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.]
I know what you are going through. I only know enough to be dangerous.
What I would do is use the TI Workbench. I'd find a part that fits the In Out and amps. Which part did not matter because once I'd hit the Back button, select "Compare All Part Type" and it gives a list of various design based on various parameters.
Then once in the selected design I'd look at the schematic they created. Like yours, 12Vin 1V Out @ 1 amp. This is the schematic :
From a glance at this screen, I can instantly see the design has
- 94% efficiency
- $10.23 BOM
- 100 khz switching Freq.
- IC Tj 37.8°C
- M1 Tj 41.1°C
- M2 Tj 45.5°C
- Vout P-P 0.012v
Then I'd play with the different "advanced options", efficiency, switching frequency and see which values changed in the new schematic.
When it comes down to it , I have no clue why they selected a 25v and 30v, 100 Amp MOSFETS
So I go into the BOM and click on the part look at the datasheet for some clues. Then click "select another part" and it will let you select another part but only devices that will work with the current parameters and design. Here you learn what the characteristics that limit their inclusion in the selection list.
I would also run a Thermal Simulation. It would show were all the inefficiencies were in the "Power Dissipation".
For this design Cin 0.02 W, Cout 4.32mW, IC 0.11W, L1 0.12W, M1 0.22W
Anyhow I thought it was an interesting way to learn and thought it may help you as well.
Best Answer
If it is confirmed the MOSFET needs a better drive then the best is to advise a suitable circuit. A simple bootstrap circuit described by Monsieur Balogh in a TI application note can do the job nicely in a cost-sensitive application. As noted in some of the comments, the UC384x was not really meant for hi-side drive - unless you make it entirely float and tie its GND pin to the MOSFET source and supply the IC via the rectified buck output - but this little circuit does the job fine:
Below is the circuit I tested with component values:
This is excerpted from a seminar I taught in an APEC seminar in 2019.