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For my astrophotography equipment, I need to construct a dew heater and a flat field generator.

A dew heater is simply a bunch of resistors connected in parallel such that each resistor outputs about 0.3W of power, and spaced roughly 1cm apart. Typical dew heaters run off 240V-12V AC-DC adapters and have controls to vary heating.

A flat field generator is a box with LEDs or LED strips mounted on one end, and a diffuser (white perspex) on the opposite end to create an even field of illumination. Typical flat field generators also run off 12V and have control to vary the brightness of the panel.

The design I came up with is this:

schematic

Here, the transistor is biased through a potentiometer. The same design would be used for the dew heater by replacing the LEDs with resistors. However, I have a couple of issues with this:

  1. I would like to have a linear (or almost linear) relationship between potentiometer value and brightness/heating (current through collector). In simulations, I can tell that for this circuit, the relationship is not linear. Is there a simple way to rectify this?

  2. For the dew heater, I require it to produce about 5W of power max. How can I calculate the values of every resistance in the diagram to ensure this? Also, since the resistors are in parallel, the sum of all currents will go through the transistor too. Is there a generic part number for a transistor that would be better suited for this task instead of having a fried transistor? Also, how much should the power supply be rated for this? I have an option to use either a 5V or a 12V power supply.

  3. For the flat field generator, the problems are even worse. Linearity issues aside, my panel will use LED strips like THIS one. I can't find a datasheet for this LED strip but it has 18 SMDs per strip and runs on 4V. I think all 18 LEDs would be wired in parallel since white LEDs have a forward voltage of about 3.3V. If that is the case, each strip would draw about 20mA x 18 = 360mA. My panel will house up to 25 of these strips. If I connect them in parallel, they will draw 9A of power! Is there a better way of doing this? If not, are there any generic transistor part numbers that will handle this amount of current, or will I have to break the circuit up into multiple drivers, each driving a few strips? Also, I doubt I will find a 12V power adapter rated 9A.

I know the design is a basic one that you'll find anywhere on the internet. If any of you has a better/more precise solution, I would love to hear about it.

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Here, the transistor is biased through a potentiometer.

There are a couple of problems here.

  1. The configuration you have chosen is called an emitter follower because the voltage at the emitter will follow the voltage at the base less about 0.6 to 0.7 V.
  2. If the wiper is turned up to, say 10 V the emitter will try to rise to 9.4 V. With R2 = 100 Ω the current will be 94 mA shared out between all your LEDs.
  3. Parallel connection of LEDs is generally a bad idea. The ones with the lowest Vf (forward voltage) will hog more current than the others and run hot decreasing the Vf further, etc. You won't get even lighting. Usually resistors are added to prevent this.
  4. Series connection of the LEDs is more efficient as the same current can power more than one LED.

I would like to have a linear (or almost linear) relationship between potentiometer value and brightness/heating (current through collector). In simulations, I can tell that for this circuit, the relationship is not linear. Is there a simple way to rectify this?

Yes.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Current control.

How it works:

  • R1 applies a voltage between 0 and 4 V to the non-inverting input. Let's set it at 3 V.
  • OA1's output will rise until the feedback voltage on the inverting input equals that on the non-inverting input. This will happen when 3 V is dropped across R3 which in turn requires 100 mA through R3.
  • The current will be divided between the two strings of LEDs and R4 and R5 limit the difference in currents between the two strings of LEDs because if either chain starts to hog the current the voltage will fall on that string.

For the dew heater, I require it to produce about 5W of power max.

Your requirements aren't clear here. If you use a dimmer then you'll have heat dissipated in the dimmer and that may affect what you're trying to do. The simplest would be to have switches for the heater resistors.

How can I calculate the values of every resistance in the diagram to ensure this?

You should be able to figure it out from my explanation above.

Also, since the resistors are in parallel, the sum of all currents will go through the transistor too.

Correct.

Is there a generic part number for a transistor that would be better suited for this task instead of having a fried transistor?

Look at the specifications for Darlington transistors. Work out at what value of current and voltage across the transistor that the power dissipation will be maximum and then figure out your heatsink requirements.

Also, how much should the power supply be rated for this? I have an option to use either a 5V or a 12V power supply.

Decide on a voltage requirement based on the worst case of the LEDs or resistors. Figure out what current is required by both loads. That's it!

For the flat field generator, the problems are even worse. Linearity issues aside, my panel will use LED strips like THIS one. I can't find a datasheet ...

"No datasheet? No sale!" or be prepared to spend a lot of time characterising them.

... for this LED strip but it has 18 SMDs per strip and runs on 4V. I think all 18 LEDs would be wired in parallel since white LEDs have a forward voltage of about 3.3V.

Correct, so it's not a good design.

If that is the case, each strip would draw about 20mA x 18 = 360mA.

You don't know that. There are no specifications.

