# Applying a voltage by DAC to a Feedback pin of a DC-DC controller

I want to design a variable output voltage in the following design: For having this variable voltage either I can use a potentiometer instead of R13 or I can apply a voltage by DAC to the feedback pin:

The controller is LM5155DSSR, and the DAC I want to use is MCP47CVD02-E/MF, which it is mentioned that is has selectable output resistor (1KOhm and 100KOhm).

Considering VOP (which is the feedback form the power stage) is 60 V then there will be 1.001v on the feedback pin:

My question is, if I want to apply a voltage from a DAC to the FB pin, considering there is already a voltage on the feedback node (1.001 V), will there be an issue? Specifically, I am concerned about the current sourcing or sinking on the FB pin and the output of the DAC.

The FB input will see the external circuitry like this:

simulate this circuit – Schematic created using CircuitLab

So as you can see, assuming the input impedance of the feedback pin is negligibly high compared to Rth and R11, when you apply an external Vdac voltage through a resistor most of the current will be drawn from or flow to the Vthfb through the Rth which is the parallel equivalent of output divider resistors.

So when the external voltage is applied, the FB pin voltage will change and the regulator will try to correct it to bring down/up to ~1 V again.

One thing I'd like to point out is the 0-Ohm R11. Note that this will completely short the output divider. If it goes high enough there's a risk of triggering the OVP.

You may want to have a look at my answer here about the importance (or role) of R11.

• Would you please let me know if the internal resistor of the DAC can play a role here? or it is not depend on this resistance value? for example I want to use MCP47CVD02-E/MF, which has a selectable output resistor (1KOhm and 100KOhm). Commented Jul 22 at 14:52
• @Andromeda it does. It's going to be in series with R11. I missed that detail in the question, sorry. Updating the answer now. Commented Jul 22 at 15:52

A most convenient way of adjusting the voltage digitally is to use a current source/sink DAC. Maxim (now Analog Devices) make a family of devices, such as the DS4424 which is controlled by I2C.

It’s also possible to use a voltage DAC. The disadvantage of this method is that the feedback node is typically a low value (e.g., 1.2V, 0.6V, or in your case 1.0V) meaning the DAC useful range is limited. An offset can be added with an op-amp to normalize the adjustment range.

If your goal is to make the regulator swing across its entire possible output range (I don't recommend this with a DC-DC by the way, it will misbehave outside a limited duty cycle range), here's a method to do that (simulate it here):

Here I've added an op-amp that 'takes over' the feedback loop. The output voltage will track Vdac, yielding an output range of 0 ~ 60V with the DAC range of 0 ~ 5V. You can adjust this by changing the feedback ratio. The op-amp doesn't need a supply any more than what it sees at its (+) input, so 5V would do for a modern rail-to-rail I/O type.

Here's a way to get a 0-60V output using a DAC input. The output is 60V for a DAC voltage of 0V, 30V for a DAC voltage of 1V and 0V for a DAC voltage of 2V. Note that the output voltage increases as the DAC voltage is reduced.

• Thaanks Carl, it is a very nice answer, would you please let me know what is the internal resistance of the V_DAC? considering my DAC will have an internal resistor, so I want to check if this value can have an effect? Commented Jul 22 at 14:54
• The V_DAC output resistance is a load to ground which should not be selected for normal operation (Table 5-4), thus would have no effect on the circuit I posted. Commented Jul 22 at 16:17
• Three issues with this. The value for feedback (249k) is quite high, which could lead to stability problems in the feedback loop. That high value also has a large tolerance influence. The DAC output resistance is another tolerance item since this would be on series with the 10k. Commented Jul 23 at 17:56

The 5515 will adjust its operation to maintain 1.001 V on the FB pin.

This means the FB pin 'looks like' a zero impedance when the 5515 is operating properly, as whatever current you feed into it, the voltage will stay at 1.001 V. You will need to check with its data sheet how much bias current that pin draws, I suspect it's not much.

If you can design the values of R11, R13 and R8 together with the DAC output resistance (let's call it Rdac), such that the output voltage is what you want when the DAC's output is connected to 1.001 V through R11, then all will work well.

You may need to think about the 5515 as a virtual ground op-amp, with the FB as its inverting input, and then choose R8 as the feedback resistor and R11 + Rdac the 'gain' resistor. Your change in output voltage will be R8/(R11+Rdac) times the change in DAC output voltage, with an offset depending on R13.

The proposed approach with the DAC is a thought in a right direction, except the DAC needs to be connected to the feedback network through a nonzero-Ohm resistor. R11 shouldn't be zero ohms.

I've used this approach with a DAC several times with different DC-DC topologies.

The same idea is described in the application note AN818 by Maxim (now Analog Devices).

$$\ V_{out} = V_{ref} \left( 1 + \cfrac {R_1} {R_2} \right) + \left( V_{ref} - V_{DAC} \right) \left( \cfrac {R_1} {R_3} \right) \$$

Carl's answer is the canonically correct way to do this - I was about to write that up but he beat me to it. The feedback pin becomes a "summing node" for the output voltage and the DAC voltage.

An alternative is to connect the DAC output directly to the low side of R13, instead of connecting R13 to ground. The DAC voltage then becomes an offset to V_FB.

Another method is to use a digital potentiometer instead of a DAC. The digital potentiometer can replace R13 directly. Digital potentiometers have a pretty wide tolerance, and lower resolution than a DAC, but it is definitely the simplest circuit to analyze.