Powerful DPSS SHG Nd:YAG Laser build

Hi all,

Since I seem to be returning to this website quite a lot, I thought I’d show you what I have been working on whilst researching and sourcing parts for a Ti:Sa Laser (anyone got a small surplus crystal or know where to find one? That has been the hardest part so far…).
It is a DPSS Nd:YAG Laser that is tranversely pumped and intracavity doubled to green. The pump diode is supposed to put out 20 W, the rod is 1.6mm x 19mm. The seller told me it should be possible to get 1-2W green out of this. My resonator is concentric with the KTP (3x3x3.5mm) in the middle and the Nd:YAG rod on the HR side. The mirrors are concave 10cm radius, one HR1064nm+532nm, the other HR1064nm HT532nm. There’s a brewster window in the resonator to get the intracavity beam polarized for the KTP.
The module is supposed to deliver the 30A for the Diode, keep it cool using a big TEC and heat up the KTP to 50°C, even though heating the KTP didn’t significantly change the output power. It does that but needs a break after 20-30 min since I underestimated the heat dissipation capabilitis of the heatsink below the main plate (not visible in the photos).
I am currently getting about 500-800 mW (can’t measure those powers…) out of the module.
Interestingly, I had to angle (plane analogy: yaw angle) the KTP, so the incidence is not 90°. I only expected having to adjust the roll angle, so I need to redesign the holder. Does anyone know why that might be the case? Maybe the crystal was cut weirdly?
Anyways, the end goal could be to q-Switch the system and engrave or even etch PCBs… But I don’t know whether I’ll ever reach the power levels to do that.

Any thoughts are welcome of course :slight_smile:


PS: The module probably is from a Lightwave 246 “Halfnote”:

Old Foto, but shows the backside:


This is impressive. I’m out and about at the moment. But I’ll take a proper look when I can.

Chris could learn a lot as he wishes to build his own 532 at some point.


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I had a think of why the crystal wouldn’t be perpendicular to the optical axis and I can’t think of much else other than the cut being off. If cut correctly the optical axis should have been perpendicular. Is it possible you got sold an off spec crystal or one that was originally cut for another wavelength?


Nice Work!

Do you have any more info on the electronics?
You have written “the module … likely from Lightwave”, do you mean the driver or some of the mechanical parts around the pump diode?

The prototyping boards dont look like off the shelf stuff, so i would assume its something you build. Therefore i am quite confused.

I might also have an idea why your tec isnt happy / able to cool the laser.
I see a huge inductor with some active devices on a heatsink next to it. Likely a PWM stage or a stepdown stage. However there is no smoothing capacitor visible.
TECs dont like PWM or high ripple current. The efficiency drops significantly.
It also looks like the tecs cold side isnt fully covered which would be really bad for efficiency.

These could be some of the contributing factors but i am shooting in the dark here as i am just guessing on what the circuit could be.

Nevertheless: very nice work.

How did you do the alignment without being able to see the 808nm and 1064nm i assume some sort of camera setup?


Thanks for the answers! I guess it is a possibility that the crystal is off spec. I’ll experiment with it a bit more, I redesigned the holder.

With the module I meant the diode and the Nd:YAG rod, they came as one “package”. The electronics are all homemade. The 30A diode driver is a bunch of LM338 voltage regulators in constant current setup with big heatsinks (in the back of the build) and a 5V 30A switch mode power supply and some caps. It is switched using a relay. Probably not the best overall solution but I am pretty far from an electrical engineer :wink:

Actually, the TEC is definitely able to cool the diode if the heatsink is cool enough, but the heat output of the TEC is so high that the heatsinks and fans on the underside of the setup are not able to dissipate it quite fast enough. Once the whole setup is about handwarm (around 26-30°C), the TEC cant do it anymore. The inductor is indeed for a PWM buck converter to do some PID controlling of the TEC voltage using the arduino but I didn’t end up using it because it simply was not necessary. Now the TEC is just on or off, no high frequency switching so the inductor just is a fancy wire :slight_smile: Actually, I designed the whole thing with a smaller TEC in mind but had to switch to this one which is why some of its cold side is showing. Also, there are smoothing capacitors, but pretty small ones I guess. Again, I never design a TEC driver like that.

