What should I pay attention to when making a boost circuit for a battery (without using a chip)?

sfcsdc

What should I pay attention to when making a boost circuit for a battery (without using a chip)? [Copy link]

 

I am currently working on a portable device that requires a 3V battery to be boosted to 9V to power the system.

The requirement is to build it yourself using discrete components and control it with MCU output PWM.

Since this is the first time I'm doing this, is there anything I should pay attention to?

The most important indicator is the battery life, so how can we achieve the highest efficiency?

topwon

It is recommended to find relevant design information about diode voltage doubler circuit

gmchen

If there are only voltage indicators but no current indicators, not only can we not do specific design, sometimes we can't even come up with a plan.

sfcsdc

topwon posted on 2019-6-6 09:15 It is recommended to find relevant design information on diode voltage doubler circuit

Can a voltage doubler circuit convert DC to DC?

maychang

sfcsdc posted on 2019-6-6 10:28 Can the voltage doubler circuit convert DC to DC?

“Can a voltage doubler circuit convert DC to DC?”

A voltage doubler circuit, also called a charge pump, can convert DC to DC.

However, without output current indicators, it is impossible to say what to pay attention to or what to do.

sfcsdc

Thanks for your reply. I am not asking you to help me design. I am just asking if there is anything I should pay attention to. For example, this question:

When I did simulation, I found that under ideal conditions (the inductor on-resistance is set to 0R), my 50% duty cycle can normally convert the 3.3V voltage to 6.6V

But in reality, the resistance of the inductor is not 0, and the battery has internal resistance.

When the on-resistance of the inductor is set to 2R, the output voltage is only about 3.3V.

My analysis is: due to the existence of inductor resistance and battery internal resistance, during the conduction period of the MOS tube, the current charging the inductor becomes smaller, so the energy storage of the inductor is less;

When the MOS tube is turned off, the output is the input voltage plus the voltage on the inductor. The voltage on the inductor is very small, so the output voltage is basically equal to the input voltage.

How can this problem be solved? In fact, the inductor must have resistance, and the battery also has internal resistance and limited discharge capacity.

captzs

Did you not pay attention to the tips on the 3rd floor? Now the battery capacity varies greatly. What battery to use? When it is increased to 9V, the load to be carried is 1A or 1mA. Only with indicators can we have a plan.

maychang

sfcsdc posted on 2019-6-6 10:43 Thanks for your reply. I am not asking you to help me design. I am just asking if there is anything I should pay attention to. For example, this question: When I was doing simulation, I found that the ideal...

Your analysis is roughly correct, because the inductor does not store enough energy, resulting in a decrease in output voltage.

The solution is to reduce the DC resistance of the inductor (the load R1 is only 10 ohms, the DC resistance of the inductor is 2 ohms, and the efficiency is too low); the second is to increase the duty cycle.

As for the internal resistance of the battery, there is nothing you can do about it.

sfcsdc

captzs posted on 2019-6-6 10:49 Didn't you pay attention to the tip on the 3rd floor? Now the battery capacity varies a lot, what battery to use; when it rises to 9V, the load to be carried is 1A or 1mA, only with indicators can you have...

Okay, let's use this lithium-ion battery.

Because it has a low self-discharge rate and can maintain a constant voltage until the battery is exhausted, but its disadvantage is that the current is small.

I would probably choose one with a capacity of 1Ah.

The output current does not need to be too large, 10mA~100mA is enough

maychang

sfcsdc posted on 2019-6-6 11:09 Well, I plan to use this lithium-ion battery because it has a low self-discharge rate and can maintain the voltage unchanged before the power is exhausted, but the disadvantage is...

To raise 3V to 9V, even with 100% efficiency, your battery needs to be able to output 300mA. 300mA is not a small load for such a battery.

maychang

sfcsdc posted on 2019-6-6 10:43 Thanks for your reply. I am not asking you to help me design. I am just asking if there is anything I should pay attention to. For example, this question: When I was doing simulation, I found that the ideal...

Your PWM signal is generated by a single chip microcomputer. From the simulation waveform, the frequency is only 5kHz. Therefore, the inductance is bound to be relatively large, and the DC resistance of the inductor is also large. If the PWM frequency is increased to above 25kHz or even above 50kHz, the inductance can be smaller, and the DC resistance of the inductor can also be reduced, so that the efficiency can be improved.

sfcsdc

maychang posted on 2019-6-6 11:20 Your PWM signal is generated by a single-chip microcomputer. From the simulation waveform, the frequency is only 5kHz. In this case, the inductance is bound to be relatively large, and the DC resistance of the inductor is also large...

