How to turn a TEG output into something I could use?

It is not a good match to have a motor ( very high start current V/Rs ) to a tiny Seeback Effect device.
EVen though solar power is up to 1kW/m2 and most of it thermal , it could be a good heat source. with flat black painted aluminum insulated with styrafoam on bottom and enclosed to prevent convection cooling except cool part on ground underneath and device mounted to bottom exposed pad, which is solar heated on top and kept hot by reflective walls to concentrate more energy such as a satellite dish and device where LNA used to be. then aim at sun and use heat pipe and CPU heatsink to keep otherside cool with silicone insulation and water plumbing from CPU cooler to a tank
.
Calc max power ,define max temp,
compute area required for heat source and direction of sun...why do I feel like a skinny rat?

Is this just a science project? Define scope of project please.

Ok sorry for the delay. I'll address your points one at the time.

Tony Stewart said:

Everyone ends up chasing skinny rats into tiny holes until you specify all the parameters of your system.

Click to expand...

I see what you mean.

input: 1V 130mA
desired output: 1.5V ~100mA

Must have components

TEG the output under the conditions it will be is 1V and 130mA
DC Motor 1.5V 60mA, it draws about 110 mA under a load.

The DC motor is a 10 rpm motor but I could make a 1:5 gearbox making the load easier to handle. The motor should run about 30 seconds.

ronsimpson said:

Is there a chance you can use two to get 2 volts?

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I really want to avoid it because there are many of this milk stirring buckets and it will be pricey to scale. Also I know more people that makes cheese with this method and I would like to sell it to them if this works. Just before anyone suggest wind power I tell you there is no actual wind there. Solar is expensive and these buckets arent all in the same room and are moved from here to there all the time so it would be hard to do all the wiring.

Dr Depper Im still looking onto the info you sent.

I still like the approach spec come up with but I need guidance to do that.

I thought I read the motor spec is 6V 60mA with a 60 RPM, but I assume

Tony Stewart said:

Everyone ends up chasing skinny rats into tiny holes until you specify all the parameters of your system.

Input power, desired converted power , efficiency, output power


For example motor rated power,, actual load vs rpm,,,, coil resistance and rated voltage all affect discharge rate. Start surge currents are often 800%* rated current

Same for Seeback device,,
Thermal,power input vs V & I output. e.g. Normally Solar cells are optimally loaded around 80% open circuit V while Seeback are around 50% of no load V= V open cct=Voc


I trust you understand this important engineering design method.

Click to expand...

You linked a 10W 4 Volt device yet only spec'd 1V 1.2W . why?
The motor is spec'd at 6V 60mA 10 RPM with gear reduction, yet you want to speed it up , which will increase the load current but maybe reduce operating time, so you need to define the work being done with values

Note the efficient load match is 50% of Voc, which I mentioned before.
So your assumptions and spces need to be validated and expanded for work to be done and heat source, Th,Tc for hot and cold side, duty cycle of output, more calculations on heat energy input, cooling heat sink thermal resistance required-will need to be done at some point.

Either define the task better or the engineering design specs.

I thought I read the motor spec is 6V 60mA

Tony Stewart said:

Everyone ends up chasing skinny rats into tiny holes until you specify all the parameters of your system.

Input power, desired converted power , efficiency, output power


For example motor rated power,, actual load vs rpm,,,, coil resistance and rated voltage all affect discharge rate. Start surge currents are often 800%* rated current

Same for Seeback device,,
Thermal,power input vs V & I output. e.g. Normally Solar cells are optimally loaded around 80% open circuit V while Seeback are around 50% of no load V= V open cct=Voc


I trust you understand this important engineering design method.

Click to expand...

You linked a 10W 4 Volt device yet only spec'd 1V 1.2W . why?

Note the efficient load match is 50% of Voc, which I mentioned before.


The motor is spec'd at 6V 60mA 10 RPM with gear reduction, yet you want to speed it up , which will increase the load current but maybe reduce operating time, so you need to define the work being done with values

I assume you tested it at 1.5V and measured current. But then the RPM would be closer to 2 RPM.

