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Avi

Mon Feb 18 2008, 05:56AM

Registered Member #580 Joined: Mon Mar 12 2007, 03:17PM
Location: Melbourne, Australia
Posts: 410

I have constructed the following circuit in order to work out how to use the IR2110 chip. Unfortunately i can not get the high side led to come on at all, even if i change the capacitors to 100uf.
Any ideas what I'm missing here or did wrong?

GeordieBoy

Mon Feb 18 2008, 11:00AM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

The circuit you have drawn with LEDs as loads won't work properly without the external MOSFETs present. The 2110 is designed to drive a bridge-leg (totem-pole) of MOSFETs. The bootstrap voltage source for driving the top device relies on the bridge swtiches being present in order to charge the bootstrap capacitor on pin 6 through the diode. (Specifically the bottom switch pulls Vs down to COM when it is turned on, this is required to charge the bootstrap cap on Vb) Also note that the 1N4007 diode needs to be a fast-recovery UF4007 if you intend to operate at more than a few hundred Hz.

Avi

Fri Feb 22 2008, 01:39PM

Registered Member #580 Joined: Mon Mar 12 2007, 03:17PM
Location: Melbourne, Australia
Posts: 410

about the diode, i realize this, but i was only using a mechanical switch at 1Hz at the time.

I put together the following tests circuits (see attachment) and here was their results:
circuit 1 - always works

circuit 2 - may work, frequency may be unstable, might skip pulses, may lock up, in all cases eventually stops working

circuit 3 - may work, otherwise stays on a single state and never changes

which brings the question, if circuit 1 is the correct one, then wouldn't that just make the vcc to com capacitor mostly useless (see it has the supply right across it, no its not for decoupling current spikes)

1203687545 580 FT39254 Ir211 Test Circuits

Avi

Wed Aug 20 2008, 12:56PM

Registered Member #580 Joined: Mon Mar 12 2007, 03:17PM
Location: Melbourne, Australia
Posts: 410

So now I have the following circuit (see attachment). My question is why when the high side is switched on (lowside switches off), it only puts a short pulse less then half a second between the high side G S junction, but yet the low side when active, provides a continuous 12v.
I have tested this with an LED, a digital multimeter, or even no load except the gate itself and watching the load being pulsed for a short amount of time after the transition of the input.
I have tested all Lin and Hin inputs and they behave correctly.

1219236983 580 FT39254 Fullbridge Driver

Steve Conner

Wed Aug 20 2008, 04:10PM

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

That's exactly what you'd expect. Bootstrapped drivers can't turn the high-side MOSFET on indefinitely, because the bootstrap capacitor that powers the high-side driver slowly drains until it can't keep the gate on any more. The high-side must be turned off periodically and the low-side turned on, to recharge it.

By the same token, the Vcc-to-Com capacitor isn't useless, because it decouples the current spikes caused by recharging of the bootstrap cap.

Avi

Thu Aug 21 2008, 01:21AM

Registered Member #580 Joined: Mon Mar 12 2007, 03:17PM
Location: Melbourne, Australia
Posts: 410

wrote ...
That's exactly what you'd expect. Bootstrapped drivers can't turn the high-side MOSFET on indefinitely, because the bootstrap capacitor that powers the high-side driver slowly drains until it can't keep the gate on any more. The high-side must be turned off periodically and the low-side turned on, to recharge it.

Oh, why is that? 0.68uF into 2.6nF should have plenty of spare charge after the current has finished charging the gate.
Does this mean there is a formula for the capacitor value to use for a given operating frequency?

wrote ...
By the same token, the Vcc-to-Com capacitor isn't useless, because it decouples the current spikes caused by recharging of the bootstrap cap.

ok thanks, wondered what if anything that was for.

Experimentonomen

Thu Aug 21 2008, 09:21AM

Registered Member #941 Joined: Sun Aug 05 2007, 10:09AM
Location: in a swedish junk pile
Posts: 497

You may want a separate floating supply for highside if you wanna put out dc.

Dr. Dark Current

Thu Aug 21 2008, 10:10AM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

Avi wrote ...

Oh, why is that? 0.68uF into 2.6nF should have plenty of spare charge after the current has finished charging the gate.

The high side driver draws its (small) supply current, this will slowly discharge the cap.

Avi

Thu Aug 21 2008, 01:28PM

Registered Member #580 Joined: Mon Mar 12 2007, 03:17PM
Location: Melbourne, Australia
Posts: 410

hmm...
well even if you put a separate supply, if it also applies charge back into it, that cant be too good for the chip.

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