# Convert bipolar square wave 30V peak to peak to 100 V peak to peak

Discussion in 'The Projects Forum' started by Tristan Demanuele, Aug 27, 2015.

1. ### Tristan Demanuele Thread Starter New Member

Aug 19, 2015
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Hi currently im working on a project and i manage to create a square wave pattern having a 30V peak to peak voltage (-/+15V amplitude).

Is there a way to boast the signal amplitude to -/+50V using an IC or analogue components please?

Thanks!

2. ### Papabravo Expert

Feb 24, 2006
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With or without an external ±50 VDC power supply.

3. ### AnalogKid Distinguished Member

Aug 1, 2013
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Your question is missing some important information:

1. What is the frequency of the wave?
2. What is the output current required?
3. What are the min and max rise and fall times required?

ak

4. ### Lestraveled Well-Known Member

May 19, 2014
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There are a lot of ways to up the output voltage, but you need to supply some more details about the application, such as load resistance, power supply voltage and current, frequency, etc. etc.

5. ### #12 Expert

Nov 30, 2010
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Plug your +/- 15 volt square wave into a 200 watt audio amplifier designed to drive an 8 ohm speaker. Adjust, "volume" knob as required.

6. ### Lestraveled Well-Known Member

May 19, 2014
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200 watt amplifier that will swing +- 50 volts........I don't think so. But, I have a Crown CE2000 (2000 watt) that will do it.

7. ### #12 Expert

Nov 30, 2010
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P = E^2 /R
200W = E^2 /8
1600 = E^2
40 = E (RMS)
40 rt2 = 56.56 Vpeak

An audio amplifier rated for 200 watts RMS into 8 ohms can present 56.56 volts peak into an 8 ohm load.
Did I slip a digit?

May 19, 2014
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9. ### #12 Expert

Nov 30, 2010
16,703
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The RMS voltage of a sine wave is .707 times the peak voltage.
If I have an audio amplifier that is rated for 200 Watts RMS into 8 ohms, it has to be able to deliver 56.56 volts peak. If the square wave of the TS is within the right frequency range, this might work.

May 19, 2014
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Agreed

11. ### AnalogKid Distinguished Member

Aug 1, 2013
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No matter what the frequency, the audio amp output has to spend 50% of its time at its peak output voltage, way more energy than a sinewave peak. Depending on the load (and the amp), it might not have the current to do that.

Although based on my experience with D300's way back, I don't doubt for a minute that a hamster in a wheel would have enough power if the cage said Crown.

ak

12. ### Tristan Demanuele Thread Starter New Member

Aug 19, 2015
9
0

These are a couple of the specs:

Frequencies: 15Hz, 100Hz and 500Hz
and max current 4ma

I have also attached a pattern of the input sine wave (the application is using the same pattern but instead of sine a square wave )

Hope this will help!
Thanks

File size:
15.8 KB
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13. ### Tristan Demanuele Thread Starter New Member

Aug 19, 2015
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If possible without an external +-50V supply

14. ### Papabravo Expert

Feb 24, 2006
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Then you first need to create a ±50 VDC supply by means of switched mode power supply or DC-DC converter. This is similar to what is done in a MAX232 chip to convert +5V into ±10 VDC so that the chip can send and receive valid RS-232 levels. How difficult this will be depends on the amount of input power and output power required.

15. ### Tristan Demanuele Thread Starter New Member

Aug 19, 2015
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How would it be possible to create a ±50 VDC supply? Is there an IC or do you have any schematics?

16. ### ScottWang Moderator

Aug 23, 2012
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AC 110V/220V → 42Vac-0V-42Vac transformer → Full wave rectifier + Caps Filters → LM317HV(+), LM337HV(-) → +50Vdc, 0V, -50Vdc.

17. ### Tristan Demanuele Thread Starter New Member

Aug 19, 2015
9
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The input source is a 3.7V DC lithium ion battery not 230V AC

18. ### Papabravo Expert

Feb 24, 2006
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I don't have a schematic for that specific application which appears challenging to say the least. The basic calculation you need to start with is the one that goes:
"power out is always less than, sometime much less than, power in"
Assuming you can make a DC-DC converter that has 80% efficiency and the power out requirement is 50 watts eg. 50 Volts at 1 Ampere.
So:
50 Watts / 80% = 62.5 Watts
So the input power requirement is 62.5 Watts
To get 62.5 Watts out of a 3.7 VDC battery requires
62.5 Watts/ 3.7 Volts ≈ 16.9 Amperes

Now you might be able to draw 16.9 Amperes from a lithium batter for a short interval, measured in seconds. This means that this step-up in voltage represents an impractical solution. What might be practical is something on the order of 100 mW of output power.
.1 Watts / 80% = .125 Watts
.135 Watts / 3.7 V ≈ .034 Amperes
This might be a practical goal. Your still going to be using up those expensive batteries.
If your battery has a capacity of 600 maH then that is about 17.6 hours @ 34 milliamps

Chips and schematics for boost converters and inverters can be had at the Linear Technology, Maxim, and TI websites.

Aug 23, 2012
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20. ### Tristan Demanuele Thread Starter New Member

Aug 19, 2015
9
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Hi all thank for the input!

I was think to amplify the signal using a transformer. I found this transformer coilcraft to amplify the signal from a +-15V to +-50V.

Do you think it will work?

Thanks