Dumb question re. simple inverter

KrisBlueNZ

Nov 28, 2011
8,393
Oh sh*t Trevor! How did you break your spine?

Dougster

Nov 8, 2014
21
The frequency is 1/(4.4*R*C) so 390k and 0.01μF is 58Hz and 390k with 0.47μF is 1.2Hz which is far too low. Is the resistor actually 390k?
No, its 390Ω. I have a frequency meter coming, and I will fine tune the output by using a 10k pot. Thanks to everyone's input, I've calculated the RC network with my .47uf cap should have about 8.2kΩ. I'm actually sort of glad the provided circuit was a joke. It forced me to learn a lot about mosfets, etc! However, it's a dirty trick to play on an unsuspecting beginner like myself.

duke37

Jan 9, 2011
5,364
The 390R with 0.01μF will give 58kHz - far too high for a laminated transformer to handle . Posh valve audio amplifiers have trouble going to 10kHz. 1kHz would be nice.

If you are going to rectify the output then the frequency is not critical and I would not include an adjuster.

Dougster

Nov 8, 2014
21
The 390R with 0.01μF will give 58kHz - far too high for a laminated transformer to handle . Posh valve audio amplifiers have trouble going to 10kHz. 1kHz would be nice.

If you are going to rectify the output then the frequency is not critical and I would not include an adjuster.
Thanks Duke, but I replaced the .01μf with a .47μf.

duke37

Jan 9, 2011
5,364
So what frequency do you estimate you will get?

Dougster

Nov 8, 2014
21
So what frequency do you estimate you will get?
About 58cps. I'm using the power my house.

Dougster

Nov 8, 2014
21
The frequency is 1/(4.4*R*C) so 390k and 0.01μF is 58Hz and 390k with 0.47μF is 1.2Hz which is far too low. Is the resistor actually 390k?
Thanks Duke. Where do you get the 4.4 in your formula?

duke37

Jan 9, 2011
5,364
I have a book, remember them?
COS/MOS B-series Devices Databook 1st edition by SGS-THOMPSON MICROELECTRONICS.
They give details of the HCC4047B.

I am sure you could find details on the web in a short time.

I doubt if the frequency is very accurate which presumably is why a trimmer is included in the original circuit. The 58Hz is close to 60Hz which is used for the mains in your part of the world.

Do you need an accurate frequency if you are generating DC?

Dougster

Nov 8, 2014
21
I have a book, remember them?
COS/MOS B-series Devices Databook 1st edition by SGS-THOMPSON MICROELECTRONICS.
They give details of the HCC4047B.

I am sure you could find details on the web in a short time.

I doubt if the frequency is very accurate which presumably is why a trimmer is included in the original circuit. The 58Hz is close to 60Hz which is used for the mains in your part of the world.

Do you need an accurate frequency if you are generating DC?

Thanks, Duke. Yes, I generate DC and store the energy in a large battery bank. I have a commercial true sine wave inverter I use before bringing the power into my house. My inverter is on its own line to power my office. It's modest right now, but it's a "passionate hobby". The frequency is important for my ceiling fan, lol!

Dougster

Nov 8, 2014
21
Can anyone recommend an IC that I can use to change (reduce) duty cycle? 12V supply, dual input, dual output. Thanks, as always. I tried to find one on DigiKey, but only confused myself.

KrisBlueNZ

Nov 28, 2011
8,393
You need to be a lot clearer with your question. You want to reduce the duty cycle of what? How do you want to control the duty cycle? For what reason? What's the background to the project?

Dougster

Nov 8, 2014
21
Thanks, Kris. I want to change the duty cycle of the output of a 4047BE astable multivibrator. I wish I had used a 555 in the first place. The outputs drive 2 mosfets in an inverter project. I have found that my inverter works great at frequencies above 125 - 150 cps (thereabouts). At lower frequencies the long duty cycle of the square wave is stressing the mosfets, as there is a period where the steady mark state creates practically no inductance, thereby briefly shorting 12v to ground thru the mosfets. A modest reduction in the duty cycle would allow me to decrease the frequency and eliminate this problem. Any thoughts?

