how to boost input voltage

I want to display a value on LCD and the circuit to be operated with AA battery.

I have a battery with a input voltage of 1.8V.

I used MAX1724 to boost the voltage from 1.8V to +5V(for other circuits). Quiescent current of MAX1724 is 1.5uA.

I have loaded a circuit with the current of 0.5mA.

I have connected a multimeter at the input side and output side of MAX1724.
Multimeter shows 1.5mA at input and 0.5 mA at output.

But I want a converter with low operating current when a circuit is loaded. Which IC can i use?
Please help. Thanks in advance.

Are you asking about lower quiscent current SMPS IC? I think ~1uA of quiscent current is required for any SMPS IC for its proper operation.

If you want still lower quiscent current then you have design your own SMPS circuit in discretes using the comparator, volt refernce, opamp and FET. It can bring down the quiscent current to about 500nA.

But if your main goal is to increase the battery life, then you have to look at options on reducing the load current. That's the one which is consuming more than 99% of the battery power, right?

I have connected a small portion of the circuit to MAX1724 i.e., a small circuit which consumes 0.5mA. I have measured the total operating current by multimeter, it shows 1.5mA.

Why is it so?

That's the basic operation of any system. The input power will be same as output power (ideal condition). that is Vin X Iin = Vout X Iout. Here with the SMPS efficiency comes into picture. The output power will usually be lower than the input power and hence efficiency will be less than 100%.

To have higher efficiency, check on the efficiency vs load current graph in DS and design accordingly.

I want my battery to withstand for more number of hours.

As with the earlier data,Vin X Iin = Vout X Iout... 1.8V x 1.5mA = 5V x 0.5mA

Efficiency = 92.5% . Am i right?

But the battery will drain faster right?

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Also, I have 6V (Two 3V batteries in series) and 3V(Two 3V batteries in parallel).

I want the output voltage of 3.3V for powering uc, sensor circuitry.

Do i have to use step up/step down for better efficiency and long battery life? my circuit total operating current is 2mA.

Yes, thats approximately correct for the system.

The battery life will depend upon the load current and the battery characteristics like chemistry, mAh rating. Battery characteristics vary with temp and load current. If you are using AA batteries, there are alkaline and lithium batteries available. May be some more chemical batteries if you search. Here are two batteries with different chemistries and varying battery characteristics.
LR6XWA : **broken link removed**
L91 : https://data.energizer.com/PDFs/l91.pdf
You can select the one which is more suitable to your application. Critical thing in battery selection will be looking into the typical discharge characteristics at particular load current. This will be defining the battery life.

If you know the load current and the required battery life, then calculate the mAh requirement from this and then select the battery type and size - and combination of series and parallel for the batteries if required.

Batteries in series will be used for increasing the voltage and paralleling will be used to improve the current delivery rating. But paralleling of batteries has to be done with caution - a faster discharge of one battery can cause rapid discharge of another and it becomes a chain reaction. I would choose connecting the batteries in series if I am using multiple batteries. Then use a buck converter as required.

Where did you find an AA battery that produces 1.8V? All the AA alkaline batteries I have used are 1.6V when brand new and the voltage slowly discharges to 1V then discharges faster to almost zero.
Oh, I remember that I tried some Eveready Lithium AA batteries one time and they were 1.8V when brand new. But they did not last any longer than Energizer AA alkaline batteries.

Thanks for the information.

I finally end up with putting the batteries in Series. @Audioguru I thought it could be around 1.8V but im not precise.

I had a search on DC DC converter IC. I found LTC3105 will be ok. Or do you suggest anyother IC?

I think sticking to Maxim IC MAX1724 would be better as it has got more than 10X better quiscent current rating both in active and standby modes. This Linear IC is better suited for photovoltaic applications, so may not be a good choice here.

In my opinion, it is better to focus on selecting the most suitable battery. If customer is going to replace the batteries on their own, then it will be better to give them a list of most suitable batteries. What do you say?

This Linear IC is better suited for photovoltaic applications, so may not be a good choice here.

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So can you suggest any IC?

I didn't get your point. Are you looking for anything else which MAX1724 is not having? What are the criteria for which you dont want to use MAX1724? If I know that, then I can suggest you some options.

Sorry I didn't read your post properly..

I have used MAX1724EZK33+T which is 3.3V output voltage. and connected a small circuit which consumes 0.6mA.

Measurements are taken:
Input voltage applied: 2.840V
Output voltage measured at MAX1724: 3.338V
Current measured after MAX1724: 0.68mA
Input current measured: 1.19mA.

So the efficiency is 67.1%??? is it right?

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But in the datasheet, for the graph efficiency vs load current, the efficiency is around 80%.

Why do i get lower efficiency. Please help.

When i increase the input voltage to 4.5V. The output voltage is not 3.3V. It is 4.035V.

Why it is happening like this?

Why it is happening like this?

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Because MAX1724 is a boost converter rather than a buck-boost converter.

You have also asked why you observed an efficiency that is lower than the value specified in the datasheet (about 80%). Correct measurement presumed, we have to assume less-than-optimal circuit parameters, e.g. unsuitable boost inductor, insufficient bypassing.

In the datasheet, for the input voltage varies from 0.8V to 5.5V, the output voltage is 3.3V for MAX1724EZK33.

Please have a look into it.

I guess, you're referring to the title page picture. It's erronoeusly mixing the general IC input voltage range with the output specification of the 3.3V type. Should be Vin = 0.8V to Vout

Just read beyond the title page. Figure 7 and 8 are giving the correct specification and the functional diagram shows the reason.

thanks FvM for pointing it out.

can you specify any DC DC converter? for the max input voltage of 5V.

I have used ADP2503 buck-boost in a higher current range, it can be operated with acceptable efficiency down to 1 mA. But there may be better suited devices from other vendors.

LTC3531, LTC3129, TPS63001, TPS63021 will be suitable for your application. There will be more SMPS from other makes also.