# Boost Converter with Vin tracking

I'm trying to design a boost converter circuit that would allow me to have the VOUT tracking to the VIN of the boost.

Could I add some circuit between VIN and the FB pin so it could regulate the FB so VOUT tracks VIN? Any tips on how to do this?

• Is Vin a continuously slow-moving variable voltage that needs to be tracked, or does it get switched from one value to another value occasionally ? Is the offset of 4V the actual offset and is this fixed and not variable ? Jun 15 at 7:26

The regulator has an internal reference of +1.2V, and this is the voltage it attempts to obtain at its feedback pin FB, by raising or lowering the the output until that condition is met. The system can be modelled like this:

simulate this circuit – Schematic created using CircuitLab

Everything in the blue box is the entire boost circuit, including 1.2V reference, error amplifier, MOSFET, inductor, and everything else needed to produce $$\V_{OUT}\$$. The only thing you have control of is R1 and R2.

You can think of this entire setup as being no different from a non-inverting amplifier, implemented using an op-amp, except in this case the op-amp has ridiculously low output impedance, and uses magnetic sorcery to produce its output voltage.

The relationship between output and input, as you would expect, will be:

$$V_{OUT} = V_{REF} \left( 1 + \frac{R_2}{R_1} \right)$$

In this case, gain $$\1+\frac{R_2}{R_1}=10\$$, giving rise to the 12V output. We can ask the opamp (regulator) to produce exactly $$\V_{OUT} = V_{REF}\$$ by turning it into a voltage follower, $$\1+\frac{R_2}{R_1}=1\$$. To do that, we set $$\R_2=0\Omega\$$ and $$\R_1=\infty\Omega\$$:

simulate this circuit

This obtains the relationship:

$$V_{OUT} = V_{REF}$$

If we try to fool the op-amp into believing the output is 1.2V higher than it actually is, it will simply respond by lowering its output until $$\V_{FB}=+1.2V\$$, as before. We can do this by introducing an offset voltage, $$\V_{OFS}\$$ here:

simulate this circuit

The relationship between $$\V_{OUT}\$$ and $$\V_{OFS}\$$ is therefore:

$$V_{OUT} = V_{OFS} + V_{REF}$$

We can introduce whatever offset $$\V_{OFS}\$$ we like, and the op-amp will always try to make $$\V_{FB}=+1.2V\$$. For instance, if you want the output to be $$\V_{IN}=+4V\$$, you simply have to create the condition that when $$\V_{OUT}=+4V\$$, $$\V_{FB}=+1.2V\$$:

\begin{aligned} V_{OFS} &= V_{OUT} - V_{REF} \\ \\ &= +4V - 1.2V \\ \\ &= +2.8V \end{aligned}

simulate this circuit

Maybe you see already where this is going. All we need to do is produce an offset between OUT and FB which varies with $$\V_{IN}\$$. Because $$\V_{OFS}\$$ is not an absolute potential (with respect to ground), the potential $$\V_{FB}\$$ we apply to FB must be the difference between $$\V_{OUT}\$$ and our desired offset voltage $$\V_{OFS}\$$:

$$V_{FB} = V_{OUT} - V_{OFS}$$

The goal, then, is to take feedback not from a resistor divider (R1 and R2), as you would do if you required a fixed output voltage, but rather from some sub-system that will produce $$\V_{FB}\$$ according to that last expression. There must be a dozen ways to achieve this, but the one that springs to my mind is this:

simulate this circuit

By controlling current $$\I_1\$$, we control the voltage $$\V_{R3}\$$ across R3, and the potential at FB will be $$\V_{R3}\$$ volts lower than $$\V_{OUT}\$$:

\begin{aligned} V_{R3} &= I_1R_3 \\ \\ V_{FB} &= V_{OUT} - V_{R3} \\ \\ &= V_{OUT} - I_1R_3 \end{aligned}

That's some way to the expression we require, but we need a voltage-controlled current sink (under control of $$\V_{IN}\$$) that allows us to vary $$\I_1\$$. One way to do this is shown here, in the green box:

simulate this circuit

R4 and R5 divide $$\V_{IN}\$$ by 100, and OA2 and M1 place exactly that resulting voltage across R6. The current through R6, which becomes $$\I_1\$$ is then:

\begin{aligned} I_1 &= \frac{V_{IN}}{100 R_6} \\ \\ &= \frac{V_{IN}}{10000} \end{aligned}

This current passes via R3, generating the offset voltage $$\V_{OFS}\$$ we are trying to obtain. That voltage will be:

\begin{aligned} V_{OFS} &= I_1R_3 \\ \\ &= \frac{V_{IN}}{10000}R_3 \\ \\ &= \frac{V_{IN}}{10000}10000 \\ \\ &= V_{IN} \end{aligned}

This circuit produces the following $$\V_{OUT}\$$ as I sweep $$\V_{IN}\$$ from 0V to +20V:

The relationship is:

$$V_{OUT} = V_{IN} + V_{REF}$$

This doesn't account for the 4V that you wanted to add to $$\V_{IN}\$$, but that's easy to do with another op-amp or a zener diode.