The alignment was done using a HeNe laser and a pinhole. It’s tricky to do with a concentric resonator since the reflected spots from the mirrors diverge fast, so the pinhole has to be close to the OC. Once the spots indicate good alignment, I turn on the diode and watch for leakage through the OC using my phone camera. It is not much, but if the laser is lasing, its visible. Sometimes it needs a few iterations and “blind fine tweaking” to get it to lase. Then I can insert the KTP and roughly align it so that some green comes out. Now I dont need the camera anymore and can fine tune everything until I reach the maximum.



Thats for explaining the alignment. I would not have thought its this “simple”. I expected fancy infrared cams and many hours of fiddling with it without seeing anything.

The diode driver is crude but should work good and dont damage the diode if you respected the LM338 data sheet and used the recommended 0,1uF output and input capacitors. They need to be close to the IC (and should not be bigger) because the IC !could! begin to oscillate if these are missing. Oscillation could cause overcurrent spikes which could damage the diode.
Also the switching relay needs to be on the primary side the of lm338, otherwise the voltage regulator would try to source the current that its set to but cant because of the open relay.
This would make it open its pass element fully (0 resistance) which would give a huge current spike into the diode if the relay closes.

For the TEC:
Ah i understand now. On and off control could be the reason why the tec-heatsink combination isnt able to keep up while the unit heats up.
While the heatsink is cold, the tec should be running with lower power to keep its efficiency high. Having it run full blast would shift its operating point to lower efficiency. This results in the heatsink getting hotter then necessary which results in more on time needed for the tec. This is a positive feedback loop which makes the heatsink much hotter than necessary which ultimately causes the tec to fail to keep up. (because the heatsink gets to hot)

There is also the fact that while the tec is off, the pumped heat is flowing back from the heatsink into the diode mount. This causes additional losses = hotter heatsink.
This is also the same reason why the tec should not be driven with high ripple currents / PWM directly. The off periods would allow time for energy to flow back.

If you want, i could explain a little more and show you how to calculate / choose the right tec / estimate if the tec is enough for your setup.
There isnt even much data neede: the heat load (diode voltage + current + temperature it should be keept at) and the tec specifications and some data from the heatsink.

Just let me know. (could also drop you a PM in german if you want)

  • Builder

Thank you so much for explanation!
Regarding the TEC: So it might be better to actually use the buck regulator (although with bigger smoothing caps) to regulate the voltage instead of doing it on-off? Although it never runs full blast, its max Voltage is 14.7V, I run it at 12V.
I am relatively sure that the TEC (datasheet of a very similar model, maybe the same) is correct, since its max cooling power is ~80W and the diode dissipates 1.7V*30A-20W optical output power=31W. Maybe a bit more since the rod also absorbs a lot of that power but still… Holding the diode at like 20°C should not be a big deal… 25°C would be alright too, as long as it doesn’t exceed 30°C and is relatively stable.
I actually dont have a datasheet on the heatsink of the setup, its this one, with fans blowing air across it.
I would be very interested in learning how to drive the TEC more efficiently, maybe the heatsink is enough but the TEC is too inefficient. Although currently I just have temp sensors for the diode and the KTP… I guess I could attach a third one to the baseplate if necessary.

I think it is a good idea to discuss this here, since the others can chip in if they want and maybe other people (than me) could learn about this topic.

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Builder has more experience on electronics. What temperature are you keeping the KTP crystal at?

The KTP is held at 45°C (I mentioned 50°C above but looked it up in the code, its 45°C). Altough to be perfectly honest, heating or no heating didn’t make that much of a difference to the output power.

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Just spotted it. First off good. KTP likes to be warm. If you have cooled the crystal down to sub 20C temps you would have definitely noticed a reduction in efficiency.

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Yep i would think the same, i just dont want to hijack your thread (without permission) about a laser and talk about tec specific stuff.
I think this will be a long post.

Lets start at the basics on how to use a tec (you did most of that right but others might need a bit of help):

  1. Thermal stuff works the same as electrical stuff, there are resistances (thermal resistance), current (heat flow) and voltage (temperature differential). If you have basic understanding of electronics, thermal stuff is easy.

  2. Tecs are very fragile, they break easily because there base material is a ceramic substrate. Make sure to have both surfaces that connect to it flat without big milling marks. Ra should be below 1,6.

