# Tag Info

20

There's a technique called a charge pump with which you can make a voltage doubler, but that will only give you 3V from a 1.5V cell, and even less from the 1.2V cell. I'm still mentioning it because several microcontrollers these days will work with voltages down to 2V. A charge pump can only supply limited current, enough to power the microcontroller, but ...

15

To make a higher power voltage from a battery like that takes a particular type of switching power supply called a "boost converter". This uses a inductor to make spurts of higher voltage. The concept is the same how a hammer makes spurts of much higher pressure than your arm can deliver to the nail directly. There are chips out there that integrate much ...

15

A motor driven by an H-bridge is also a boost converter. Here's an H-bridge: Replace the motor with an inductor, resistance, and voltage source (back-EMF): Let's just consider that we are driving the motor in one direction, and S3 is always open, and S4 is always closed: Rotate V1, S1, and D1 (same circuit): flip the whole thing left-for-right ...

12

Your boost is operating in discontinuous conduction mode or DCM (inductor current goes to zero each switching cycle). The duty cycle becomes a function of load as well as the duty cycle. If you increase the load, the inductor value, or switching frequency, you'll reach a point where you'll see your regulation where you expect it - this is called CCM, or ...

11

That's an average power of Power = 5 A $\times$ 10.5 V $\times$ 100 $\mu$s / 10 ms = 0.525 W. Average power is easy for almost any battery. You just need a store to accomodate the pulse. A capacitor that will "droop" say 0.5V in 100 $\mu$s needs to be C = I $\times$ t / V = 5 A $\times$ 100 $\mu$s / 0.5 V= 1000 $\mu$F. A supercap ...

10

This is the most efficient way I can think to do it. There's a MAX1682 charge pump to give you 6.6v at the super capacitor. The voltage doubler is pretty efficient, probably more then 90%, but they can't supply huge currents. But what's the average current? 5A * 100us / 10s = 0.05mA. That's well within the MAX1682's 45mA spec. From a brief look at the ...

8

Your inductor probably have high serial resistence. You might want to take lower value with thicker wire. Increase frequency if needed (but not too high, switching losses increases). Use ulta-low ESR caps, tantalum ones. Add some 0.1 + 22-66uF of ceramic cap in parralel. If frequency is high enough just leave ceramic. Implement synchronious rectification to ...

7

In general, in both boost and buck switching regulators, a higher switching frequency allows the use of physically smaller inductors and capacitors. However, a high switching frequency can also eat into the overall efficiency of the regulator, through switching losses both in the switch itself and in the gate-drive circuit. Yes, the diode contributes ...

7

Note: This solution can not provide 1 Ampere output while boosting from 5 Volts, as pointed out by @markrages. Also, the product mentioned is not suggested to be operated beyond a step-up ratio of 1:3, i.e. up to 15 Volts from a 5 Volt source. Hence, this answer does not meet the criteria specified in the original question. With the amended 12 Volt source, ...

6

Buck-boost converters are typically used when an input voltage source varies and may be either above or below the required output voltage at different times. One example would be a small wind turbine charging a battery bank, where the turbine voltage output is above the battery voltage in a high wind and below it in a very low wind. A very common example ...

6

Like Dave says your inductor can be smaller. A switcher first stores the supplied energy in the coil's magnetic field and reconverts that to an output current, a number of times per second. The higher the frequency, the shorter the period and the time for which it has to store the energy. Energy = power $\times$ time, so a 10 times higher frequency means ...

6

What Phil said 2. This is not the EMF you are looking for. One problem is in your equating the voltage with the back EMF. This is not back EMF - this is energy stored in the system "demanding to be given a new home. I say demanding" because the energy WILL be transferred elsewhere and it will be delivered at a rate that the system wishes it to happen at. ...

6

The nearest I can find is this device: - With a 0.5V input it will produce 5V but only with probably 50mA output (Green graph for TPS61202): - Hopefully it'll work for your application

6

Look at the spec for the device. There are two types; one switches at 1.6MHz and the other switches at 600kHz. Let's say, just to make life easier on folk reading this answer that it switches at 1MHz. How much energy can it charge the inductor with - again the spec has the answer - maximum duty cycle is (on average between the two devices) about 90%. For ...

5

Not really a whole off-the-shelf part, but... A small step-up converter with an autotransformer sounds just right (flyback converter without isolation). Also, you could try something like figure D1 on page 93 of Linear Tech's app'note 47 (without Q1 and the three resistors and the capacitor around it, just to the point that says "avalanche bias"). Adjust ...

