Jan 20, 2013 The IR2110 circuit uses VCC (not VDD), the bootstrap diode and the bootstrap capacitor to create the additional required voltage. The capacitor charges to (VCC - Vb)V when the high side MOSFET is on and VS is pulled to ground. When VS is no longer pulled to ground, the capacitor has a potential difference of (VCC - VB)V across it. Following figure shows the bootstrap capacitor in IR2110 from its datasheet. The bootstrap capacitor holds enough energy to supply the 'HO' Driver circuit. An analytical calculation of this current which is attached as PDF (Attachment-2:.
I am currently learning how to calculate the bootstrap capacitor for IR2110 Application. So far I have identified every parameters apart from one:
Qgate 27E-06
Frequency 21600
Icbs(leak) 0
Qls 5.00E-09
Vcc 14
Vf 0.78
Vls 4.5
Iqbs(max) 230E-06
Vmin ??
I use the forumla from Application Note AN-978 on page 6. At the moment I am not certain on how to obtain Vmin, I use a half bridge configuration and the voltage is about 100V and the IGBT is IRG4BC20UD.
Please explain me how to obtain Vmin?
Regards
MM
Some of you may remember that I posted a question where I requested some feedback on my PCB design. The design was for a Half-Bridge driven by a IR2113 chip. I went ahead and etched the PCB but unfortunately, only the low side MOSFET seems to switch. In my initial calculation, the bootstrap capacitor's value came out to be 1uF or so. Initially, since the circuit didn't work, I decided to change the capacitor to a higher value (100uF).
That didn't cause any change in the circuit. However, if I short out the capacitor the High-Side does turn on. The gate voltage is around 11V. If I don't short out the capacitor, the gate voltage is around 5.5V.
The low side seems to switch just fine. There is a bit more weirdness to the circuit: the SD pin is supposed to low in order to enable the chip - yet, having it high makes no difference to the output. I am still able to switch the Low-Side MOSFET. Furthermore, it seems that somehow the Lin and Hin pins are switched. If I connect the Lin pin to High, it doesn't switch on/off the Low-Side. Infact, if I take the Hin pin and take it High, it causes Lo output to turn on the Low-Side.
The following is my circuit. C2 is 0.1uF and C1 (bootstrap) was 1uF (as per calc.) but I've changed it to a larger value (100uF) incase I was wrong somewhere. For the load, I have a 1K resistor connected. Vcc is 12V.
Previous Schematic: http://i.stack.imgur.com/T3fRn.png
I have updated the schematic. I hope it looks better now. if there if still some confusion, please let me know. I sincerely apologize for wasting the time of folks here. I understand that the schematic was hard to follow. I drew it in haste and I was also not familiar with ERC and DRC checks and I've just started to read up on them.
Getting back on topic: the PCB layout hasn't changed so I'm not sure if if I should just etch another PCB altogether or not. One of my main concerns is that in the datasheet, the source of the upper MOSFET and the drain of the lower MOSFET are not connected. Instead, they are connected to the load. However, in the various schematics I've seen online (case in point), they are connected. Would this make any difference?
Also note that the datasheet has two bypass caps. One is connected between Vcc and Gnd. The other is connected between Vdd and Vss. In my case, the Vdd is tied to Vcc therefore I'm only using one bypass cap.
I again apologize for the poorly drawn schematic that I posted previously.
4 Answers
$begingroup$There are a number of uncertainties in what you say you are doing. You need to stabilize the patient and understand what is really going on in order to make progress. At present there are several things wrong which are obscuring each other. Being able to deal with each ne in isolation makes life much easier.
(1) Your sentence below does not make sense. Can you please explain more clearly what you mean. 'The MOSFETS are connected' and 'drain and source ... connected' can both mean the same thing. If they DON'T mean the same you need to explain what you mean. You said:
- ' My main concern is that in the datasheet schematic, The mosfets are connected. Whereas in my circuit (and others that I've seen over the web), the drain and source are connected together.'
(2) This would be fatal:
- My Vdd is the same as Vss, and so I'm using the same cap. for both of them - assuming thats ok.
As Oli says, presumably you mean Vdd = Vcc. You need to read what you write before sending. We all need to and we all get it wrong sometimes BUT when you are asking the questions and want help then confusing the assembled masses with typos is a very bad idea.
(3) This probably means that your IC is dead or walking wounded OR you have a disconnected line - possibly ground. When things go this wrong you need to carefully measure everything - voltages when on and ohmic connections when off. look for shorts AND opens.
- ... the SD pin is supposed to low in order to enable the chip - yet, having it high makes no difference to the output. I am still able to switch the Low-Side MOSFET. Furthermore, it seems that somehow the Lin and Hin pins are switched. If I connect the Lin pin to High, it doesn't switch on/off the Low-Side. Infact, if I take the Hin pin and take it High, it causes Lo output to turn on the Low-Side.
(4) If you were using 'real' voltages your IC and random other things would now be dead. Shorting the capacitor connects the isolated high voltage 'island' in thje IC to the drain of the lower FET which gets connected to ground when the lower FET is turned on. Shorting this cap could be an extremely exciting and non productive thing to do in many situations. Hopefully you determined before you did this what you expected to happen and didn't just do it to see what happened. When you are dealing with power rather than signal the magic smoke is never too far away. You may have had some already without being aware of it.