  3. Use a thermal interface material. Small voids from surface irregularities would lead to the tec

  4. The Tec needs to have good contact on its whole surface. The Tec base material is a good conductor but if the Tec isnt contacted on the whole surface you get parts that dont contribute to the heat pumping action which lowers the performance.

  5. Keep thermal shorts as high resistance as possible. The tec needs to be mounted in some way. This most of the time means clamping (using bolts) it between surfaces. Doing that will create a bridge between hot and cold side with a defined thermal resistance. The resistance is responsible for how much energy is flowing from hot to cold. Having long bolts with low thermal conductivity and insulation washers is key to keeping that resistance high. So stainless steel bolts and plastic insulation washers.

  6. A Tecs is more efficient if its temperature differential is low and its overall temperature is high.
    Getting 10K differential from -80°C to -90°C needs way more power than getting the 10K from 60°C to 50°C.
    Keep that in mind while looking at the graphs on your tec data sheet. Most of them have curves for 50°C hot side temperature and for 27°C hot side temperature.

Your calculation for the dissipated power of the diode is perfectly correct. However as we are doing a “back of the napkin” / “ballpark” calculation, i would rather be a little pessimistic than optimistic.
I would take the full 30A*1,7V = 51W instead of subtracting the optical power. You all ready said it: the crystal thats cooled will also absorb some of the optical power and convert it to heat.
You also got additional heat that needs to be pumped because your mounting hardware is essentially a “thermal short” (similar to an electric short) because it connects the hot and cold side of the tec together.
Some amount of power also gets into the cold side because of convection and conduction of the surrounding (hot) ambient air.

Its hard to estimate how much these factors play a role, so i would just go all in and have too much tec power that too little.

So our first fixed value is: we need to pump around 50W

A tec essentially just creates a temperature difference across both surfaces. This difference depends on current and heat load (=pumped thermal power).

So to estimate current usage and choose the tec, we need to know the ambient conditions and the temperature we want to keep our load at.

You said the diode is smack on 808nm at 25°C. So that gives us our fixed cold side temperature of 25°C.

To get the actually differential across the tec, we would need the hot side temperature. However the hot side is connected to a heatsink which isnt fixed in temperature. The heatsink temperature depends on the thermal load.

A heatsink has a value “thermal resistance” which describes → how much temperature rise it will have at a given dissipate power level.
The value is for example 0,75K/w which means the heatsink would get 1,5K hotter if we put 2W of power into it. The formula is Power * Resistance = Temperature rise.
Your heatsink does not state values and you added fans so its likely not accurate. The value is easily measured. Drop some power into the heatsink and measure the rise above ambient temperature. The resistance is then just: Temperature Rise / Power = Resistance.

So we need the load of the heatsink and the maximum ambient temperature to get the heatsinks temperature. (which is equal to the hot side temperature)
However there is a problem: The load into the heatsink is dependent on the tecs thermal output power on the hot side.
Which itself is dependent on the hot side temperature.

To explain this, I will use a different data sheet of a slightly different tec because its clearer.
Datasheet here

Lets assume for a moment we know that the heatsink will get to say 55°C.
That would mean we need a differential temperature of (55°C-25°C) = 30K across the tecs sides.
With these variables, it would be possible to see how much power the tec needs.

A look into the data sheet lets us use this curve to see if thats even possible with our tec candidate.

We use the Th = 50°C (temperature hot side = 50°C) graph because our imaginary heatsink reaches 55°C which is close enough. Qc is our heat load of the diode (50W) which gives us a estimated drive current at DT=30°C of around 6,8A. The tec is almost fully loaded and cant take much more power.
Scrolling a bit down gives us:

We need just over 12,7V to pump 50W @ DT=30K.

Everything is good so far?
We know our tec needs 12,7V @ 6,8A to pump 50W so our total heat load into the heatsink would be around 87W of power for the tec with an additional 50W of pumped heat = 137W into the heatsink.

NO! our tec is struggling hard and is just barely able to pump the heat which results in very bad performance. Our heatsink needs to be much bigger than necessary with the right tec choice.

Having a look at the next graph will tell us the whole story:

The COP is a measure of how efficient a heat pump is. having a COP of 2 would mean pumping twice as much heat as electrical power is spent.
The COP on our imaginary setup is around 0.57. So we need to spend around 90W to pump 50W.
The graph shows If we could use the tec at 6V, we would get a COP of over 1, meaning spending 50W to pump 50W.

