i found this circuit on the internet to convert 12v dc to 220v ac

i found it in many sites and in all the sites it had the same values of the capacitors and resistors but i didn't understand how it works (i didn't find an explanation for its behavior )
can u please explain to me how it works and if i want the output to be equal to the input but it's ac what should i do ? and if it won't work then can u please give me a circuit that coverts dc to ac without increasing the voltage ?
i know i can use a 555 timer but i wanted to try this circuit as a new way to convert dc to ac

 

The cross-coupled transistors make a classic astable multivibrator (oscillator) in which, when one switches 'on' it switches the other one 'off'. Collector current flows alternately in the two halves of the primary winding of the transformer and induces a voltage in the transformer secondary winding. The turns ratio of the transformer determines the output voltage. For an output voltage equal to the input voltage the number of turns in the secondary winding and each half of the primary winding should be the same (assuming a lossless transformer).

 

For an output voltage equal to the input voltage the number of turns in the secondary winding and each half of the primary winding should be the same (assuming a lossless transformer).

well then there is no need to the transformer right ? i thought the circuit itself has a voltage gain and its output is 220v { didn't notice that the transformer is connected in the opposite way } so if i want Vout = Vin can't i just cancel the transformer ??

 

i found this circuit on the internet to convert 12v dc to 220v ac

i found it in many sites and in all the sites it had the same values of the capacitors and resistors but i didn't understand how it works (i didn't find an explanation for its behavior )
can u please explain to me how it works and if i want the output to be equal to the input but it's ac what should i do ? and if it won't work then can u please give me a circuit that coverts dc to ac without increasing the voltage ?
i know i can use a 555 timer but i wanted to try this circuit as a new way to convert dc to ac

Astable Multivibrator using Transistors is a good read and detailed explanation of how the circuit actually works. The resistor and capacitor values determine the free running frequency. In a circuit like you posted to make an Inverter the values are selected to make 50 Hz which is a standard mains power frequency in most parts of the world. That is why you see the same values. Also, keep in mind with an inverter of this design you get a square wave out.

If you just want to for example blink a few LEDs then all you need is a simple two transistor multivibrator circuit with a low enough frequency that you can visually see the LEDs blink alternately. A Google of "led transistor multivibrators" will bring up a few dozen hits and circuit ideas using transistors.

Ron

 

well then there is no need to the transformer right ? i thought the circuit itself has a voltage gain and its output is 220v { didn't notice that the transformer is connected in the opposite way } so if i want Vout = Vin can't i just cancel the transformer ??

Yes indeed.

Ron

 

Astable Multivibrator using Transistors is a good read and detailed explanation of how the circuit actually works. The resistor and capacitor values determine the free running frequency. In a circuit like you posted to make an Inverter the values are selected to make 50 Hz which is a standard mains power frequency in most parts of the world. That is why you see the same values. Also, keep in mind with an inverter of this design you get a square wave out.

If you just want to for example blink a few LEDs then all you need is a simple two transistor multivibrator circuit with a low enough frequency that you can visually see the LEDs blink alternately. A Google of "led transistor multivibrators" will bring up a few dozen hits and circuit ideas using transistors.

Ron

thank u so much i actually want it to use its output as an input for a multiplier circuit

 

Then you can take the output from either collector. You get two square waves 180 degrees phase shifted. Should work out just fine.

Ron

 

Just be aware that running an output from the collector will not give you pulses of the supply voltage in. It depends on where you take your "common" signal from.
If from Gnd, the AC wave form will have peaks from 0V and supply x 2 Volts.
If common = +Vin, then you get pulses of Vin amplitude.
That may not matter unless you need AC that is centered around the Gnd level.
A bit more info on what you are actually going to use it for would help.
Do you have an application circuit?

i actually want it to use its output as an input for a multiplier circuit

If you just want to multiply the supply volts by 2, a couple of diodes from the collectors, Anode to collectors, and the cathodes connected together with a cap to Gnd, will give a Vin x 2 output DC.

It could be capacitive coupling will work for your application too.

 

What is it you are trying to do? The circuit in post #1 is intended as an AC power supply. Note that it makes a squarewave output, not a sinewave, and some devices do not handle this well. If you do not need the voltage change and you don't need transformer isolation, then the circuit in #1 probably is not the best one for your application. What is your application?

ak

 

What is it you are trying to do?

i wanted to try this circuit as a new way to convert dc to ac

i actually want it to use its output as an input for a multiplier circuit

Like in this one:

 

Then you can take the output i thought i should take the output from the two collectors

I first thought that i should take the output from the two collectors but when i tried it was 0v then i realised that the two ouputs are 180 ° out of phase

If from Gnd, the AC wave form will have peaks from 0V and supply x 2 Volts

Why ??
Acuatlly this is what happened the output wasn't equal to Vcc it was more than it but i didn't understand why and how could it be possible to go more than vcc ? and what is better as an input for a voltage multilpier ?
and i wanna multiply it more than 2 but thanks it's important to know that

What is it you are trying to do? The circuit in post #1 is intended as an AC power supply. Note that it makes a squarewave output, not a sinewave, and some devices do not handle this well. If you do not need the voltage change and you don't need transformer isolation, then the circuit in #1 probably is not the best one for your application. What is your application?

ak

Yes i know it produces a square wave i just wanna take its output to use it as an input for a simple voltage multiplier of capacitors and diodes it'll work as an input right ?