My panel will house up to 25 of these strips. If I connect them in parallel, they will draw 9A of power!

Amperes are a measure of current, not power.

Is there a better way of doing this? If not, are there any generic transistor part numbers that will handle this amount of current, or will I have to break the circuit up into multiple drivers, each driving a few strips? Also, I doubt I will find a 12V power adapter rated 9A.

Your maths may be OK but you still have no specifications. 9 A won't be a wall-wart. It will be a proper power-supply.

Finally,

... about 0.3W of power, and spaced roughly 1cm apart.

We don't say "0.3 watts of power" in the same way we don't say "1 cm of distance" or "100 kph of speed". The watt is a measurement of power so we just say "0.3 W".

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  • \$\begingroup\$ Thank you so much for those valuable insights! Your points #3 and #4 were very solid. I vaguely remember reading that somewhere on the internet while researching for this. =) I also managed to test the circuit you drew, and it works like a charm! The current output is very linear indeed. I remember studying OpAmps way back in my university years (almost 12 years back) but at the moment have gotten very rusty since my profession turned out to be different than what I studied for. =P I appreciate the placement of R3 and R4 to keep the chains balanced like you said. Thank you! \$\endgroup\$ – AweSIM Feb 7 '20 at 19:15
  • \$\begingroup\$ Elaborating on the dew heater, I require that the heating elements (the resistor ladder) should dissipate a maximum of 5W. The end result of this project will be a controller that will be encased in a box, and a heater element made up of a resistor ladder that will be covered in duct tape and wrapped around the telescope near its objective lens. The idea is to heat the lens just enough to keep it a degree or two above the dew point. For this reason, I cannot have switches for resistor heaters. If I understood you correctly, you're implying that I make different heating elements and switch... \$\endgroup\$ – AweSIM Feb 7 '20 at 19:19
  • \$\begingroup\$ ...and switch them using switches. It needs to be one single heating element with a controller that can control the current flowing through the resistance ladder to vary the power dissipated by the heating element to the lens. However, as you have been kind enough to provide me with a very practical circuit using OpAmps, I think I can use the same for this too. By removing the LEDs and simply connecting 14 resistors in parallel between Vcc and collector, I can tune the resistance values so that the heating element dissipates 5W. I can then use the potentiometer to vary the current through... \$\endgroup\$ – AweSIM Feb 7 '20 at 19:23
  • \$\begingroup\$ ...through the heating element to reduce the heating as required. You're also right about having no datasheets being bad. I'll try looking for a better replacement for LED strip. My bad I used amperes as a unit of power. That was a typo. =) Thank you for your invaluable guidance. =) I'm think I'm now headed along the right path now =) \$\endgroup\$ – AweSIM Feb 7 '20 at 19:26
  • \$\begingroup\$ I'm glad it helped. (1) Can you explain what the flat field generator is for? I would have thought you would avoid ambient light like the plague. (2) Is your dewpoint + 2°C just to save battery power or something else? \$\endgroup\$ – Transistor Feb 7 '20 at 19:54
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I would like to have a linear (or almost linear) relationship between potentiometer value and brightness/heating (current through collector). In simulations, I can tell that for this circuit, the relationship is not linear. Is there a simple way to rectify this?

enter image description here

  • Opamp and a transistor based simple constant current can be used for LED section. You will have better control of the current than the circuit above.
  • You can have a potentiometer instead of R3 for varying LED current manually.
  • Take a note of current flowing through the R3. Upper limit for current will be the current carrying capability of R3. One can also time the resistive divider network as ell as long as the load(LEDs) see sufficient voltage.

For the dew heater, I require it to produce about 5W of power max.

  • Connecting resistors in parallel as a load. Can the resistors be directly connected to the load? With individual switches to control at discrete levels to generate different levels of heat.
  • Beefier transistors can be used along with heatsink and then the 1W rated 10 ohm resistors can be connected in parallel. The current can be set such a way that each resistors are dissipating about 0.8W or so. Resistor value or numbers can be modified to adapt for different current or heat necessity.
  • The temperature rise depends on the ventilation around, heat flow, volume or teh shape of the vessel/holder, way of mounting the resistors to the panel etc. Needs experimentation.

I doubt I will find a 12V power adapter rated 9A.

The 9A rated LED drivers aren't uncommon. My personal choose is to go for a dedicated LED driver in this case. Will drive softener sets of LED strips together with three or four LED drivers. Others might have better solutions too.

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  • \$\begingroup\$ Thank you so much for providing a better solution than the one I had. I can see how the presence of an OpAmp produces a linearly controlled output. =) I didn't quite understand your third point, perhaps you could elaborate on that if that's not too much trouble. =) Also, I will look for a suitable LED driver to ease my work as you suggested. Thank a lot! \$\endgroup\$ – AweSIM Feb 7 '20 at 19:34

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