I can't say how using a current sink like this will affect the control loop of the regulator, without studying the system (with the TPS61178 boost regulator instead of OA1) as a whole. You'll almost certainly need to provide loop compensation to prevent this setup from oscillating or ringing under certain loads, but that's another story.

What you are refering to is termed "envelope-tracking". To add "envelope-tracking" capability to the DC/DC step-up/down converter that you refer to may provide high efficiency to devices that require dynamic adjustment of power supply's voltage(s), such as used by audio amplifiers, RF amplifiers, or other types of circuits whose efficiency depend on the difference between power supply voltage that is present across the output device at variable signal levels (the more this difference is, the less efficiency.) So adding an envelope tracking signal to the FB pin of the TPS61178 device or similar devices, the output voltage can be changed and tracked in accordance with the envelope of the converter's input signal, thus dynamically adapting the DC/DC converter's supply voltage and optimizing the power consumption of the circuits that it supplies power to.

The addition of such "envelope-tracking" circuitry is usually quite a chalenge in design, for minimising slew/rates, offsets etc so as to provide a tracking output supply voltage with minimal delays and response times etc. Some are simpler but with limited accuracy, others are very complex and with high performance.

I suggest you do some reading on this "envelope-tracking" as it can vary in degree of complexity and this is not what you may want to dwell on. It would seem beyond the scope of this forum to provide you with "a circuit" that may not be suitable for you. You'll need to asses, and trade-off complexity for cost for ease of implementation vs accuracy. Take a look at TI's design application notes for the part in question for suitable ideas and adapt them to suit your application requirements.

Additional Note : PMP15036 Envelope-Tracking Power Supply Reference Design for Audio Power Amplifiers application note from TI is a good starting point. There is a proven enevelope-tracking circuit used in the feedback path of a TPS61178 for an audio PA. This may get you started and you may be able to adapt it to suit your application.

• Additional Note : PMP15036 Envelope-Tracking Power Supply Reference Design for Audio Power Amplifiers application note from TI is a good starting point. There is a proven enevelope-tracking circuit used in the feedback path of a TPS61178 for an audio PA. This may get you started and you may be able to adapt it to suit your application. Jun 19 at 9:07

The following circuit might be a useful place to start in designing a boost converter whose output voltage tracks the input voltage, but offset by a given amount.

simulate this circuit – Schematic created using CircuitLab

The ratio of R1 to R2 should be the ratio of the desired difference between Vout and Vin to the feedback voltage.

$$\frac{R1}{R2} = \frac{V_{out}-V_{in}}{V_{feedback}}$$

For example:

$$\frac{R1}{R2} = \frac{16V-12V}{1.25V} = 3.2$$

Assuming the base current is negligible, the current through R2 will equal the current through R1, and the voltage drop across R1 will be $$\\frac{R1}{R2}\$$ times the voltage drop across R2. So, when Vout correctly tracks Vin, Vfeedback will be at it's appropriate value.

Because the bases of Q1 and Q2 are tied together, we will assume the voltage at the emitter of Q1 is close to the voltage at the emitter of Q2.

Obviously, because we made some approximations, and because there will not be perfect matching of components values, this circuit may not provide the accuracy you require in the tracking between Vout and Vin. But depending upon your requirements, it may be adequate.

One issue that may need to be dealt with is reverse bias of the Vbe junctions of Q1 and Q2. Under normal operation, they will be forward biased. However, when your converter is starting up, Vin will appear before Vout. Similarly, when power is turned off, Vout may (or may not) fall before Vout. (Things like charge stored in output capacitors, load resistance etc will affect whether this occurs.) Typical BJT transistors will not tolerate a reverse Vbe much more than 5V, so care needs to be taken to avoid such an occurence.

One approach is to use the input voltage as the reference voltage for the regulator.

Part 1, the reference. Let's say the regulator normally has a 12 V input and a 6 V reference. Connect the input to the reference input through a 6 V zener diode, with a pull-down resistor to GND. Now, when the input increases to 14 V, the reference input will increase to 8 V. The input is subtracted down to the reference voltage, rather than divided down; the reference is a constant 6 V below the input.

Part 2, the output. Do essentially the same thing at the feedback (FB) input. Using another zener, subtract the output down to the feedback input. For example, with a 10 V zener diode between the Vout and FB, the output will be regulated to be 10 V greater than the reference. In this example, when the input increases from 6 v to 8 V, the output increases from 16 V to 18 V.

Two things:

1. This circuit will need compensation to prevent oscillation. Or more accurately, will need modifications to the compensation circuit the switching boost converter controller already needs.

2. The TI part you linked will not work for this, because its reference voltage is derived internally, with no way to influence it with external components. This is true for the vast majority of boost converter chips, both controllers and ones with the power switch integrated into the chip.