CD4047 Inverter, solar powered

Thread Starter

Summit23

Joined Oct 21, 2021
7
I am building this solar powered inverter: https://ibb.co/58m5L8B using a 12V, 0.8 Ah battery as the input source and a CD4047 IC for the inverter. However, the resistors connected to pin 2 are rated at 0.5W and the IC has a maximum power dissipation of 500 mW. I want to keep the efficiency of the system as high as possible. Would I need to limit the input current to the CD4047 IC?
 

MrSalts

Joined Apr 2, 2020
730
I am building this solar powered inverter: https://ibb.co/58m5L8B using a 12V, 0.8 Ah battery as the input source and a CD4047 IC for the inverter. However, the resistors connected to pin 2 are rated at 0.5W and the IC has a maximum power dissipation of 500 mW. I want to keep the efficiency of the system as high as possible. Would I need to limit the input current to the CD4047 IC?
The cd4047 controls the MOSFETs, there is little current passed through the cd4047. Essentially all the current just flows through the Mosfet. I haven't looked at the circuit on details. It seems very simple. I hope it works out for you.
 

Papabravo

Joined Feb 24, 2006
17,296
The thing to be concerned about is the available power from the +12VDC source and the time of the rising and falling edges. Your schematic is incomplete because it does not have values for the resistors and the capacitors. We don't know what frequency you will be operating at. Keep in mind that a power transformer designed for operation at 50 Hz. won't be very happy if you run it at a higher frequency. If you have sloppy edges on the gate drive the MOSFETs may spend too much time in the linear region, get hot, and fail. As with all power conversion schemes the power out will be less than the power in, sometimes it will be much less. You should have at least a rough idea of power in and power out. Flying blind with power electronics can be dangerous and/or fatal.

One more thing. The output is not likely to be a sinewave.
 

Thread Starter

Summit23

Joined Oct 21, 2021
7
The thing to be concerned about is the available power from the +12VDC source and the time of the rising and falling edges. Your schematic is incomplete because it does not have values for the resistors and the capacitors. We don't know what frequency you will be operating at. Keep in mind that a power transformer designed for operation at 50 Hz. won't be very happy if you run it at a higher frequency. If you have sloppy edges on the gate drive the MOSFETs may spend too much time in the linear region, get hot, and fail. As with all power conversion schemes the power out will be less than the power in, sometimes it will be much less. You should have at least a rough idea of power in and power out. Flying blind with power electronics can be dangerous and/or fatal.

One more thing. The output is not likely to be a sinewave.
It's for a project. I am planning on having the circuit schematic completed by the end of today. As soon as I do, I will post a new thread.

Thanks.
 

ronsimpson

Joined Oct 7, 2019
1,657
Would I need to limit the input current to the CD4047 IC?
No, The power lost in the Gate of the MOSFET is small compared to the loss in the D-S of the MOSFET. If you increase the resistance value to try to save power the FETs will switch slower and thus greatly increase the power loss.
 

Audioguru again

Joined Oct 21, 2019
3,849
The datasheet for the CD4047 says the oscillator resistor value should be 10k to 1M. If you use 10k then with a 12V supply and it being powered for half the waveform then its heating is only 12V squared/10k= 0.014W divided by 2= 0.007W.
The max current drawn by the CD4047 is only 0.5mA, then it heats with 0.5mA x 12V= 0.005W.

The output will be a squarewave with poor voltage regulation. Many modern electrical products need a sinewave and do not work from a squarewave.
 

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Ian0

Joined Aug 7, 2020
3,761
Use a SG3525 not a CD4047. The SG3525 is designed to drive MOSFETs but the CD4047 has very limited output current. The SG3525 automatically inserts dead time so there is no overlap when both MOSFETs are one together due to the CD4047's limited drive current being unable to switch them off quickly.
 

Thread Starter

Summit23

Joined Oct 21, 2021
7
One more thing. The output is not likely to be a sinewave.
The datasheet for the CD4047 says the oscillator resistor value should be 10k to 1M. If you use 10k then with a 12V supply and it being powered for half the waveform then its heating is only 12V squared/10k= 0.014W divided by 2= 0.007W.
The max current drawn by the CD4047 is only 0.5mA, then it heats with 0.5mA x 12V= 0.005W.

The output will be a squarewave with poor voltage regulation. Many modern electrical products need a sinewave and do not work from a squarewave.
Thank you for pointing that out. I was considering using a 3-stage low pass filter, but it did not meet my requirements and investing more effort in trying to figure it out would result in me missing my deadline.
 

