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The LM358 Op Amp PWM to Voltage Converter. The circuit presented uses an LM358 Op Amp in a Non-Inverting Voltage Follower configuration. If you’re unfamiliar with Op Amps, you can read an easy to follow description HERE. It also uses resistor and capacitor in series to form an RC time constant that charges to the desired voltage. Then check out this outstanding single chip PWM motor speed controller circuit that will give you a complete 360 degrees of continuously varying motor speed control right from zero to maximum. The speed is controlled through an externally applied varying DC voltage source. How to Build a High Torque DC Motor Speed Controller Circuit. Convert 12V PWM circuit to 5V PWM using opamp [closed]. I can convert a 12 V PWM signal into 5V signal, by using voltage divider on 12 volt signal, then directing it to base of NPN common emitter transistor with collector connected to 5V. Simulate this circuit – Schematic created using CircuitLab. The op-amp's '-' input is held at 2.5.

I can convert a 12 V PWM signal into 5V signal, by using voltage divider on 12 volt signal , then directing it to base of NPN common emitter transistor with collector connected to 5V

^{simulate this circuit – Schematic created using CircuitLab}

How can i do same using Op amp ?

M.Ferru$endgroup$

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5 Answers

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It depends upon your PWM frequency and the shortest mark/space pulses resulting from the nearest PWM positions to 0 % and 100 %.

This would carry out the general function:

^{simulate this circuit – Schematic created using CircuitLab}

The op-amp's '-' input is held at 2.5 V (half of 5 V).

The op-amp's '+' input switches between 0 V and around 4 V (about a third of 12 V, more precisely 3.8 V).

You will have to choose a suitable op-amp for your PWM frequency. If your PWM frequency is too high or the mark/space pulses too short, R1 and R2 should be lowered to ensure a fast rise/fall time at the op-amp pins while charging/discharging the stray capacitance there.

TonyMTonyM$endgroup$$begingroup$

This is a simple circuit to get the job done. You can use a simple voltage divider followed by a buffer.

Ppm To Pwm Converter

^{simulate this circuit – Schematic created using CircuitLab}

Choose the op-amp, R1 and R2 to suit your application's requirements. I have chosen R1 = 1K, R2 = 1.5K. When the input is 12V, the output is 4.8V. You may find more suitable values to get 5V on the output.

If you use 5V to power the op-amp you may not get 4.8V on the output because op-amp output voltage is never the same as the op-amp supply voltage. Using 12V supply for the op-amp can give you better results.

MacitMacit$endgroup$$begingroup$

Many circuit using opamp can fit your need. There is some point you have to consider :

• The original circuit on your post change the polarity of the PWM (70% duty become 30% duty cycle PWM)
• Opamp have slew rate, this mean, at high frequency, the PWM's waveform may change. Choose your part wisely ;)

M.FerruM.Ferru$endgroup$$begingroup$

How can i do same using Op amp ?

One inverting amp will do. Or two inverting amps if you want to maintain the same polarity.

But there are much simpler ways for you to do that.

dannyfdannyf$endgroup$$begingroup$

If all you are doing is feeding the attenuated signal to an Arduino digital port then just use a resistor potential divider: -

Let's say you make R2 = 1 kohm, then R1 has to be 7/5 * 1 kohm = 1.4 kohm

You might choose to make R1 = 1.5 kohm (in case the 12 volt rises a little bit). Check your logic high levels on the Arduino to see how high R1 could be before it fails to register a logic one when 12 volts is at the input.

For instance, a lot of IOs are TTL compatible meaning that 2 volts is the minimum logic 1 so, make a potential divider from 12 volts to 3 volts. This would make R1 = 1.8 kohm.

Using fairly low resistance values means rise and fall times will be largely unaffected at the output (unlike an op-amp or a comparator which will likely worsen the rise times and add a delay). Stay simple if you can.

Andy akaAndy aka$endgroup$

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How does a sine choke exactly function? In most high power inverter systems, the primary side of the output transformer is always driven by a PWM signal. The secondary output which is sent to a load should also come out to be PWM. How does a pure sine wave inverter exactly convert this PWM into a pure sine wave?

VishalVishal$endgroup$

4 Answers

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Pwm To Voltage Converter Circuit

Here's a PWM signal fed to a transformer and superimposed on this diagram is the sinewave that the PWM represents: -

The secondary of the transformer usually has an inductor and a capacitor that form a 2nd order low pass filter thus converting the PWM signal into (more-or-less) a fairly decent sinewave.

For instance, if you take the high frequency content of the PWM waveform it looks like a square wave with varying duty cycle and, you can low-pass filter this quite easily to get this: -

On the left is the original square wave. In the middle a little bit of filtering has happened and on the right the filtering is far greater.

Thus, the high frequency edges of the PWM signal can be greatly reduced leaving the low frequency content that represents the sinewave. In effect you get something that typically looks like this: -

You can still see that the waveform has a little bit of the PWM signal but, in the main, it is a sinewave.

If your PWM frequency is 60 kHz and your AC is 60 Hz you could position a filter to have a cut-off of 600 Hz and there would be 2 decades between it and the 60 kHz. A 2nd order filter would attenuate the 60 kHz by 80 dB (40 dB per decade): -

You might note that I mentioned a filter having a cut-off of 600 Hz and wonder why it is position ten times higher than the AC 60Hz. You might ask why not have it at 60 Hz and this would be a good question. The reason it isn't at 60Hz is two-fold: -

1. There would be a 3dB attenuation of the AC
2. If the filter was extremely resonant it would consume vast amounts of current because, in effect, it is also a series resonant circuit across the line.

It has to be positioned as far away from 60 Hz as possible to avoid large circulating currents in the L and the C of the filter BUT you don't want it up close at 60 kHz because it won't filter out the high frequency content very well. Minimum is 100 Hz I would say and it should be at least 1 decade away from the lowest PWM frequency (generalism alert!).

Andy akaAndy aka$endgroup$$begingroup$

The basic answer is by low pass filtering. The slower 'average' of the PWM waveform is the desired line frequency (like 60 Hz) signal. Superimposed on this signal is the high frequency PWM chopping.

The inductance of the transformer together with some capacitance on the output form a L-C low pass filter. This filter passes the 60 Hz component with little attenuation, but greatly attenuates the many kHz PWM chopping frequency.

Switching power supplies in general work on this concept. The pulse signal contains a deliberate low frequency component, which is ideally all that makes it to the output.

Olin LathropOlin Lathrop$endgroup$$begingroup$

You have two options:
1. Low voltage DC is step-uo converted with push-pull PWM and step-up ferrite transformer at high frequency, it can be also a flyback converter or phase shifted PWM converter rather than push-pull. At the secondary of HF transformer is a rectifier with caps and you get the HV DC voltage. The second stage is H-bridge with sine wave PWM that has to have an output lowpass LC filter to filter out high frequency.
2. DC voltage is inverted with sine wave PWM, then filtered with lowpass LC filter, then a classical step up transformer.

Marko BuršičMarko Buršič$endgroup$$begingroup$

Just a small addition to the answer by Andy. If you carefully craft the PWM waveform it's possible to move the energy in the harmonics to higher frequencies to make them easier to filter out. Have a look for Harmonic Elimination Pulse Width Modulation. I've done a small write up about it. http://www.grant-trebbin.com/2013/10/harmonic-elimination-pwm-comparison-and.html

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