So your assumptions and spces need to be validated and expanded for work to be done and heat source, Th,Tc for hot and cold side, duty cycle of output, more calculations on heat energy input, cooling heat sink thermal resistance required-will need to be done at some point.

Either define the task better with all energy inputs or the engineering design specs.

I'm not an expert on the 'thief, efficiency depends on the supply voltage, quality of the transformer, gain and vce sat of the transistor etc, the reason why i mentioned it here is the o/p says he's a bit of a noob with electronics so I wanted to keep it simple, after all we are only stirring milk to make cheese, if the system was only 25% efficient but worked with minimal effort that may well be enough for a prototype.

dr pepper said:

I'm not an expert on the 'thief, efficiency depends on the supply voltage, quality of the transformer, gain and vce sat of the transistor etc, the reason why i mentioned it here is the o/p says he's a bit of a noob with electronics so I wanted to keep it simple, after all we are only stirring milk to make cheese, if the system was only 25% efficient but worked with minimal effort that may well be enough for a prototype.

Click to expand...

I started to wonder how many liters you actually stir with such a motor...

Matienzo said:

input: 1V 130mA
desired output: 1.5V ~100mA

Click to expand...

I don't have a data sheet on the TEG you have.
I think it will output 1V at no current, 0V at 130mA, but is is best run at 0.5V at 65mA. That is the max power point. It can not do 1.V at 130mA.

I have a circuit that loads the cell down to (0.75V) and outputs 1.6 volts. It is not quite right yet but I will work more on it tonight.

Does anyone have a spice model for a TEG cell? I think I will use a 1V source with 7.7 ohms internal resistance.

Tony, thank you for helping me streamline my messy approach to this problem. Let me try again.

Tony Stewart said:

I assume you tested it at 1.5V and measured current. But then the RPM would be closer to 2 RPM.

Click to expand...

As you said, I made the motor run on 1.5v and 100mA and the torque was strong enough to move the blades. RPM decreased indeed

Tony Stewart said:

Either define the task better with all energy inputs or the engineering design specs.

Click to expand...

Power input: 130mW -I measured volts and amps directly from the TEG
Power output: 150mW
Duty Cycle of output: 25% -Does this need a time value too?
Th: 210°C -This is the temperature of the wall of the heater
Cooling heat sink thermal resistance: I don't know. It is an old PC 40x40 heat sink.
The ambient temperature on the room may peak at 26 °C .
The motor runs at 150mW under a light load.
The gear ratio was to slow it down not to speed it up. The correct ratio is 5:1. That is why I said it might make the load easier to handle.

Is this a proper description of the task and some of the engineering design specs? I hope the task at hand starts making sense to everyone.
As Dr Pepper said I'm a noob with electronics. Most of the questions you guys have I never even thought of but are making me learn a lot, not only about electronics but how I approach a problem and explain it to someone else.

Tony Stewart said:

So your assumptions and spces need to be validated and expanded for work to be done and heat source, Th,Tc for hot and cold side, duty cycle of output, more calculations on heat energy input, cooling heat sink thermal resistance required-will need to be done at some point.

Click to expand...

I agree with all of the above, I'll just need the guidance.

What would be the problem with charging a battery or a capacitor and then release the energy in a controlled way?

Matienzo said:

Spec I found this IC to use as a step up. Do you think it is adequate? I'm not sure how I would connect the battery and such after the IC but I'm looking into that

Click to expand...

Good find Matienzo,

That is the kind of chip (LTC3108-1) you can use to convert the output from the Peltier device into a voltage to charge the battery. There are quite a few chips of this kind; some are designed specifically to charge LiIon batteries. I will investigate further.

With this kind of application it is best to work in stages. One stage will be to get the Peliier device charging the battery. Once that is achieved the rest of the electronics should be pretty straight forward, especially as you have now told us what the application is: stirring milk.

Straight away you could get a CR123 LiIon rechargeable battery (used in many cameras) and once it is charged see how it drives your motor. You will need a LiIon charger. I am thinking about not bothering with the step down converter and driving the motor directly from the battery.

If necessary later we can possibly change the motor to suit the battery and the load of stirring the milk.

spec.