The next part is a little “gusstimation”.

To make sure we stay inside the maximum efficiency band of a tec, we need to choose it according to our heat load. A good rule of thumb would be around 3x as much Qcmax as we need to pump. This almost always gives us a COP of around 1 for the usual laser applications.

We need to pump 50W → our tec needs to be able to pump >150W.
Lets use a Tec that respects this rule of thumb:

This rule of thumb also makes our hen and egg problem vanish. We can choose the actual heatsink resistance based on our expected heat load.
It would be around twice the power we need to pump. So our heatsink needs to be able to dissipate around 100W without getting hotter than around 60°C.

A heatsink hotter than 60°C would make the tec struggle as its differential would increase and therefore reduce the COP.

If we have a normal indoor device, its expected that is should work in temperatures of up to say 35°C ambient. Some applications require this to be higher, others can get away with lower max temperatures.

If we stick to the 35°C max ambient, our heatsink needs stay below 60°C. This gives us a differential of 25°C for 100W of power or a heatsink with a thermal resistance at or below 0,25K/W.

We can now just use the tables inside the data sheet to see how much voltage and current the new tec will need to keep our load cool.

Reading them would give us around 6A @ 7,25V which gives around 44W of power usage with a chop of over 1.
The heatsink needs to dissipate around 100W which gets us to a resistance of 0,25K/W.

The first that that also could cool the load would use 140W and wasnt able to be run at heatsink temperatures of 60°C. It could only pump 50W at 55°C or below.
So the heatsink would need to be way bigger because the differential was just 20°C @ 140W which is around 0,14W/K, so much much bigger.

Long long text :smiley:
If something is unclear, just tell me, its hard to write everything down into a coherent explanation.

On your setup, there are at least two facts that kill the performance of the tec.
You said your tec struggles at just over 30°C heatsink temperature. So its essentially just able to create a differential of 5°C which is very low.
I suspect a bad tec (cracked or a dodgy source) or that your mounting is subpar.

The tec is build out of many thermal couples, each is contributing its own share of pumped power. You tec is a 127 type, meaning it contains 127 pairs of pumping elements. Having the tec surface not covered completely leaves say 10% of the couples with high thermal resistances to your load which makes them run at way higher differential temperatures resulting in much lower heat pumped.

Thats certainly a factor but cant explain just 5k of differential.

The L-type clamp that holds the module onto the tec certainly contributes another 5-10% performance loss because of the thermal short.

However i think you got a dodgy ebay tec. I seen my fair share of tec from crappy sources, they are cheap but contain non working junctions that waste power and dont pump heat:

(darker = colder ,two tecs on a heatsink)

This thermal picture shows two elements, one good one and a dogdy one.
You can clearly see that the cheap one has a complexly failed row of couples that let heat flow back into the cold side. Something like that is killing performance.

(grammar and typing mistakes for your pleasure, i dont want to go over this again :smiley: - i am not a native speaker)

Way to long but i struggle to boil it down further.

  • builder

Thank you so much for taking the time and writing this out! I can see that I definitely did make a few mistakes when setting this up but I guess thats the learning experience. Didn’t even think about the thermal shortcircuits for example!
In my eyes I have a few things to do now. I need to figure out what thermal resistance my heatsink has and if that is enough for my setup. I also need to check whether the TEC is working correctly, it certainly is possible that my driver doesn’t actually put out enough current.
The more I think about it, I probaby need to increase the heat dissipation capabilities (aka replace the heatsink) since the diode driver heatsinks also produce a lot of heat as well. I actually sized them very similarly using heat resistance and corresponding equations. Part of that gets blown away but the other part flows into the housing of the setup, heating up the heatsink. But still, as you say, the TEC should easily be able to handle a 5-10°C temp difference… I’ll look into it!

Again, thank you for the post, I’ll definitely bookmark it for later reference.



No problem!

I hope i could get some information that you did not know across.

Measuring the thermal resistance is easy. You could just use your current setup.
Let your setup run until it reaches steady state, as you currently have overheating problems, just run the diode at a lower power level. Its not needed that the system lases. Just make sure it draws a constant amount of power.
Measure the complete power your system is drawing from the grid, the ambient and heatsink temperature and calculate heatsink thermal resistance from there.