Like in this one:
View attachment 155983

yes this is what i wanted thank u but what is the point of using bridge configuration ? wouldn't two transistors be enough and they'll make the required wave ?

 

Why ??
Acuatlly this is what happened the output wasn't equal to Vcc it was more than it but i didn't understand why and how could it be possible
and i wanna multiply it more than 2

The only way you will accurately see the output(s) is with a scope. Keep in mind the output of a multi-vibrator is a square or rectangular waveform. If you are using an inexpensive DMM which is average responding you will not get a reliable reading. The peak value of the square wave should be close to your applied voltage.So how are you measuring output voltage?

Never mind, I see things have been resolved as I waddled around.

Ron

 

yes this is what i wanted thank u but what is the point of using bridge configuration ? wouldn't two transistors be enough and they'll make the required wave ?

1. Every half wave, when transistor is OFF, plus from power supply flows to load through collector resistor and it dissipates most of energy.
2. When transistor is ON, then collector resistor connected directly to power supply and dissipates maximum energy.
In summary, efficiency of such circuit is very low, few percents, usually.

 

Why ??
Acuatlly this is what happened the output wasn't equal to Vcc it was more than it but i didn't understand why and how could it be possible to go more than vcc

Basically, if you look at the coil as an auto-transformer, the center tap is the +12V supply.
Then one end is grounded (by a transistor) so current flows and a rising magnetic field is produced. This rising magnetic field induces a voltage in the other half of the winding and that voltage adds to the supply volts so the open end of the winding goes to supply volts x 2.
Then the first transistor turns off and the second one switches on and the same thing happens in the other end.
There will be short duration switching spikes of a lot higher voltage produced too, so most circuits have spike suppressors of various sorts included.
Do you have an oscilloscope to look at the waveform?

 

1. Every half wave, when transistor is OFF, plus from power supply flows to load through collector resistor and it dissipates most of energy.
2. When transistor is ON, then collector resistor connected directly to power supply and dissipates maximum energy.
In summary, efficiency of such circuit is very low, few percents, usually.

And ofcourse since it's efficiency is very low then it's not commonly used right ? It's the first time i see it but thank u i learnt about a new circuit

The only way you will accurately see the output(s) is with a scope. Keep in mind the output of a multi-vibrator is a square or rectangular waveform. If you are using an inexpensive DMM which is average responding you will not get a reliable reading. The peak value of the square wave should be close to your applied voltage.So how are you measuring output voltage?
Never mind, I see things have been resolved as I waddled around.

Ron

So it's because of the DMM not like the circuit itself is making this increase in voltage
I tried two DMM so i had different results

Basically, if you look at the coil as an auto-transformer, the center tap is the +12V supply.
Then one end is grounded (by a transistor) so current flows and a rising magnetic field is produced. This rising magnetic field induces a voltage in the other half of the winding and that voltage adds to the supply volts so the open end of the winding goes to supply volts x 2.
Then the first transistor turns off and the second one switches on and the same thing happens in the other end.
There will be short duration switching spikes of a lot higher voltage produced too, so most circuits have spike suppressors of various sorts included.
Do you have an oscilloscope to look at the waveform?

I understand thanks but i actually didn't use a trabsformer since i want Vout = Vin but i had a value of Vout = 1.5 Vin and i knew then that the problem is from the DMM i used another one and it worked

 

And ofcourse since it's efficiency is very low then it's not commonly used right ? It's the first time i see it but thank u i learnt about a new circuit

Yes, efficiency is low with two transistors and collector resistors circuit, but circuit, contains bridge commutator and multiplier is widely used as high efficient switched-capacitor Dickson charge pump.

 

So it's because of the DMM not like the circuit itself is making this increase in voltage
I tried two DMM so i had different results

That does not surprise me. When measuring a voltage using a DMM there are two types of DMM. We have what we call Average Responding RMS Indicating on the low cost end and then we have RMS Responding RMS Indicating on the more expensive end. If I measure a nice 240 VAC or 120 VAC true sine wave both meters will read the same. However if I measure a square wave or modified sine wave only the true RMS Responding RMS Indicating DMM will give me accurate results.

Here is an example of what I am getting at. The meter on the right is a more expensive Fluke 87 true RMS responding meter while the meter on the left is an inexpensive basic DMM. The meters are measuring a true sine wave, the mains power in my house.



Both meters above read within 1% of nominal mains voltage and each other. Now in the below image both meters are measuring a 120 VAC 60 Hz MSW (Modified Sine Wave) coming from an inverter circuit (close to a square wave). Note the difference in the measured value for the very same waveform.



The 117 Volts is the correct RMS value but note the meter on the left. If I were to look at the signal on an oscilloscope I would see a 164 Volt peak MSW and about a 330 volt peak to peak signal. That is what I was getting at earlier.

Ron