0-10V adjustable power from 12V source

[alexchannell]

Needs to be solid state and can do 10W (or 1 amp at any voltage).

I was looking at DC-DC converters but they all seem to not be able to do less than 1.25 volts. If I can get one that could go down to .5V that would probably be good enough. Any ideas?

 

[3v0]

Needs to be solid state and can do 10W (or 1 amp at any voltage).

I was looking at DC-DC converters but they all seem to not be able to do less than 1.25 volts. If I can get one that could go down to .5V that would probably be good enough. Any ideas?

You may be able to find a LDO (Low drop out) linear adjustable linear regulator that can do it.

 

[alexchannell]

If I'm not mistaken, the LDO part only specifies that the output voltage can be close to the input voltage, not that it can output very low voltages.

 

[dknguyen]

If I'm not mistaken, the LDO part only specifies that the output voltage can be close to the input voltage, not that it can output very low voltages.

An LDO just allows your Vin be a lot closer to your Vout (and still do it's job) than a regular linear regulator. THe input voltage doesn't need to be close to the output. It is not a requirement for proper operation.

However, you didn't specify your Vin (or your maximum Vout for that matter). A regular linear regulator would also work if your Vin is above 10V+Vdropout.

Your main problem would be heat. Heat dissipation in a linear regulator increases as current increases (no need to explain why that is right?) and as the difference between input voltage and output voltage increase (since it's burning off the extra voltage as heat, and more extra voltage means more heat).

A DC-DC converter's minimum output voltage is limited by the minimum on-time (similar to it's resolution). Below this minimum-on time, it would just be 0V pretty much. THe higher the input voltage the smaller this minimum on-time must be to achieve the same voltage. There would be a step between 0V and the minimum non-zero voltage...it's mainly an issue of the DC-DC converter's resolution (for lack of a better term). Unless you're going to be powering the converter from 1.25V...the minimum operating voltage of the converter is not a problem. The minimum operating voltage is for the input, not the output. Output is limited by minimum on-time/duty cycle (and input voltage).

 

[alexchannell]

Well a linear regulator can output any voltage that is less than Vin-Vdropout. Heat dissipation does increase though as the difference between Vin and Vout get larger (and as current gets larger too). If you can find a way to dissipate the heat more effectively, you can do it.

However, you didn't specify your Vin. A regular linear regulator would also work if your Vin is above 10V+Vdropout.

An LDO just allows your Vin be a lot closer to your Vout (and still do it's job) than a regular linear regulator. THe input voltage doesn't need to be close to the output.

Yep, getting the regulation below 1.25V seems to be the issue (dropout of all regulators I can find) Vin will be 12-15V, Vout 0-10 (if reasonably possible)

 

[alexchannell]

I don't think you are understanding the concept of dropout voltage...

Thats entirely possible
My understanding is that it is the difference between the input voltage, and the closest output voltage you can have. If you have in input of 10V, a dropout of 1.5, the highest regulated voltage you can have is 8.5V. That is incorrect?

 

[dknguyen]

Nah, you understand it just fine. I just couldn't figure out why you thought a a linear regulator or converter couldn't go down to near 0V so I thought maybe that was the case. But it wasn't and that post was deleted.

I see your problem now. It's because the 1.25V is being used as a reference for linear regulators and DC-DC converters and the resistive divider feedback normally used only lets you step down the the output voltage for comparison to the 1.25V. But you need to "step-up" the output voltage and compare it to the 1.25V in order to output a voltage less than that. The problem is getting around that 1.25V band gap reference.

Here's how to do it...
1. Use an power op-amp (due to your high power/currents) wired as a voltage follower (aka buffer).
2. Have a variable voltage divider (potentiometer) output feed into the input of the op-amp voltage follower
3. Have a linear regulator feed into the input of the divider (and best to also use it to power the op-amp).

Really...any input voltage for the divider (ie. switching, linear, or a voltage reference IC) will work as long as it's regulated because the divider is stepping down the input voltage and telling the op-amp voltage follower to output that voltage. If the base voltage that is being stepped down is inaccurate...well the output voltage will be too). The purpose of the op-amp voltage follower/buffer is for the current capability that you require since a voltage divider cannot output variable currents without also fluctuating it's output voltage.

You will want a rail-rail input and output op-amp in order to get very close to 0V. If you can't find a rail-rail power op-amp you might have to use bipolar supplies so the op-amp can output 0V (since with bipolar supplies, 0V is now in the middle of the range instead of the extreme).

 

[alexchannell]

Ohhhh....I see your problem. It's because the 1.25V is being used as a reference for linear regulators and DC-DC converters and the resistive divider feedback normally used only lets you step down the the output voltage for comparison to the 1.25V. But you need to "step-up" the output voltage and compare it to the 1.25V in order to output a voltage less than that.

Could you just use an op-amp?