Thread Starter

Summit23

Joined Oct 21, 2021
7
Here is my complete circuit:
circuit.png
This was the only solar panel I could find at my local electrical wholesaler. The rating is Vp = 18.36; Ip = 0.54A. I use the voltage regulator to provide a constant 12V to the battery. The inverter then provides 12Vrms to the transformer, which is stepped up to 220Vac on the output. The lamp is a 10W lamp. Would this work out as expected?
 

Thread Starter

Summit23

Joined Oct 21, 2021
7
Use a SG3525 not a CD4047. The SG3525 is designed to drive MOSFETs but the CD4047 has very limited output current. The SG3525 automatically inserts dead time so there is no overlap when both MOSFETs are one together due to the CD4047's limited drive current being unable to switch them off quickly.
Will have a look at the SG3525
 

Audioguru again

Joined Oct 21, 2019
3,849
You use a 270ohm resistor with the LM338 which has 1.25V across it then its current is 1.25V/270 ohms= 4.63mA. The 2.2k resistor also has the 4.63mA in it producing 2.2k x 4.63mA= 10.19V. Then the output of the M338 is 10.19V + 1.25V= 11.44V which is too low for a 12V leak-acid battery that needs 13.7V to 14.5V.

The squarewave signal to the transformer when the output current is low is 13.7V peak, not RMS. Then the output from the transformer is 220/12 x 13.7= 251V peak. A 220V RMS sinewave is 220 x 1.414= 311V peak but an old and slow incandescent light bulb will be brighter than normal and burn out sooner when the battery is fully charged.

If you filter the squarewave then the Mosfets will be linear all the time and make a lot of heat which wastes the battery power.
A modern inverter circuit uses Pulse-Width-Modulation at a high frequency to produce a 50Hz sinewave with low heating in the switching Mosfets.
 

Ian0

Joined Aug 7, 2020
3,761
You can choose the resistors on a SG3525 to give a mark-space ratio Of 64%, then you can have the correct peak voltage AND the correct rms voltage. This is euphemistically known as a “modified sinewave”
 

Thread Starter

Summit23

Joined Oct 21, 2021
7
You use a 270ohm resistor with the LM338 which has 1.25V across it then its current is 1.25V/270 ohms= 4.63mA. The 2.2k resistor also has the 4.63mA in it producing 2.2k x 4.63mA= 10.19V. Then the output of the M338 is 10.19V + 1.25V= 11.44V which is too low for a 12V leak-acid battery that needs 13.7V to 14.5V.

The squarewave signal to the transformer when the output current is low is 13.7V peak, not RMS. Then the output from the transformer is 220/12 x 13.7= 251V peak. A 220V RMS sinewave is 220 x 1.414= 311V peak but an old and slow incandescent light bulb will be brighter than normal and burn out sooner when the battery is fully charged.

If you filter the squarewave then the Mosfets will be linear all the time and make a lot of heat which wastes the battery power.
A modern inverter circuit uses Pulse-Width-Modulation at a high frequency to produce a 50Hz sinewave with low heating in the switching Mosfets.
For a square waveform, I thought that Vrms = Vp. If not, then how would I be able to calculate it?

For the lamp, I do plan on using a resistor across it. I forgot to include it in the diagram above.

I will definitely have a look at PWM to further improve my inverter in the future. I've read up on it, but didn't pay much attention to it.
 

Audioguru again

Joined Oct 21, 2019
3,849
A 220V RMS sinewave voltage varies from 0V peak to 155.5V peak.
A squarewave with 110V peaks produces the same power.

Why do you want a resistor across the lamp?
 

RDBatBSSL

Joined Jan 17, 2011
8
I am building this solar powered inverter: https://ibb.co/58m5L8B using a 12V, 0.8 Ah battery as the input source and a CD4047 IC for the inverter. However, the resistors connected to pin 2 are rated at 0.5W and the IC has a maximum power dissipation of 500 mW. I want to keep the efficiency of the system as high as possible. Would I need to limit the input current to the CD4047 IC?
You need to make sure that there are gate resistors of between 4k7 and 10k grounding the gates, otherwise the o/p fets could be destroyed it the 4047 created spurious pulses. (I speak from experience, having used 4047 ic's in the past). You also need to make the RC network adjustable so that you can trim the frequency to 50Hz for the transformer to operate correctly. The 0.8ah battery will only give you a very short duration and you must allow for low voltage shut down or the 4047 might again produce unexpected results. Put heatsinks on the output devices too. The frequency of the chip needs to be 100hz because it divides the output into two outputs.
 
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