- That didn't cause any change in the circuit. However, if I short out the capacitor the High-Side does turn on. The gate voltage is around 11V. If I don't short out the capacitor, the gate voltage is around 5.5V.
(5) Making the capacitor 100 times larger than you calculate in a circuit that may be switching at 10's to 100's of kHz is highly likely to produce interesting results. These could include an interesting emulation of lot's wife. But may not. The capacitor has to charge in the time that the low side FET is on. It discharges when the high side FET turns on. There will be charge & discharge time constants controlled by the resistance in the IC power paths. This may still work OK OR the voltages may only rise to a fraction of what is intended in the time available. Which may be consistent with what you are seeing.
I could add a bit more but that should do to start :-). Sometimes we all have a bad day - you need to try and not let too many things at once get out of control as then you can't asily analyse what is going wrong.
As others (and I) have suggested, measure everything you can and see if it makes sense. If you have an oscilloscope see what it can tell you. If you don't have a'scope, start putting lunch money aside for one. Even a relatively cheap scope can be a mightily powerful tool. An oscilloscope is possibly the most effective and powerful debugging and fault finding aid you will ever have for analog circuits.
- In my initial calculation, the bootstrap capacitor's value came out to be 1uF or so. Initially, since the circuit didn't work, I decided to change the capacitor to a higher value (100uF).
Practical aspects:
If hand driving at low speed, connect Hin and Lin and SD low with pull down resistors. Then, if you hand switch them and they 'bounce' they will bounce from low to applied signal level and not to some unknown state.
BUT the bootstrap powering circuit for the upper gate completely relies on there being an AC signal at VS to supply AC at Vb which draws power from Vcc and delivers it to Vb. If there is no AC your =upper gate signal will decay 'rapidly'. How long it takes depends on upper driver power consumption and was part of your capacitor calculations (from memory). This is an example of where an oscilloscope will help you see what is happening. The 100 uF you are using is a large value and decay time may be long enough to see what is happening 'by eye', but maybe not.
If you want to drive with a microcontroller you could reduce Vcc to Vdd rather than raising Vcc - as long as IC minimum voltage spec is still met.
Re-etch of PCB MAY be a good idea BUT you should be able to carefully go over circuit and check that what you have is what you intend. Do it pin by pin.Talking to yourself about it as you go can help :-) (really). Describe what you expect to see and what you really see and why they are or aren't compatible. [Watch for men in white coats observing you suspiciously when you are talking to yourself - or do it in your head or, works well, get a knowledgeable friend and explain it to them. The very act of explaining often works wonders.]
This is a great example why neatness and attention to detail matter. Sloppiness like drawing nets right thru other parts leads to mistakes or at least makes it less likely mistakes will be noticed (because it already looks like a mistake so the brain tunes out the real ones).
Look carefully at the schematic. Note that the anode of the diode and one side of C1 do not have a dot showing that the software thinks there are connections there. Maybe it does, but then the schematic is misleading. If the software (looks like Eagle?) doesn't think those nets are connected, then they won't be connected on the board. Both those parts create the power supply for the high side drive, so these issues match your symptoms.
Once again folks: Neatness and attention to detail matter!
This should be self-evident, but apparently there are a lot of sloppy people out there. There is no place and no excuse for this in engineering. This schematic is sloppy in various places. I would never dream of letting a customer or anyone else see a schematic from me that looked like this. I would be embarassed to let anyone see this, and you should be too. The fact that you're not says a lot about you. Doing it right is really quite easy and hardly takes any time. Just avoiding one such error makes it worth while. How long have you spent diagnosing the problem and will spend fixing it, compared to maybe 5 minutes to clean up the schematic and run the ERC and DRC checks? Don't draw nets thru or right up against parts, if you rotate a part fix the text so that it is normally readable again, make sure the text doesn't collide with other things, etc. Again, this should all be self-evident. Hopefully you have learned a lesson here and will be more careful in the future.
As Olin mentions, the schematic looks like it has a few (unintentional most likely) mistakes that will be propagated through to the PCB. Looks like there may be a (according to datasheet diagram) bypass cap missing too.
I wouldn't worry too much if this is the first attempt at doing this stuff, and it's a personal project.
Take care with drawing the schematic, learn about and get into running ERC/DRC checks regularly, and these kind of mistakes become a lot harder to make.
Substituting a 100 uF capacitor for a 1 uF capacitor is not valid at higher frequencies because the ESR and ESL of a tantalum or electrolytic capacitor come into play and become dominant at higher frequencies.
Stray inductance caused by layout can cause problems. Make sure these loops enclose a minimum of area:
- IC1 VB to C1 and back to IC1 VS
- IC1 HO to R1 to Q1 gate, and Q1 source back to IC1 VS
- IC1 LO to R2 to Q2 gate, and Q2 source back to IC1 COM
- HIN driver OUT to X3 pin 1 to IC1 HIN, and IC1 VSS back to HIN driver GND. Yes, you probably need to route a ground line with the signal; expecting GND to magically stay at 0 V via X1 pin 2 is probably not enough
- LIN driver OUT to X3 pin 1 to IC1 HIN, and IC1 VSS back to HIN driver GND.
Check VIH and VIL for IC1 HIN and make sure you're driving the right levels.
VC (Q1 drain) doesn't seem to be hooked up to a power supply.
Also, the 1N400x series diode might be too slow for the application. Try a Schottky or ultra-fast diode.