Try this on for size: ditch the Peltier device and power your milk stirrer by one rechargeable 18650 LiIon battery which would give a continuous motor run time of 68 hours. Suppose you stirred the milk once every ten minutes for one minute, that would mean that the battery would last for 680 hours or roughly one month before it needed recharging, You could add as many batteries as you like: each battery would add another month duration.

spec

spec said:

ditch the Peltier device

Click to expand...

The idea is to use the heat as a source. I would like to use the peltier not only to avoid recharging the battery but also for what I will learn trying to do it that way. I would like to use this circuit (and/or what I learn while making it) for future peltier projects.

Matienzo.

Back in post #28 you said that you had measured the TEC's output voltage and current. How were these measurements done?

Power measurements, as a product of a voltage and a current measurements, are highly dependent on loading conditions. So a power number without the test conditions is almost meaningless.

ChrisP58 said:

Back in post #28 you said that you had measured the TEC's output voltage and current. How were these measurements done?

Click to expand...

I put it against a 200°C with a heat sink on the cold side and took the measurements straight from the peltier terminals.

Matienzo said:

I put it against a 200°C with a heat sink on the cold side and took the measurements straight from the peltier terminals.

Click to expand...

OK, but how?
Did you measure the voltage at no load? Just put the meter across the terminals?
Did you measure the current by putting the meter across the terminals?

If so you measured the voltage at no current. AND You measured the current at no voltage. THEN You measured both voltage and current at zero watts.
If you measured the Voltage and Current across a 7 ohm resistor then that is very different.
This is why I/we keep asking how was it measured.

oh I didn't know that.

If you have to start this from zero and the only components that are a must are the peltier and the motor on a 25% duty cycle (15 sec on, 45 sec off) where would you start?

OK,
I have a LT3526L-2 boosting the cell output to 1.6volts and holding the cell near the maximum power point.
BIG BUT
To store power on a cap you must have the voltage moving upward to put power into the cap and voltage dropping to remove power. Just holding the voltage at 1.5V is not possible.

Cell----makes power
10uF cap Need to store power for mS.
LT3526L-2 or equivalent to charge up a capacitor to 3.3 or 5.0 volts.
Big capacitor to hold power for a long time.
Buck PWM to drop the 5V to 1.5V at 100mA
100uF cap to store power for the start up time of the motor.

Then we need a simple circuit that starts the motor when the storage cap reaches 5V and stops the motor when the cap reaches 2.0V.

What do you think?

At first I thought of using a LT3526 to boost to 1.7 volts.
Start the motor at 1.7V and stop the motor at 1.0V.
This is simple but I need to see what size of cap we need.
With 15 ohm resistor as a motor, not good but a reasonable load.
4700uf 50mS motor run and 180mS charge time.
47,000uF 500mS and 1.8 seconds.

This is why I want more voltage on the storage cap. Instead of 0.7 volts delta, If we has 7 volts change on the cap then the times would be 5S and 18seconds.
-----edited-----
The reason for the low duty cycle is:
I am certain you can not get 130mW from your cell.
I am not exactly at max power point on the cell. I am loading down the cell to 0.67 volts @ 40mA.

If the cell works the way I think it does:

Then two cells in series will output 1volt at 50mA. (max power point)
I tried two cells that match the curve above.
I get the motor running for 1.5 seconds, resting for 0.8seconds, using a 47,000uF super cap.

I think I have a 0.47F 3.3V capacitor on my bench. $4.00usd

Hmm, the fact that you didn't check the power output from the Peltier device is surprising but, by definition, a Peltier device does output some power, so what I have said still stands. It just means that the ratio of motor on to off time will be greater. Also, in view of the graph that Ron shows for the Peltier device, some form of power point optimization will probably be required.

But if Ron's circuit works without the use of a battery, all to the good and job done.

spec

After shooting off my mouth about super caps I need to choose one.
Many caps I looked at have very high ESR. They can not handle current. (resistance inside the capacitor)
Attached is cap data. I can get for $4.00. This one has 0.2 or 0.3 ohms resistance. 0.22F or 0.47F.
There are about 10 kinds that will work with the 100mA current.

0.47F cap, turn on at 1.7V and off at 1.0V.
Red trace is voltage. Green trace is current.
Not using a real motor but a 15 ohm resistor.