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uC controlled DC-to-AC inverter

Discussion in 'General Electronics Discussion' started by pidzero, Mar 27, 2016.

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  1. pidzero

    pidzero

    4
    0
    Mar 27, 2016
    Greetings, forum.

    I'm trying to build a DC-to-AC inverter. I set a course for Google, YouTube, and websites of well-informed engineers. Days of my personal R&D, and sifting the good information from the bad, I have developed the following schematic.

    (file attached, also available at https://drive.google.com/file/d/0B8PLvKnJdHK2QzhnRVhYaG9wREE/view?usp=sharing)

    My original need for such a circuit was to create a 90VAC 20Hz signal--which would actuate an old Western Electric 500M rotary telephone's ringer. I quickly realized, however, that such an inverter could be suitable for many other uses/designs.

    By setting only pin A HIGH, I expect a normal signal (not inverted) at LOAD.
    By setting only pin B HIGH, I expect an inverted signal at LOAD.

    By alternating pin A and pin B, an alternating current SQUARE wave is generated. I believe, as this project advances, that I can use PWM to achieve an imperfect but usable SINE wave (or TRIANGLE, or SAW, etc (useful, perhaps, for music making)). The only LOAD that i have applied yet is my oscilloscope and a multimeter--which is hardly a load at all.

    If V+ is greater than 0V, but less than 4V4, the AC peaks similar voltage (actually about a volt less than V+ (and of course the inverted V+).
    If V+ is greater than 4V4 however, the AC voltage does not rise with increased V+. In addition, my power supply indicates "Current Limited". That seems to confirm that I'm doing something wrong.

    It seems that V+AC does not exceed 3V3 provided by either pin A or pin B. This indicates to me that I lack a thorough enough understanding of transistors, and how they can amplify a signal. Perhaps also, nearly every resistor set at 1Kohm is inappropriate.

    In anticipation of helpful responses, I extend my thanks for your reading of this topic.
     

    Attached Files:

  2. Alec_t

    Alec_t

    2,844
    760
    Jul 7, 2015
    Your amps A and B act primarily as level-shifters.
    The collector resistors R3 in amps A and B prevent the bases of Q1 and Q2 in the H-bridge from being pulled up to V+. As a result, both Q1 and Q3 (likewise Q2 and Q4) conduct simultaneously, causing shoot-through.
     
  3. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    If you drive the transformer at too low a frequency, then the core will saturate and the input current will rise as voltage rises but the output will not rise in sympathy.

    You may be able to get away with a 240V 60Hz transformer operating at 90V and 20Hz but it will be close.
    If you have a centre tapped winding of suitable voltage, then a CMOS 4047 and a couple of n channel FETs would be all that is necessary to get a square wave.
     
  4. pidzero

    pidzero

    4
    0
    Mar 27, 2016
    I replaced R3 in both amps A & B with various resistors and measured at LOAD:

    R3 (in both amplifier circuits) | maximum output voltage measured at LOAD
    R3=jumper | nearly 0V
    R3=10Ω | nearly 0V
    R3=100Ω | ± 7V78
    R3=220Ω | ± 8V03
    R3=330Ω | ± 6V08
    R3=1KΩ | ± 3V39
    R3=open circuit | ± 1V95​

    There appears to be a sweet spot at or near 220Ω, where I measure the highest AC output voltage at LOAD so far.

    Still, I cannot drive my power supply past 9VDC at V+; as I increase the variable DC power supply, the supply refuses to go beyond 9VDC. A "Current Limited" LED on the power supply will also turn on at this threshold.

    "Shoot-through" is a new term for me. I believe you are saying that each transistor in H BRIDGE is somewhere between ON and OFF, allowing some current to pass (and therefore short circuit). Is this correct? How might I observe this with a scope?

    I believe I understand, that for a circuit of this nature, MOSFETs would be better than Darlington transistors. My earlier research indicated that. Unfortunately, my local vendor and I don't always keep the same hours, which complicates my middle-of-the-night electronic urges. I definitely want to get my hands on enough MOSFETs to replace the Darlingtons in H BRIDGE. Perhaps I may be able to score a couple of CMOS 4047's as well. I have never used this type of component before.

    Thanks for your responses, Alec_t and duke37! I have a little ways to go, but I'm eager for success!
     
  5. Alec_t

    Alec_t

    2,844
    760
    Jul 7, 2015
    Both Q1 and Q3 will be ON when PinA is high. Both Q2 and Q4 will be ON when PinB is high.
    Connect a low-value resistor (e.g. 1 Ohm) in series with Q1 emitter and use the scope to monitor the voltage across it. When PinA is high the monitored voltage will be non-zero because of the shoot-through.
     
  6. pidzero

    pidzero

    4
    0
    Mar 27, 2016
    Forgive my density; I'm not certain what you mean by "Connect in series with Q1 emitter", and "monitor the voltage across it." I have positioned a 1Ω resistor where it is that I think that you have said (RED resistor in attachment) and measured across the resistor (GREEN points in attachment). Unsurprisingly, my meter reads 0V, leading me to believe that I have misunderstood.

    I beg your pardon; would you kindly clarify?
     

    Attached Files:

  7. Alec_t

    Alec_t

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    760
    Jul 7, 2015
    That's the right place. With the resistors R1-R4 all equal 1k, the load disconnected and PinA at 'logic high', you should see a non-zero (<1V) voltage when the meter +ve probe is on the V+ point and the meter -ve probe is on the Q1 emitter.
     
  8. pidzero

    pidzero

    4
    0
    Mar 27, 2016
    Here are my readings at H BRIDGE's Q1's emitter, Pin A 'logic high', 1Ω resistor in-place, and no load:

    When AMP A's R3=220Ω:
    +Ve and -Ve show only a few millivolts (<9mV), the same sort of readings I get with multimeter's probes disconnected from any circuit at all. (Could we then call this 0V and without shoot-through?)​

    When AMP A's R3=1KΩ:
    +Ve and -Ve show nearly 0V1 (100mV), a non-zero and <1V reading.​

    The take-away here is that both AMP's R3 ought to be 220Ω rather than 1KΩ, as the 220Ω configuration produces almost no (if any) shoot-through. Is this correct? How then shall I drive more voltage to LOAD? Have I approached limitations of my 3A, 30VDC variable power supply? Limitations of the components of my circuit?
     
  9. Alec_t

    Alec_t

    2,844
    760
    Jul 7, 2015
    So, with R1-R4 all equal 1kΩ you're getting about 100mA of shoot-through current. With the Amp'sR3 = 220Ω that reduces to ~9mA. Ideally you should reconfigure the bridge and its driver circuits so that Q1-Q4 have a 'dead time' in which it is ensured that Q1 and Q3 can't possibly be 'on' at the same time. This might require 4 separate 'AMPs' with push-pull outputs. Googling should find you various examples.
    Depends what load current you are trying to draw. Presumably it's now too much for the supply if the supply self-limits to a 9V output.
     
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