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IamSmooth

Fri Oct 08 2010, 03:44AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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I have been reading about the TL494 for pwm. It seems the duty varies from 0 to 45%. Whatever happened to 0 to 100%? Why isn't the chip designed for this? If this is the case, how does one ramp up a buck converter to unity? Does this mean one has to use some other means for pwm like a 555 and a comparator?

Patrick

Fri Oct 08 2010, 04:23AM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
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IamSmooth wrote ...

It seems the duty varies from 0 to 45%. Whatever happened to 0 to 100%? Why isn't the chip designed for this?

i believe the 0-100% may imply no switching takes place, also 0-45% applies to each output of the tl494, so you can get up to 90% switching power-time through a switching half bridge, the missing 10% is usually calculated to be used as a convientent deadtime between switch transistor times. so 45% power conduction + 5% deatime =50% for the A output, with the same 45%+5% = 50% for channel B. thus the time base = 100%. i think i have said all this right.

Steve Conner

Fri Oct 08 2010, 12:00PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
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If you connect both outputs of the TL494 together, you get 0 to 90% duty and double the switching frequency. Some of the old TL494 datasheets have an example circuit for a buck converter made in this way.

The duty is limited to 90% so that you can use a bootstrap circuit, like I explained in the other thread.

Other ICs are available more suitable for buck converter control, like the UC384x. (There are half a dozen versions, I can never remember which is the right one

)

Mads Barnkob

Fri Oct 08 2010, 12:15PM

Registered Member #1403 Joined: Tue Mar 18 2008, 06:05PM
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As the datasheet states in the start the dead time control comparator have a fixed offset that gives about 3 - 5% dead time (varies with different makers), so running it as push-pull will give you 0 - 45%.

You could run a totem pole or darlington that drives a single mosfet for a higher duty cycle? I know the tl494 can be run single ended but I have forgot all about if that gives 0 - 90% duty cycle with just that option...

IamSmooth

Sun Oct 10 2010, 04:58AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Do the error amplifiers on pins 1/2 and 15/16 cut off the pwm? If I want to use the chip to control a buck/boost converter, what is the best feedback mechanism to use when I achieve the voltage? Do I compare a divided output voltage to a reference on the error amplifier, or do I use the FEEDBACK pin?

Steve Conner

Sun Oct 10 2010, 08:37AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

IamSmooth wrote ...

Do I compare a divided output voltage to a reference on the error amplifier

Yes. The FEEDBACK pin is for connecting the compensation network to. RTFDS. The new TI datasheet isn't that good, but there are older TL494 datasheets with application circuits in them.

IamSmooth

Mon Oct 11 2010, 01:59AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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A bootstrap is needed with the high side driver because the source voltage rises with the drain when the igbt opens. The bootstrap raises the gate to still keep it open.

But what if the high voltage going to the drain comes from a separate isolated power supply such that there is no potential from the line to the 15v power source to gate drive chip? Would not the source voltage, with respect to the gate, connected to the chip, stay the same without the need of a bootstrap?

Steve Conner

Mon Oct 11 2010, 08:36AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Yes. This is how it's done in applications that need to run 0-100% duty cycle. That's why we told you about DC-DC converters and so on in that other thread.

Edit: I see you said "drain" in that sentence above. You're taking about the low-side device in a bridge, or the switch in a boost converter.

But, this isolation condition can't hold for BOTH devices in a half-bridge. One of them must have an isolated gate drive.

You can rearrange a buck converter to make it hold, in fact this is how cheap PWM motor controllers without reverse or regen braking are made. But then the output voltage is referred to the input rail, not ground.

IamSmooth

Mon Oct 11 2010, 10:50PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Clearly, there is something I don't understand. I have a pwm signal going to the TC4421. This chip has the necessary bypass capacitors and is powered from its own 12-15v supply. The OUT pin goes to the gate on the IGBT. The emitter goes back to the TC4421 ground to complete the circuit. I can scope a pulse wave on the gate.

I now take an isolated 50v dc voltage and put it on the collector. A resistor from the emitter back to the 50v source completes this pathway.

What happens is that as I increase the collector voltage from 0 to 50vdc, the voltage across the resistor rises as a DC value. That is, the igbt does not switch - it is always on.

Why did this fail? This worked when I had a gate drive transformer secondary completing the Gate-emitter pathway, and it still switched an isolated voltage. When the igbt completes the circuit I know the emitter voltage goes up, but it should not have any potential when referenced to the gate as it is from an isolated supply.

EDIT: If I disconnect the lead going to the gate and I deliberately take the lead to the common low point the isolated voltage on the resistor goes to zero; if I put the gate lead high then the isolated voltage on the resistor goes up. It just does not do it through the TC4421. If I disconnect the isolated voltage and connect the 15v supply going to the TC4421 to the igbt collector, then everything works. So, the problem is with the isolated voltage, or maybe not enough time for the igbt to turn off. Do I need the gate voltage to go negative to turn it off?

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