You could also try different fan speeds as there is also a “break even point” aka more airflow trough the heatsink does not decrease its thermal resistance much. If you have enough flow, the tip of the heatsink fins will be at ambient temperature. This means that you get a significant reduction in dissipating surface area which will result in almost the same thermal performance as lower air speeds. (but with lower noise and lower power usage)

Many of the ebay tecs i measured did not adhere to their specifications. Some 12706 tecs drew less than 4A at 12V.

Laird Thermal Systems have a very very good calculator for this. Only useful for there modules but still nice to for understanding the behavior of the a TEC on a heatsink.


Thats a 150W tec that could suit your laser, they are available at DigiKey.
They are not as good as Thermonamic ones, (lower cop) so you would need around 0,2K/w of heatsink.

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You found my post on those 532 nm “2 watt” DPSS heads I had posted years ago when active in that forum under the member name of Alaskan. I no longer live in Alaska, so now using this ID here. Later learned these heads typically produce closer to 1.75 watts CW, some less. I haven’t worked with them since leaving Alaska, but still have several heads in storage up there in the great white north.

I have a manual on that laser as well as recently purchased some control boards for it from Starlight Photonics, an ebay seller. I’d like to someday build a high power green DPSS, as Curtis mentioned. Have collected a lot of surplus KTP and YVO4 crystals from ebay as well as YAG rods and optics to do so. Don’t know if I will get to that project as time passes quickly and I’m on to other things now.

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Thanks builder, I’ll definitely do that.

Actually, pretty much all of the 12706 TECs I bought off of ebay drew like 3A at 12V, either its a systematic scam or all of these sellers really don’t know what they are selling.
Again, I am confident that my TEC (if it is indeed working correctly), should be able to handle it , since “the laser only dissipates about 40W max and requires a heatsink about the size of a CPU cooler” (source). But we’ll see.

Regarding building the high power 532: The guy I bought my module off of did mention that he had two of them inside a simple confocal resonator with a brewster plate and the KTP in the middle and achieved almost 4.7 W of green. Never saw proof of that but man that would be awesome.

Another question: would a longer KTP be helpful to extract more green? I may be able to get a 10mm long one, which would be almost 3x as long as my current one. This paper says it does but I wonder if anyone has practical experience on that topic.

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Hey Nik,

Thats also my experience.
A good -to spec- 12706 Tec should draw around 5-6A at 12V, depending on the differential and absolute temperature.
At powerup (so no differential temperature and cold heatsink) you should absolutely get 6A otherwise its not a TEC1-12706.

I suspect that these cheap tecs are some sort of rejects.
A single bad solder joint inside the tec (there are at least 254 in there!) could introduce enough resistance to limit the current to lower levels.

Or they systematically use smaller junctions = less material = cheaper = higher resistance = lower current and performance.

CPU coolers are quite good. The heat pipe type thats also used by gamers is usually capable of dissipating more than 150W while keeping temperatures low enough to use tecs. (<65°C).
Sadly no manufacturer publishes real values, only marketing stuff.

Even Noctua does not (even on request) state the thermal resistance of there NH-D15. They just go ahead and invent a different measurement system that makes it impossible to calculate real temperatures…

For the KTP: i have no expertise there.
(thumbs up for using sci-hub!)


I have finally gotten to measure the TEC module I bought off of eBay.

As you can see, I am not getting the expected 9A at 14V, only about 6,5A. I suppose there might be something broken in there, just as you said. So I’ll have to replace it.
I found a much bigger heatsink on ebay which I already attached to the baseplate. It didn’t come with a resistance value but a very, very similar one on rs-online has got ~0,5K/W with no forced air and 0,2K/W with forced air. In the spirit of pessimistic calculations, I proceeded with 0,5K/W.
Again, Tcold = 25°C and Qc=50W. Ambient is 20°C on average in my basement. So, Th = 20°C+50W*0,5K/W=45°C. deltaT is therefore 20°C.
I found this TEC on rs-online (never want to go to ebay again for TECs). The price is no joke but I really want this thermal issue to go away. At Qc=50W and deltaT=20°C it draws 5A. Thats 6,5V (at I=5A and dT=20°C). I used Th=50°C in the datasheet btw. COP for this point is around 1,5.

