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I set up an H-bridge inverter with this circuit.

enter image description here

Some say it is not a proper schematic (I don't know how to produce an more proper schematic as it is quite exactly how I did the experiment), so here is an actual photo of the circuit:

enter image description here

  • Top left, under the book, AL10, 6V
  • Top right GBF, e(t) square 'logic voltage' 0V / 5V
  • Left: logic inverter
  • Bottom left: resistors, transistors
  • Bottom right: adjustable R and L for the charge.

Transistors are MOSFET IRL2203N controlled by a 0V / +5V square voltage (e(t)). The charge is a 1kohm resistor.

Here is what I get (dark blue is e(t), cyan is -e(t) and red is proportional to ic(t)):

enter image description here

Notice the transient in the charge (by transient I mean that I'm expecting iC (red curve) to be a squared signal but it seems to be more an RL-ish I0(1 - exp(-t/tau)) signal with tau << 1/f).

With a higher frequency, here is what I get (sorry for changing color background):

enter image description here

The transient is even more noticeable.

Questions

  1. Do you know where does the transient come from?
  2. Did I choose the right transistor to achieve what I want? (I do need a H-bridge with 4 commanded switches but should I go for this transistor, an other, a thyristor, a triac) ?
  3. The charge I'd like to use in real life is not a purely resistive one but a L+R association. Do I have to change something?

Thanks.

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  • \$\begingroup\$ E(t) is shown as a voltage source so it can't have transients on it. Draw a proper schematic please. \$\endgroup\$ – Andy aka Mar 11 '17 at 11:35
  • \$\begingroup\$ @Andyaka See my edit. \$\endgroup\$ – cjorssen Mar 13 '17 at 12:05
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    \$\begingroup\$ A proper schematic and one that doesn't imply a voltage source as the input. \$\endgroup\$ – Andy aka Mar 13 '17 at 12:09
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    \$\begingroup\$ We communicate circuits with schematics here. Cartoons need not apply. \$\endgroup\$ – Olin Lathrop Mar 13 '17 at 12:12
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    \$\begingroup\$ I apologise for the down votes. We hate everyone here ;-) Seriously though, please take this as a genuine question. Given the amount of test kit in your photo (including a 4 channel 'scope with all the channels in use), how is it that you can't produce schematics? I did physics A Level and even then we drew transistors as schematic ones, not as the physical package they come in. \$\endgroup\$ – Paul Uszak Mar 13 '17 at 13:00
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Do you know where does the transient come from?

I don't see what you call a "transient".

Did I choose the right transistor to achieve what I want?

Since you tell nothing about intended load, current, switching frequency, etc, we can't answer.

This schematic will not work, since you use all NMOS, the gates need to be driven a proper voltage above the sources. You need a real MOSFET driver.

Additionally, this has negative dead time, which means both top and bottom MOSFETs will conduct simultaneously and short the power supply. Again, a real driver IC will include dead time management.

Here is an example. Notice charge pump to generate gate drive for high side MOSFETs.

http://www.nxp.com/assets/documents/data/en/data-sheets/MC33883.pdf

This one does not include dead time management, so you will have to do it yourself in PWM settings. There are plenty of chips that would fit the bill, some include other features like overcurrent/short protection, which is also nice to have.

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  • \$\begingroup\$ Thanks for you time (+1). See my edit regarding transient. Please forgive my ignorance, but what do you mean by "the gates need to be driven to a proper voltage above the sources"? Good point about the dead time management. \$\endgroup\$ – cjorssen Mar 13 '17 at 12:10
  • \$\begingroup\$ Since you use breadboard and long wires you'll get all sorts of transients and noise, so don't worry about that. However to drive this NMOS, you need to have Vgs>5V. You will never be able to drive the top 2 NMOS properly with this circuit, you need proper driver. In this circuit they simply act as voltage followers. \$\endgroup\$ – bobflux Mar 13 '17 at 12:21
  • \$\begingroup\$ Great. So now I quite understand whet went wrong. Thanks. \$\endgroup\$ – cjorssen Mar 13 '17 at 16:12
  • \$\begingroup\$ Good! Just to make sure: When the top MOS conducts it is a very small resistor, so both its D and S pins are basically at your +6V power supply voltage. However, to make it conduct, it needs at least 5V Vgs, so its gate should be driven at +11V minimum (relative to your GND of course). You need a driver circuit which does that. Either a bootstrap FET/Bridge driver which makes its own supply, or a driver supplied from a +12V supply that you provide. \$\endgroup\$ – bobflux Mar 13 '17 at 16:31
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Controlling a H-bridge without any driver is kind of...tricky ! So resolving your problem is going to be tough.
You really need a H-bridge driver for your application, and the HIP4081 is exactly what you need :
http://www.intersil.com/content/dam/Intersil/documents/hip4/hip4081a.pdf
I did use it quite a lot of time and it is really great !

If you don't want to purchase one, you have to make a driver with discrete components like so (this is an example there is a lot of discrete drivers for MOSFET) :
enter image description here
The low side driver is the same as the High side driver with the two bipolar transistors. Of course this is only for one branch of the H-bridge. (The voltages depends of your application of course..)

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  • \$\begingroup\$ You can actually really do a +1 :D \$\endgroup\$ – Tagadac Mar 13 '17 at 15:37
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    \$\begingroup\$ Sorry, I totally forgot :) \$\endgroup\$ – cjorssen Mar 13 '17 at 15:58
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The spikes at the edges of the control pulses are easily achieved by long probe gnd routes - maybe the other totally disconnected and by having badly adjusted probes. A little long wires in the circuit does the same.

You really should have a proper schematic as already the others have stated. The oscilloscope measurements are so common cause for errors that we should know how the wires or probes are connected - a detailed knowledge, other is useless.

ADDENDUM due the comment and updated question content:

Your wires are not long, they are VERY LONG. 20kHz square pulses suffer radically and you spread huge disturbing fields like a radio transmitter. If you had a 100uF bypass capacitor between the poles of the supply voltage about max. 2cm off your mosfets, the situation would be easier.

Other: Driving the gates through 1kOhm resistors give no hope for fast switching and even slightly squarewave resembling output at 20 kHz. There are big internal capacitances in the mosfets that need high gate driving current pulses. The beginners very commonly think that mosfets are controlled by voltage only and totally ignore what is needed for rapid state inversions.

And worse: Your mosfets seem to be all of the same polarity - not impossible - but your mosfet gate pulses do not go high enough for the upper pair. I just today noticed that they are only 5V over the GND. The others seem to have noticed this much quickier than me. Sorry for that. Too low drive voltage prevents proper current. In your case it's quite a luck. The slow switching would otherwise cause a serious short circuit to your supply voltage.

About the partially inductive load: Nothing bad because your mosfets have the reverse diodes. But the driving of the mosfets really should leave for the inductive kickback pulses some time to die. this also helps the overlapping ON times directly from +supply to minus. The inverter in not a good way to make the driving pulses. The wait for dying needs some on-turning delay.

If you one day decide to drive some electric motor, you should notice that a running motor is a generator, too. It may not be the best idea to switch the motor at the full speed to reversed voltage.

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  • \$\begingroup\$ Thanks for your time (+1). See my edit. As a matter of fact, I do have long wire in my circuit... \$\endgroup\$ – cjorssen Mar 13 '17 at 12:11
  • \$\begingroup\$ @cjorssen still added some new data \$\endgroup\$ – user287001 Mar 13 '17 at 15:18

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