Engineer It: Can't Take the Heat? Share the Current
In this training video, TI's Aaron Paxton demonstrates how to source 5-A or more with Current Sharing Linear Regulators and discusses the benefits of using two low-dropout linear regulators versus a single LDO to source 6-A.
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Hey there, Aaron Paxton here, from the Linear Power Group at Texas Instruments. Here to talk to you about how to source 5 Amps or more with current-sharing linear regulators. Typically when we think about DSPs or FPGAs we know that the power requirements are quite hefty.
And in order to meet those power requirements, when we are designing our power scheme, we typically use switch-mode power supplies. Now, the reason for this is obvious. Switch-mode regulators are very efficient.
As a result of that, there's not going to be a whole lot of power dissipated across the IC itself and, consequently, not that much heat generated. However, there are drawbacks to using a switch-mode power supply. For one, we know that there is a certain amount of ripple that will be induced on the output voltage of the regulator.
And this ripple, if unimpeded, has the potential to cause adverse effects on the system. So typically, designers like to try to filter this out as much as possible. Now, we can do this with capacitors and inductors.
But this can be cumbersome. It can be difficult. And quite frankly, it can be bulky depending on which frequencies we're targeting. But another elegant solution that we have is to use a linear regulator as a sub regulator in order to not only reject that ripple, but to provide a clean DC regulated output.
But you might be thinking, how exactly do we source 5 Amps or more with a linear regulator? We know that linear regulators are generally lossy. So let's take a look at that equation.
We can see here that the simplified power equation for a linear regulator is going to be equal to the Voltage drop across the linear regulator itself-- V in minus V out-- multiplied by the current being sourced by that linear regulator. And consequently, we know that with power loss comes some heat generation. Here's one of several equations that you can use to find out how much heat is being generated.
In this case, I'm looking at the heat transfer to the board as a result of power dissipation. And it says that heat transfer is equal to this R theta JB value, which is a value that is unique to each LDO. Multiply that against the loss of power incurred across the linear regulator, and we get the amount that is transferred in terms of heat-- degree C-- to the board.
So why exactly does heat matter? Well, we know that if a linear regulator is heating up on a board in a single spot and it's quite a bit of heat, it's not just heating up itself. It could heat up surrounding components. And that can degrade performance.
The other thing that we need to consider is that as the linear regulator heats up, it's capable of going into thermal shutdown and stop regulating as a result. We typically want to avoid those two things. But how exactly do we go about doing that? Let's take a look at the power equation again.
We know that there's two terms here-- the first term being the voltage drop across the LDO. If we reduce V in minus V out, bring V in closer to V out, we can reduce the amount of power loss. However, for this example, let's say that that's not possible. The other thing we could look at is I out, or the current being sourced by the regulator.
We typically can't reduce this either. Because we know that the DSP or FPGA requires a certain amount of current. And we want to meet those demands.
However, one thing that we can do is use two linear regulators operating in parallel. That changes the power loss equation to the one you see at the bottom here. P loss is now equal V in minus V out times I out over 2. That's because when two linear regulators are operating in parallel, you're splitting the current being sourced by each one in half.
So how exactly do we go about accomplishing this? Well, this is the ideal scenario. We see that there is two LDOs here. And they have their inputs tied, their outputs tied, as well as the feedback nodes, are all tied together.
Now, I mentioned that this is ideal. Because typically, linear regulators do not have the same output voltage. When they have the same exact output voltage, there is no problem. Current will be split 50-50.
However, there are usually subtle but slight differences in the output voltage between two regulators. When this happens, one of the LDOs becomes dominant. And the other is non-dominant.
As a way of trying to compensate for being a non-dominant regulator, this LDO will typically close it's pass device. As a result, the current distribution between the two regulators is not going to be equal anymore. And it's not going to be 50-50. It's going to be something closer to 20-80, 90-10, or even worse.
So how do we go about actually implementing it in a way that we get that equal distribution of current? Well, what we need to do is actually add a control loop. And that's what we're going to do here.
You can see that there are two sense resistors right here. These are a means of measuring the current floating through each regulator. And you can see that these are the inputs to the op amp. The op amp measures the differential voltage that it sees at these nodes, and will correspondingly drive one of the LDOs at its feedback node.
We see that this LDO on the bottom is the master LDO. And that means that it's stagnant. Its output voltage is fixed.
And that is what the other LDO is being compared to. This op amp will continue to drive until it sees equal current distribution through each of the linear regulators. So now that we've talked a little bit about the theory, let's go through an example, a real-life example, and how it actually looks.
Right here I have a thermal camera to measure how much heat is being generated as a result of regulation. I have a digital multimeter. I have a programmable DC power supply.
I have an additional DC power supply. And then I have a load box because we're going to be sourcing quite a bit of current. And lastly, this is the actual board that exhibits current-sharing functionality.
You can see that there's two LDOs on here. One LDO is the master LDO. And the other is the slave LDO.
So the way I have it currently set up is one of these LDOs is sourcing 3 amps by itself. That means that the master is enabled and the slave is disabled. You can see here that we're only going to enable one LDO, the master LDO, and try to source 3 Amps through it.
And when we do that, what we see is we get a temperature rise up to 68 degrees Celsius. Now what we're going to do is we're going to enable the second LDO. We will put the board into current-sharing configuration.
You can already see that with the second LDO enabled, that the heat is starting to spread not just at one LDO, but an additional LDO. You can also see that the max temperature on the board measured is only going to be 54 degrees, which is a significant reduction from what we saw before. So this is showing that current is actually being distributed equally between two LDOs.
But I mentioned before that these are 3-Amp LDOs each. So together they're capable of sourcing up to 6 Amps. So let's do that here.
It's important to point out a few things on this board. As I mentioned before, there are two LDOs on this board. But there are also two sense resistors.
These resisters are responsible for sensing the current that is flowing through each LDO. But in addition to those sense resistors, there is also an op amp that we talked about that is part of the control loop. That's featured right here.
And it's actually sensing the voltage differential from those two sense resistors. Lastly, there is a large input capacitor that we put here. The reason that it is so large is because we're dealing with quite a bit of inrush current at 6 Amps. In order to avoid the LDOs going into Drop Out during startup, the input capacitor will allow the input to remain stable.
Now that we have the LDOs operating up to 6 Amps, let's take a look at the thermal camera again. Now that we've increased the load current to 6 Amps, we see that both LDOs are still operating in current-sharing operation. And they're each sourcing 3 Amps each. We also see that the hottest temperature recorded on the board has now reached 65 degrees Celsius and is rising.
What we've seen today is that linear regulators are great as part of a filtering strategy for switch-mode power supplies. And although we're sourcing up to 6 Amps, this doesn't necessarily rule out linear regulators as a viable solution. For more information and technical conversation on current-sharing, please go to the following links. Thanks for watching.
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This video is part of a series
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Engineer it - Linear & low drop-out regulators (LDOs)
video-playlist (6 videos)