RF Circuit Battery Efficiency

[kellys_eye]

You can still do that with MOSFETS - look up the 'h-bridge'.

 

[(*steve*)]

Sure, it's an H bridge. What is its quiescent current? Can you do better?

 

[mcasey]

I bought a new switch to replace the LH1505 to try out. I CAN'T figure out why the COM & NO pins connect when I apply power to the either/or. It doesn't matter what I do with the IN (digital control). What am I doing wrong?





http://www.ti.com/lit/ds/symlink/ts12a44514.pdf

 

[mcasey]

Got it! No digital controls can be floating.

 

[mcasey]

parts wise, its a very minimalistic circuit. So as such, I doubt there is much you can do and still have it work properly

the only thing you could look at would be to replace the linear regulator with a switching buck regulator
the linear reg. will be very inefficient at dropping 6V down to 3.3V

So I bought (what I think is) a more efficient regulator. However, I can't figure out how to wire it to get 3.3 volts. I do not have an inductor; is it necessary? Input is 5 volts. Please help.

http://www.ti.com/lit/ds/slvsbk5/slvsbk5.pdf

 

[BobK]

I do not have an inductor; is it necessary?

Uh, yes. It is a switch mode regulator and the inductor is a critical part of the circuit.

However, I can't figure out how to wire it to get 3.3 volts.

It is a fixed regulator, wire it exactly as in the datasheet, and you will get 3.3V.

Bob

 

[davenn]

So I bought (what I think is) a more efficient regulator. However, I can't figure out how to wire it to get 3.3 volts. I do not have an inductor; is it necessary? Input is 5 volts. Please help.

It is a fixed regulator, wire it exactly as in the datasheet, and you will get 3.3V.

as Bob said wire it up per the data sheet. you are missing a bunch of components......

2 resistors, 4 capacitors, a special diode, and the inductor ..... you MUST use all the correct parts and their values else it wont work

 

[kellys_eye]

wire it exactly as in the datasheet, and you will get 3.3V.

Bob

Indeed - the datasheet tells you everything you need to know about using the device - how to connect it, how to calculate the component values etc.

You can also 'cheat' and Google the device and search the 'image' results for examples of working circuits others have built. Indeed there are members on here who have attempted the very same:

https://www.electronicspoint.com/threads/1st-buck-converter-design-help-advice.280000/

 

[(*steve*)]

Also be aware that building it on a breadboard may cause some problems.

Fingers crossed.

 

[mcasey]

Ok, now I'm terrified! Although I'm sure this circuit seems very simple for most you on this forum, it's very difficult for me. Is there a more simple device (other than an inefficient linear regulator) that will get me down to 3.3 volts, or am I going to have to venture down this rabbit hole ?

 

[kellys_eye]

There are basic 3.3V modules available on eBay

http://www.ebay.co.uk/itm/DC-5V-to-...191447?hash=item1c57818997:g:oREAAOSw3ydVqeff

OK, they are rated at 5V input but a series diode from your 6V supply will fix that.

 

[davenn]

Is there a more simple device (other than an inefficient linear regulator) that will get me down to 3.3 volts,

why didn't you continue with the one you posted a datasheet for in post #3 ? I said it looked good
a much lower component count

 

[mcasey]

I haven't given up on it yet, but it is a little overwhelming for me. I need to order more components (like the inductor) and play around with it.

 

[mcasey]

I have decided to tackle this regulator, but need some help identifying the components needed... The items circled in GREEN are obvious to me. The items circled in RED have me confused.





http://www.ti.com/lit/ds/symlink/tps5403.pdf

My objective is to supply the RF receiver and decoder with 3.3 volts without the use of an inefficient linear regulator.

 

[(*steve*)]

Imagine you want 3.3V at 100mA. You have a battery which is currently reading as 6.6V.

A linear regulator would have as it's input 6.6V at 100mA, and the excess energy would be expended as heat. This regulator would be 50% efficient at best. As the difference between the input and output voltage changes, the efficiency varies.

A switch mode regulator in the same circumstance will input 50mA at 6.6V and output 3.3V at 100mA. It is (theoretically) 100% efficient.

There switch mode regulator electrical energy in a magnetic field, then converted that magnetic field to electrical energy. This is done in such a way as to change the output voltage in a controlled manner.

