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I'm working on a design which needs a 9V supply. This is currently provided by a 9V battery, but I'm looking to change to a lithium battery (3.3 to 4.7V).

This supply has a pulsed load of up to 2A, with a pulse width of 10 to 300 us pulsing at a frequency of 2 to 200 Hz. The maximum average current is only perhaps 40 mA or much lower, but it's coming in huge spikes.

I don't have much experience with boost converters. I've been considering the TI LMR62014, which switches at 1.6MHz.

My first thought is to use a resistor and a large capacitor to smooth the average current through the converter.

schematic

simulate this circuit – Schematic created using CircuitLab

Is this the right approach? Is there any glaring problems with it?

My other thoughts would be to implement a constant-current device (which would probably increase the part count quite a lot), but I'm open to any thoughts or suggestions.

Thanks!

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  • \$\begingroup\$ Have you measured the actual load voltage during your pulses? I suggest you do so. \$\endgroup\$ – WhatRoughBeast Nov 19 '15 at 18:59
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Using a large smoothing capacitor like C1, isolated from the boost output as with R1, may be the only way to use the LMR62014 for this. The LMR62014 has a switch current limit of about \$2A\$, which means that for \$L_o\$ of \$10 \mu H\$, \$V_{\text{in}}\$ of \$3.3 V\$, and \$f_s\$ of \$1.6 {\text{MHz}}\$, the most current delivered to a \$9 V\$ output will be ~ \$0.65A\$. So, it can't support high current pulses on its own. This may be a blessing, as there would be no need to try and get much bandwidth out of the control loop. Just let the bulk capacitance do the job.

Of course, C1 of \$470 \mu F\$ may not be enough depending on how much droop you can stand during the pulse. Using the equation provided by ThePhoton, shows that maximum expected droop for a \$300 \mu sec\$, \$2 A\$ pulse would be about \$1.3V\$, for a part with negligible esr.

It will be required to have some kind of load at the output of the boost at all times, otherwise \$V_{\text{out}}\$ will rise until something limits it by breaking down. A load of \$40 {\text{mA}}\$ (\$R_o\$ of \$225 \Omega\$) would keep operation in CCM if \$L_o\$ = \$10 \mu H\$ and \$V_{\text{in}}\$ < \$4.7V\$.

For best dynamic performance it's undesirable to change conduction modes, which will mean in this case stay in CCM. The CCM/DCM boundary is defined as:

\$L_{\text{crit}}\$ = \$\frac{(\text{dc}-1)^2 \text{dc } R_o}{2 f_s}\$ ; Where \$\text{dc}\$ = \$\frac{V_{\text{out}}-V_{\text{in}}}{V_{\text{out}}}\$

A good reference is "Understanding Boost Power Stages". If you're going to be designing a boost, that's a good place to start.

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For modeling purposes, I'd model the load with a current source. Or possibly a resistive load representing the load when the pulse is not occuring, plus a pulsed current source in parallel to represent the load's transient current demands.

It will help you out to know the rise and fall times of the pulse when the current demand increase.

Your approach of adding a large output capacitance on the power source is the usual way of dealing with switching loads. However you do need to be careful of a few things.

To really know how your boost converter will react, you'll need to know how its control loop is designed. With 1.6 MHz switching, it's typical to design a control loop with ~200 kHz bandwidth, so after about 5 us of increased demand the boost converter should be able to respond with increased duty cycle and provide the additional needed current.

However your added output capacitance could affect the response. So it would be a good idea to model the control loop and be sure it remains stable with good phase margin after adding your output capacitor. I am not as sure about boost converters, but in buck converters, the output capacitor ESR has important contribution to the loop response, so be sure to include that in your model.

It's also possible there will be some ring in the load response. So if you need a really clean response, you should model the converter's control loop and simulate the transient response.

If the converter response time is indeed about 5 us, then considering

$$V = Q/C = I\cdot{}t/C$$

for a 2 A transient from a 470 uF capacitor for 5 us, you can expect about 20 mV voltage ripple in response to the transient. If this is not acceptable then you will need to consider even higher capacitor values.

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  • \$\begingroup\$ Remember that a boost converter has a right half plane zero, so it's very difficult to get high converter bandwidth (like 200kHz). A buck doesn't have that limitations so 200kHz would be achievable there. \$\endgroup\$ – John D Nov 19 '15 at 18:10

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