I think this leaves enough breathing room for stable operation especially since the thermal resistance is most likely lower than the used value, possibly even half that. I didn’t have the opportunity to measure it yet since I had to disassemble the whole module to replace the heat sink. Didn’t think I’d need to… But I suppose this will work.

What do you think?


I will write something more when I get home.

On a quick glance, I see a mistake regarding Qc.
You calculated the heatsink temp with Qc=50W.
But that’s only the thermal power that’s flowing into the cold side of you module (power dissipation from the diode).

Your heatsink however needs to dissipate the pumped power of you diode (Qc=50W) AND the power that’s needed to pump the 50W.

If you need 6,5V at 5A to pump 50W at dt=20K, you essentially using 32,5W of electrical energy to pump 50W across the thermal differential.

So your Tec dissipates: Qc + the power needed to pump Qc.

This power gets dumped into the heatsink. So you need to calculate heatsink temperature with 82,5w, not 50W.

(i am home now, so longer text) EDIT:
For your TEC, yes it seems to be a cheap underperforming Tec, sorry about that :frowning:

The Tec from RS should be ok. I cant tell if it lacks a critical “feature”, RS advertises it as “with sealing option” but the datasheet does not state if it is in fact sealed.
Having a tec without epoxy / silicon seal will make it perform a little better because it got less “thermal short” tough the seal.
However condensation can be an issue that will cause corrosion over time.
Internally, the tec will get a little colder than on its surface, in moist environments on hot days, this moisture in the air will condense and corrode the tec.

Otherwise, its fine, maybe a little small Qc wise but still good enough!
A little big (55mm) for your load? Remember, having the tec not fully covered will result in bad losses as the outer junctions will get colder than your load. This causes more Dt therefore reduces the COP and Qc.

Its also quite expensive.
I would still recommend the Laird Tec.
Its cheaper and got more Qc, so a little better COP.
Its still 53mm long, so could still make problems fitting.

Otherwise, a much cheaper option (1/2 price) could be a Tec1-19906. Phillip has some from a reputable source.
That would result in a little lower COP (around 1,2) and would need around 13V or so.

I am not sure whats your current supply voltage is that you got available in your system.
If you only got 12V, thats a nogo.

Ambient temp:
You calculated with 20°C ambient temperature on average. However you should use the highest temperature that can occur/that you want your tec to still keep up with.

I would go for at least 30°C otherwise, you might get problems in the summer months

That should work. Most of the time data isnt available, so a educated guess is needed. I also check around to see if someone is selling something similar. That works most of the time.
The difference between 0,2 and 0,5k/W is big, so you should have enough headroom.

Just one thing: you said you got other thermal loads on the heatsink, like your laser driver or electronics.
You can easily account for them if you know there power dissipation: just add them to your heat load calculation like so:

Th= (Qc + Pn + Pe) * Tr + Ta
Qc is your heat load on your tec,
Pn is the Power thats needed to pump Qc
Pe is Powerdissipation of your electronics (or other loads)
Tr is thermal resistance of your heatsink
Ta is ambient temperature

Its the same formula but you load the heatsink with everything thats dissipated into it.

  • builder

Ah, right, forgot about the electrical power. I would actually prefer the Laird TEC but I only have 12V available and the shipping time is 8 weeks. I’ll use the rs-online one then. At 30°C and 0.5K/W COP is pretty low (0.8) and the heatsink is at 70°C but those are very pessimistic values. I’ll measure the heatsink this week using power resistors to see what it can do. The heatsink is actually as big as physically possible for the module so I can’t really just put a bigger one on there.

Also the heatload by the drivers is unknown. I use some fans to blow air across the LM338 heatsinks and out of the module and the drivers are thermally isolated from the casing but some of the hot air is going to make contact with the casing. Honestly, I didn’t think this would be such an issue but it is.

Worst case, it’ll only be able to run 1h or so at a time. We’ll see.

Try to mount your heating resistors exactly where the tec would go. This way, you will account for heat spread issues and thermal resistances from your outing point to the heatsink.

Thermal stuff is easy but a challenge to get right. It can make or break a device.

I think your setup with the bigger heatsink and better tec should work out good.
Starting with the actual heatsink measurement will be great!
That should give you all the data you need.

Unknown heat load: you could try to add insulation inside your case to keep the hot air away from your “cold” heatsink.
Something easy like paper to funnel the air along its path works great.