A linear regulator has an input current equal to the output current, dissipating the difference between the input and output voltage times the current as heat.

The switch mode regulator has an input power equal to the output power, so if the output voltage is lower than the input voltage, the input current will be lower than the output current. The input current times the input voltage equals the output current times the output voltage.

The inductor, diode, and capacitor in the top right corner of the circuit are three of the necessary four parts of a switch mode regulator (the 4th part is something which switches a current path on and off)

Without those three components on the top right corner, you can't regulate voltage with this chip.

Also remember that there are additional losses which I've not mentioned. Nothing is ever 100% efficient.

 

[davenn]

I have decided to tackle this regulator, but need some help identifying the components needed... The items circled in GREEN are obvious to me. The items circled in RED have me confused.

My objective is to supply the RF receiver and decoder with 3.3 volts without the use of an inefficient linear regulator.

the info is in the datasheet
here's one example from it for calculating the inductor value ....

Inductor Selection


The higher operating frequency allows the use of smaller inductor and capacitor values. A higher frequency

generally results in lower efficiency because of switching loss and MOSFET gate charge losses. In addition to

this basic trade-off, the effect of the inductor value on ripple current and low current operation must also be

considered. The ripple current depends on the inductor value. The inductor ripple current (iL) decreases with

higher inductance or higher frequency and increases with higher input voltage (VIN). Accepting larger values of iL

allows the use of low inductances, but results in higher output voltage ripple and greater core losses.

To calculate the value of the output inductor, use Equation 3. LIR is a coefficient that represents inductor peakto-


peak ripple to DC load current. It is recommended to set LIR to 0.1 ~ 0.3 for most applications

continued on pages 10 and 11

and following that is info on the capacitor values

 

[BobK]

A switch mode regulator in the same circumstance will input 50mA at 6.6V and output 3.3V at 100mA. It is (theoretically) 100% efficient

On average.

The typical buck converter topology has all the output current coming from the source during the on phase and none during the off phase. So the source would be outputting 100mA when the switch is on and 0 when the switch is off, for an average of 50mA.

Bob

 

[mcasey]

Imagine you want 3.3V at 100mA. You have a battery which is currently reading as 6.6V.

A linear regulator would have as it's input 6.6V at 100mA, and the excess energy would be expended as heat. This regulator would be 50% efficient at best. As the difference between the input and output voltage changes, the efficiency varies.

A switch mode regulator in the same circumstance will input 50mA at 6.6V and output 3.3V at 100mA. It is (theoretically) 100% efficient.

There switch mode regulator electrical energy in a magnetic field, then converted that magnetic field to electrical energy. This is done in such a way as to change the output voltage in a controlled manner.

A linear regulator has an input current equal to the output current, dissipating the difference between the input and output voltage times the current as heat.

The switch mode regulator has an input power equal to the output power, so if the output voltage is lower than the input voltage, the input current will be lower than the output current. The input current times the input voltage equals the output current times the output voltage.

The inductor, diode, and capacitor in the top right corner of the circuit are three of the necessary four parts of a switch mode regulator (the 4th part is something which switches a current path on and off)

Without those three components on the top right corner, you can't regulate voltage with this chip.

Also remember that there are additional losses which I've not mentioned. Nothing is ever 100% efficient.

Thank you for this explanation. It clarifies the main difference between the linear and switch mode regulators. Unfortunately (for me ), it doesn't clarify how to solve the equations listed on the datasheet.

 

[mcasey]

the info is in the datasheet
here's one example from it for calculating the inductor value ....




continued on pages 10 and 11

and following that is info on the capacitor values

I'm sure the datasheet explains in detail, how to calculate the inductor and all capacitor values. However, I do not have the skills required to decipher these formulas. I have no idea how to solve these formulas:



Furthermore,
I do not know what type of diode to buy:




I do not even have a basic understanding electronics. I just know my battery only lasts 26 days, and I want it to last longer, AS LONG AS POSSIBLE. I certainly appreciate all of the selfless help I have received on this forum, but I don't know what I don't know .

 

[mcasey]

I found this online. Nice and simple. Thoughts?

Model: TSR 1-2433

http://www.mouser.com/ds/2/687/tsr1-537631.pdf

I think I can handle this one