5

The 31WOC is probably not the actual part number, rather some type of code described in the datasheet for different variants of the chip. I think I would forget about what the actual part number is, and just locate the feedback pin and adjust the resistor divider as necessary (assuming it's an adjustable regulator). Most use an internal reference of ...

5

With the components values that you have selected it is indeed more suitable to run with the 200kHz frequency. Even at 200kHz I find that a more suitable output capacitor may be more like 33 or 47uF. If you are using an ideal inductor with no equivalent series resistance specified then I would suggest that you try one of the realistic inductors from the ...

5

The traditional way of getting this for low charge levels is known as a capacitor diode bridge circuit AKA voltage multiplier. You'd use the batteries to generate a AC waveform which you then feed into the end of this multiplier circuit. The higher the AC voltage the fewer stages you need to get to 10KV. Be careful though, this circuit can store charge ...

5

How can I design a circuit that can be charged to around 10kV (somewhere between 5 and 20kV is fine) from two AA batteries (~3V)? The difficulty in this question is understanding some of the requirements so I'll address this first because without answers it's debatable that this can be answered properly. Firstly, is the load going to be applied once the ...

5

I think you'll have problems with your layout. C3/C4 MUST be closer to pin 1 (EDIT this should read pins8/9 not pin1). When I say closer I mean living on it! Ditto C7 - it needs to be camped right on pin 7. Now I've never used this part but this is standard procedure for this type of device. Think about the current pulses flowing from pin 7 to C7 and the ...

5

General Information and Current Flow Buck converter has two stages depending on the state of the switch. The switch, in your case, is in the IC and is in between pins 5 and 6. Let's draw the first stage where the switch is ON and D1 is reverse biased. As you can see, the loop is big, because the ground island limits where the current can flow. In order to ...

5

Seriously, this is well over your head if you have to ask here, especially if you think the only issue is "which component can I used?". 48 V at 5 A is 240 W. That is not a trivial amount of power. 90% efficient for a boost converter is not impossible, but you really have to know what you're doing and be aware of all the little tricks to not lose a ...

5

To charge one of your capacitors as specified requires an energy of: \require{cancel} \begin{align} W &= \frac{1}{2}CV^2 \\ &= \frac{1}{2}1500 \mu F (0.95 \cdot 400V)^2 \\ &= \frac {1}{2}0.001500 \frac{J}{\cancel{V^2}} 144400 \cancel{V^2} \\ &= 108.3J \end{align} You want to do this to two such capacitors, 16 times per second. The ...

4

I've also had problems with this circuit in LTspice. I don't think my problem was exactly the same as yours but this is the only decent result when searching for "ltspice boost converter" so I'll put my answer here. Here are the things I did wrong: I used the generic "nmos" model. It doesn't work. I don't know why but it seems like it has a really high ...

4

I went into AN19 http://cds.linear.com/docs/Application%20Note/an19fc.pdf and it has a section for Frequency Compensation on pin VC. I simply increased the capacitance to 1uF and kept the resistor on 1K and I no longer have this problem. I am going to talk to Linear Tech to better fine tune this circuit and see if there is a more optimal value for that ...

4

Yup, opamps will pick up some of the signal on their power and interpet that as input. This is exactly what the power supply rejection spec is there to quantify. A ideal opamp would have inifinite power supply reject, but actual opamps have rather less than that. Even with the power supply rejection spec, you can't take it at face value. It is very rare ...

4

If the circuit connections don't adhere to reasonable practise then no-amount of filtering will really help that much. Any connections between the microphone and the input of the amp must not share common points with supply rails. This means pin 2 MUST be directly connected to pin 4 and the bottom end of the pot must be connected to pin 4. Ditto the ...

4

Since Dave Tweed has pointed out the flaw in the other answer, I have basically copied my answer to Single transistor level up shifter ... Note also the interesting solution by Nicolas D in the question. I have a few solutions (some solutions provided by Microchip HERE): 1) Direct connection: If Voh (high-level output voltage) from your 3.3V logic is ...

3

My experience (across four different cellular modems, both GSM and CMDA), is that they are intended to be run directly off the battery. For the AirPrime SL808x series (as an example as I haven't worked with the SL6087), VCC_3V6 (which is used for the power amplifier), is rated at 3.3v min to 4.3v max, with a typical value of 3.6v -- this closely matches a ...

3

Strange. This is a single supply module, yet it has two switchers on board. Anyway. Yes, connect a voltage between 3.5V and 28V to the input and measure the output voltage, which you indeed should be able to adjust with the multiturn potmeter. Check the switchers' datasheets for the maximum output current. The right IC seems to be an LM25005, that can ...

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