I improved the circuit so that the voltage of the output capacitor is charged to a specified value before supplying power to the load.

The simulation uses an op amp for comparison. The actual circuit is to use the MCU's ADC to sample the output voltage and then open the load path after reaching the required value.

sfcsdc

If I do this, no matter how large the resistance of the inductor and the internal resistance of the battery are, I will just charge the output capacitor more slowly and keep charging until it reaches the value I need.

After reaching the corresponding value, power is supplied to the load.

maychang

sfcsdc posted on 2019-6-6 12:48 If I do this, no matter how large the resistance of the inductor and the internal resistance of the battery are, I will just charge the output capacitor more slowly and keep charging until I reach the value I need...

The problem is: after you start to supply power to the load, your Boost output voltage may drop. After the output voltage drops, do you still supply power to the load?

sfcsdc

maychang posted on 2019-6-6 12:57 The problem is: after you start to supply power to the load, your Boost output voltage may drop. After the output voltage drops, you still supply power to the load without...

Yes, look at the simulation picture:

The 10R resistor is the equivalent load.

After the output voltage VF2 reaches the specified value,

The comparator output voltage VF4 is continuously switched to maintain the output voltage at a specified value.

The VF2 in the simulation diagram remains basically unchanged.

topwon

1. When using lithium-ion batteries, you should pay attention to choosing power-type ones to support high current output. (In addition, lithium-ion batteries have passivation phenomenon, so pay attention when using them at low power consumption)

2. If the output is 100mA, I think it should be feasible to use a diode voltage doubler circuit. The circuit is also simple. The premise is to see what your requirements are for the output voltage accuracy, ripple, efficiency, etc.

3. If you use a switching circuit to boost the voltage, you should do feedback control, right?

sfcsdc

topwon posted on 2019-6-6 13:23 1. When using lithium-ion batteries, you should pay attention to choosing power-type ones to support high current output. (In addition, lithium-ion batteries have passivation phenomenon, so pay attention when using them at low power consumption) ...

It would be better if it could be simpler. In fact, after the voltage is boosted, it is only used for a short while and then turned off. It does not work all the time.

But I have never come across voltage doubling circuits before, and most of the ones I found online are about AC to AC conversion.

I don't know how to design the DC to DC converter.

Can you recommend some information?

topwon

sfcsdc posted on 2019-6-6 13:32It would be better if it were simpler. In fact, the voltage is turned off after a short time, and it does not work all the time. But I have never been exposed to voltage doubling circuits before, and there are many online...

The principle of DC to DC conversion is the same as the AC voltage doubling circuit. A switch is still required to control the on and off of the DC input voltage, which actually becomes an AC input.

The basic idea of the voltage doubling circuit is that the capacitor is charged during the positive cycle of the open tube operation, and during the negative cycle, the capacitor is isolated from the ground due to the unidirectional conductivity of the diode, and the charge is transferred to the next level. At the same time, since the power supply and the isolated capacitor are in series, the voltage is doubled.

btty038

The voltage alone is OK. There is only so much energy. There is loss in the conversion. Tesla is very good.

topwon

sfcsdc posted on 2019-6-6 13:32It would be better if it were simpler. In fact, the voltage is turned off after a short time, and it does not work all the time. But I have never been exposed to voltage doubling circuits before, and there are many online...

You can search for PWM drive voltage doubler circuit.

For example, this circuit, but you already have a PWM signal, so you don't need a 555 to generate an oscillation source.

This circuit can boost DC voltage. It is simple and practical. IC1 (NE555) is connected as a multivibrator with an oscillation frequency of about 8.5KHZ. The output square wave signal drives T1 and T2. D1, C3, D2 and C4 form a voltage doubler rectifier circuit. The voltage across C4 is close to twice the amplitude of the square wave voltage output by T1 and T2, which is about 20V in this circuit. The maximum output current of the circuit does not exceed 70mA. When the output current is 70mA, the output voltage is 18V, and the conversion efficiency is 32%.

If the voltage output requirement is high, the output end can be connected to a 78LXX system voltage regulator IC.