Maybe, unfortunately I don't have a great deal of knowledge about op-amps, basically none really... time for some reading.

 

[dknguyen]

Maybe, unfortunately I don't have a great deal of knowledge about op-amps, basically none really... time for some reading.

Ahhh, you're replying too quick. I edit my posts for like 10 minutes after I type them. See my previous post again.

 

[alexchannell]

Nah, you understand it just fine. I just couldn't figure out why you thought a a linear regulator or converter couldn't go down to near 0V so I thought maybe that was the case. But it wasn't and that post was deleted.

I see your problem now. It's because the 1.25V is being used as a reference for linear regulators and DC-DC converters and the resistive divider feedback normally used only lets you step down the the output voltage for comparison to the 1.25V. But you need to "step-up" the output voltage and compare it to the 1.25V in order to output a voltage less than that. The problem is getting around that 1.25V band gap reference.

Here's how to do it...
1. Use an power op-amp (due to your high power/currents) wired as a voltage follower (aka buffer).
2. Have a variable voltage divider (potentiometer) output feed into the input of the op-amp voltage follower
3. Have a linear regulator feed into the input of the divider (and best to also use it to power the op-amp).

Really...any input voltage for the divider (ie. switching, linear, or a voltage reference IC) will work as long as it's regulated because the divider is stepping down the input voltage and telling the op-amp voltage follower to output that voltage. If the base voltage that is being stepped down is inaccurate...well the output voltage will be too). The purpose of the op-amp voltage follower/buffer is for the current capability that you require since a voltage divider cannot output variable currents without also fluctuating it's output voltage.

You will want a rail-rail input and output op-amp in order to get very close to 0V. If you can't find a rail-rail power op-amp you might have to use bipolar supplies so the op-amp can output 0V (since with bipolar supplies, 0V is now in the middle of the range instead of the extreme).

This looks like it would work great, except wouldn't this blow most op-amps due to heat? Based on what I looked up, the excess energy is turned into heat just like a linear voltage regulator. If I have a regulated 12V supply going to the op amp, and have a 2 volt input on Vin... I will have an output at 2V (and I'll be using 1 amp), and the op-amp will have to dissipate 10W in the form of heat?

 

[dknguyen]

This looks like it would work great, except wouldn't this blow most op-amps due to heat? Based on what I looked up, the excess energy is turned into heat just like a linear voltage regulator. If I have a regulated 12V supply going to the op amp, and have a 2 volt input on Vin... I will have an output at 2V (and I'll be using 1 amp), and the op-amp will have to dissipate 10W in the form of heat?

Exactly, that's why you need a power-op amp and a good heatsink...the best ones I know of for PCB components are 15C/W for D2Pak ICs. You might need to figure out some cooling or something, or go use a buck converter and some other circuitry to get around the minimum voltage limitation caused by the internal bandgap reference and the voltage divider in the feedback loop.

 

[kchriste]

Another solution is to use a LM317 (Or LDO if the Vin-Vout is too small for the 317) but reference it to a regulated negative supply rail. Here is an example of this from the datasheet:

There are probably better ways of supplying the -6.9V or whatever.

 

[dknguyen]

Another solution is to use a LM317 but reference it to a negative 1.25V supply rail. Here is an example of this from the datasheet:

There are probably better ways of supplying the -1.25V.

I take it using a negative rail for the feedback resistive divider might also work for a switching converter? If that's the case, things could be made to generate much less heat. I'm just not sure if the feedback loops used in those get unstable if the output voltage is below the internal reference voltage inside the converter IC.

 

[Hero999]

You'll struggle to get 1A at 10V with 12V in with an LM317.

Besides where are you going to get the negative rail from?

Does the input have to be 12VDC? Is and AC supply available?

If so it would be easy.

 

[alexchannell]

You'll struggle to get 1A at 10V with 12V in with an LM317.

Besides where are you going to get the negative rail from?

Does the input have to be 12VDC? Is and AC supply available?

If so it would be easy.

Unfortunatly, no. It's on a motorcycle so I have a 12-14.4V DC source. (going off the stator is not a good idea for reliability due to varying output - I think).

 

[Roff]

Have you seen LT3085? It is adjustable to zero volts. You will still need to get rid of a lot of heat if you want 1A at all voltages.

 

[Hero999]

This will work but you'll need to supply the op-amp with a higher supply voltage; a voltage double such as a 555 circuit can do this.

You can use a p-channel MOSFET if you swap the + and - inputs to the op-amp but you'll need to add a 100nF capacitor from Vout to the -input, the dropout will be higher and it won't be so stable.

For 0V output, you also need to select an op-amp with inputs that work down to 0V, unless you want to add a negative supply.

This isn't current limited, current limiting can be added at the expense of a higher dropout or you can just add a fuse or PTC resistor (polyfuse/switch).

 

[GonzoEngineer]

You can do it with a Philbrick Operational Amplifier, Model K2-W!