what Rjc usually refers to.

• ###### Re: Question regarding thermal resistance of a UA7805 regulator

Working at it!

Just realised a big rookie mistake: forgetting to take into account the current requirements of an Arduino which is going to cause big thermal problems for my LM7805.  I've gone from negligible power dissipation to extremely significant with PCBs due for delivery on Thursday and no room for a heatsink - minimum of 8C/W so quite large.  Thinking of pros and cons of either:

1. redesigning and re-ordering a PCB with room for a heatsink
2. off-boarding the LM7805 and connecting with short wires and the LM7805 on a heatsink
3. redesigning and re-ordering 2 PCBs, providing a 5V@0.5A PSU and dropping the LM7805 altogether.

Option 2 is the easiest and cheapest but option 3 is the best but most expensive.  Once I start soldering though, option 3 becomes VERY expensive!

What I think I'll do is option 2.  At the moment, I can envisage other problems surfacing that will require reworking so it would make sense to wait until that happens.  I just need to watch the impact of doing it.  Also V2 of my project would be a lot cheaper to create as I'll have the vast majority of the parts already.  I'm prototyping that part of the circuit now to verify the impact.

Hey ho.  I can commend myself for spotting it before I fry everything whilst kicking myself for missing it when I spent so long on the design and thermal calcs in the first place!!

• ###### Re: Question regarding thermal resistance of a UA7805 regulator

Hi Andrew,

How much power needs to be dissipated? I saw a value of 2W in an earlier comment, is that an example, or the actual value?

If it's 2W, then it shouldn't need a large heatsink, e.g.:

A 15 degC/W heatsink will be quite small, so might fit if there's a bit of space around the TO220.

2 of 2 people found this helpful
• ###### Re: Question regarding thermal resistance of a UA7805 regulator

No, that was an example.  I need to get rid of 6w.  In my original calcs, without Arduino, current draw is in small mA so power dissipation was tiny.  If I’d have realised in time, I would have gone with a different approach: the issue is the voltage drop from a rectified 24V input to 5V output which I initially was going to do something about just because of inefficiency but didn’t because, well, I had an LM7805 to hand and it was going to work.

• ###### Re: Question regarding thermal resistance of a UA7805 regulator

Could you add an external DC-DC converter module to get the voltage down where you want it?

3 of 3 people found this helpful
• ###### Re: Question regarding thermal resistance of a UA7805 regulator

There's several that are designed to replace a 7805 almost directly (taking up almost the same amount of space), e.g.:

3 of 3 people found this helpful
• ###### Re: Question regarding thermal resistance of a UA7805 regulator

Doug / Shabaz,

that's brilliant thanks.  I didn't realise such things existed.

The Arduino takes 5v input, 5.5V max, 320mA (average use of features) and inverting charge pump (LTC1683) 3.5V to 5.5V max, 14mA so a 5V 1A converter should be sufficient - possibly 500mA, although the datasheet for the Arduino mentions the provision of a 5V 1A supply.  Is there anything in the datasheet I need to be picking up on?

The Traco Power component, for example, gives 50mV pk-pk; 0.4% regulation load variation.  It doesn't state any thermal characteristics except "Allows full load operation upto +70C ambient without heatsink"

Can't thank you enough!

• ###### Re: Question regarding thermal resistance of a UA7805 regulator

Hi Andrew,

You might find the Arduino takes much less power on average (but I'm not sure.. also depends on which Arduino) in which case the 7805 may be fine with a smaller heatsink, but in any case that DC-DC converter should be ok without a heatsink.

• ###### Re: Question regarding thermal resistance of a UA7805 regulator

Measuring current for reading three thermistors and displaying text values on the LCD is drawing approximately 0.235mA - although it does vary between that and 0.224mA.  I could get away with a 15C/W heatsink which can be quite small (and I think I might squeeze in just) but I much prefer your idea especially as it's a pin equivalent, fits and a lot more efficient.  I also don't know if it will work a lot harder once I get the graphics going and the voltage/current monitoring.  Got one on order so I will be able to test it out tomorrow hopefully.

• ###### Re: Question regarding thermal resistance of a UA7805 regulator

Just following up.  I ordered a Traco Power part and it fits comfortably on the PCB I designed without obstructing anything.  Very pleased with that.