XR-8038 square wave generator

Discussion in 'The Projects Forum' started by Tate, Oct 27, 2010.

  1. Tate

    Tate Thread Starter New Member

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    I am wanting to use a XR-8038 as a signal generator to control an IGBT transistor. I need a positive bias to turn the transistor on and a negative bias to turn the resistor off. I need to do this at 20KHz, 28KHz and 40KHz.

    I have the XR-8038 wired to a +/-12V power supply, and get very nice square wave output. The output is 0V to+12V when the ground of the oscilloscope probe is grounded to the common, and -12V to 0V when the probe is grounded to the negative power supply. The circuit is wired according to the Generalized Test Circuit on the datasheet.

    http://users.ece.utexas.edu/~grady/EE362L_Spec_Inverter_triangle_wave_generator.pdf

    The question is, how do I get a square wave that is +12V to -12V, or shift the wave to +6V to -6V?

    Thank you for your time.

    JT
  2. SgtWookie

    SgtWookie Expert

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    The attached will get you close for the +12v/-12v. It's basically an emitter follower configuration.

    Attached Files:

  3. DonQ

    DonQ Active Member

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    Something is not right here. If your scope is referenced to the -12, and you get a -12 reading, you are getting -24V from somewhere. I think you are just having a problem interpreting your measurements. (For example: If you put the probe itself on the -12 with the ground clip on the output, it would not be called "grounded to the negative power supply", but this, along with an inadequate pull-up, would explain your readings.)

    If you have the 8038 powered between the -12 and +12, and a proper pull-up on the output transistor tied to +12, your output should have the resistor pulling up to +12 rail when the transistor is open, and when the transistor is conducting, it will pull to the -12 rail.

    If you measure this relative to 0V (the center point between your supplies), your output will be +12/-12. I think this is what you are wanting. End of story...

    If you measure relative to the -12V rail (scope ground clip on the -12, probe center on the output), you should get a square wave from +24V to about 0V. If you have a very large valued pullup (or none at all), you may not get all of the 24V to show.

    Start with about a 10K for RL and try again.
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  4. SgtWookie

    SgtWookie Expert

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    Thanks for your post DonQ; you made me re-visit the datasheet.

    Figure 1 shows the output transistors' emitter grounded. However, there is no "ground" pin on the IC! This is cause for some confusion/ambiguity, as "ground" is normally defined as the 0v reference. Since there is no "ground" as an input to the IC, it is an unknown whether they are creating an "artificial ground" by taking the average of Vcc and Vcc and using that point as an artificial ground, or if the ground point in the diagram actually means Vee.

    I actually have an NTE-labeled version of the 8038 kicking around here somewhere, if I can only find the darn thing. :rolleyes: It's an unknown whether it is an ICL8038 or an XR-8038, but they should be equivalent.

    Anyway, if RL is of too low a value (<12k for +12v/-12v) even with no load, the saturation voltage of the square wave output can exceed 0.5v from whatever they're calling ground. The output will be very sensitive to additional loading. It will not drive a capacitive load well without aid.

    I've attached a schematic of what the output transistor configuration might be, along with a voltage follower/buffer made from a couple of transistors. It's not ideal, but it's quite simple.

    [eta]
    Here's the NTE864 datasheet, which is the IC that I have:
    http://www.nteinc.com/specs/800to899/pdf/nte864.pdf

    They show pin 11 as GND instead of Vee, which clears up that whole ground/virtual ground issue.

    So, the schematic I've attached should work just fine, unless Qout has been unduly stressed due to RL being of too low of a value.

    Attached Files:

    Last edited: Oct 28, 2010
  5. DonQ

    DonQ Active Member

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    Yeah, it's like a 24V supply for the 8038, with a built-in center-tap on the power supply.

    I guess my point was that if he is looking for +12 to -12 from the center-tap, he should be able to get that without any extra transistors. That is as long as he doesn't need to draw much current. With current, the transistor in the 8038 takes the place of your Qout and RL is the external resistor already there. Only Q1 and Q2 are added for the current.
  6. Tate

    Tate Thread Starter New Member

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    Thank you genglemen for your assistance. Yes, I trying to get +12 to -12 from the center tap. I just looked at my power supply that I will be using when I take this project off the bread board and the supply will be +6 to -6 so I may have to make some changes.

    I am using an RL of 10Kohms and it is working well at this point. I plan on amplifying the signal before putting it into the IGBT.

    SgtWookie, I will be trying your solution in the morning, I will post the results. I am still very green behind the ear when it comes to electronics. Thank you for your responces.
  7. SgtWookie

    SgtWookie Expert

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    The XR-8038 is specified to operate on a single supply as low as 10v, and dual rail supply as low as +5v/-5v, so you should be OK with that.

    Keep in mind that the 8038 square wave output has a very low current sink capability; try to sink more than ~2mA and the output saturation voltage starts getting too high, and may burn out the square wave output due to power dissipation.

    For +6v/-6v rails, or +12v and GND across Vee, use 6k or higher for RL.

    The transistor emitter follower configuration I posted will multiply the 2mA output by the gain of the transistors.

    You might consider using a bjt (transistorized) 555 timer instead of the XR-8038. The XR-8038's are scarce and expensive, as they are out of production, along with having meager source/sink capability on the square wave output. You would be better off to save it for a signal generator project.

    555 timers are cheap, plentiful, and can source/sink up to 200mA on pin 3, and require very few external components to set up a PWM output. They will operate just fine with a single-ended supply from 5v up to 16v; some are rated for up to 18v.
  8. Tate

    Tate Thread Starter New Member

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    Since I am having trouble getting the square wave to be cross zero volts, I am seeing that a 555 and the emitter follower circuit you posted earlier in post #2 should get me everything I need. I have already built the 555 signal generator but I did not know how to get a +6 to -6 signal with it. Any other suggestions would be greatly appreciated.

    Again, I want to say thank you for all of your help.
  9. SgtWookie

    SgtWookie Expert

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    I ran across my NTE864's, which are equivalent to the x8038's - but I've been occupied today fiddling around with car problems (finally got my EGR problem resolved - yay!)

    You wouldn't need an emitter follower with a bjt (bipolar junction transistorized) 555.

    Have to run for a bit.
  10. SgtWookie

    SgtWookie Expert

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    I'm really too tired to do this thread justice this evening. Sorry about that, but it's the truth.
  11. Tate

    Tate Thread Starter New Member

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    SgtWookie, Thank you for suggesting the 555 to provide the signal. I did not know that you could get a negative signal from a 555, but it worked. Now I am going to further research how to drive an IGBT transistor. The load I am working with has a resonant impedance of 50 Ohms and it was suggested I try to use an IGBT transistor.
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