KrisBlueNZ

Nov 28, 2011
8,393
You could use two monostables (aka one-shots). It depends what you want to set, and what will be changing.

Inserting a fixed-duration monostable between each output and each MOSFET gate will give you a fixed ON-time. If the 4047's frequency is fixed, this corresponds to a fixed duty cycle. If the 4047's frequency is variable, the duty cycle will be higher at higher frequencies, because the ON-time is fixed.

Inserting a fixed-duration delay between each output and each MOSFET will give you a fixed OFF-time (within each half cycle). If the 4047's frequency is fixed, this also corresponds to a fixed duty cycle. If the 4047's frequency is variable, the duty cycle will be lower at higher frequencies.

If you want to set the duty cycle to a constant value regardless of frequency, things get more complicated. You could use a different kind of oscillator that generates a sinewave and pass that through some comparators, or if you want a simple duty cycle like 25% for each MOSFET, you can run the oscillator at a multiple of the main frequency and use a frequency divider to divide the cycle into pieces.

The 4047 seems convenient because it has complementary outputs, but that's the only advantage it has over other methods, because it saves you one inverter IC. That advantage becomes less important if you add monostables to those outputs. And you should already be using a MOSFET driver to ensure quick switching - the 4047's outputs can only source and sink a few mA. So I think you could reconsider your timing source.

For example an oscillator based on a 32000 Hz watch crystal will give you a nice stable frequency that you can divide down by varying amounts. You could also use a small microcontroller, with a watch crystal for its second oscillator, to generate both control signals. The microcontroller could monitor the output voltage using an analogue input, and adjust the duty cycle accordingly. This could be a good next step in the development of your inverter.

Dougster

Nov 8, 2014
21
Wow. Thanks for all the effort! I think I'll try the monostable on each output.

KrisBlueNZ

Nov 28, 2011
8,393
Here's another way to do it, using only one monostable. This means that you can adjust the duty cycle for both sides in just one place. It adds a CD4027 dual JK flip-flop (only one half is used) and a CD4081 quad AND gate (only two of the four gates are used). Use a decoupling capacitor on the 4027.

The oscillator needs to run at twice the inverter's frequency. The 4027 divides this frequency by two, and produces two opposite output signals. One monostable pulse is directed to output A by the gating, then the next pulse to output B, then back to A, then B, alternating.

Dougster

Nov 8, 2014
21
Thanks Kris - I like the idea of adjusting the duty cycle in just one place so both gates are identical! Where should the decoupling cap go? Steve educated me on MOSFETS, Duke gave me the correct freq. formula for the 4047, and you've given me some outstanding design advice on several different circuits as well as suggesting the perfect low side gate drivers! Thanks to you guys, I'm closing in on a finished project. Otherwise, I would have trashed the whole thing a few weeks ago. Agony and ecstasy!

KrisBlueNZ

Nov 28, 2011
8,393
Every IC should have a decoupling capacitor, typically 0.1 µF multi-layer ceramic, connected as closely and directly as possible between the positve supply pin and the 0V pin. On logic ICs these are pins 14 and 7, or 16 and 8.

Dougster

Nov 8, 2014
21
Thanks Kris - I used 2 4047's as one shots on the outputs of the oscillator to shorten the duty cycle of the gates, and my inverter works great!

I also have a "commercial" inverter - it's rated at 2000 watts, but only weighs about a pound. My transformer alone in my inverter (500 va) weighs about 15 pounds). How is this possible? Don't commercial inverters use transformers? Sorry to ask such a basic question, but I'm going crazy wondering how this is possible!

BobK

Jan 5, 2010
7,682
The low weight inverter probably works by using a high speed switching boost converter that produces about 170V DC and a MOSFET bridge to invert it to AC. The transformer (or just inductor, depending on the circuit) can be much smaller because it runs at a frequency of probably 100KHz or more.

Bob

Dougster

Nov 8, 2014
21
Thanks, Bob. The problem with your explanation is that the inverter output is only 60 cps.

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