PM,
V2 is a DC PS 0-15.5V / 40A (Manson SPS9400), so should have enough output capacitance IMO.
The problem is that i have no understanding in what you are trying to do with this Vload circuit and how it supposes to work.
Perhaps you can (again?) explain in some detail what your goal is with this circuit, so i could try to find the problem in it.
I measure 153mA average through R2 during the 250us on time (every 100ms) when M1 switches to ground which is as expected for a 100 Ohm resistor at 15.5V.
I measure 317mA average right behind D2 coming from C1 (7 pulses) during that same on time every 100ms.
Itsu
OK, it is my mistake to assume that everyone understands the circuit operation! I will reference the circuit used in my post #519 and have listed both the schematic, SP1, and SP4 below.
Now, look at the SP1 scope shot. First, before we start the seven pulse sequence, we have the potential at Vc1 of 21.55v that is equal to the potential at Vload due to R1.
At the first positive pulse from our SG to the 3/4 bridge, we create a positive voltage on VL1 while VL1a is ~ 0v. This immediately creates a positive voltage rise on C1 via charge separation that is ~1 V/t (~3.2v in this case) above Vload minus the voltage drop across D2. This voltage increase across C1 creates the half sine looking current seen on CH4(grn) that is flowing
into Vload. CH3(pnk) connected to Vload shows the resulting increase in Vload during the time the SG pulse is positive.
Then, when our SG pulse goes to 0v, VL1 now goes to ~0v and VL1a goes positive. This immediately creates a negative voltage drop on C1 that is again ~1V/t or ~-3.2v. We don't see this negative voltage drop on C1 at this time because CH3 is connected Vload which has enough internal capacitance to hold Vload relatively constant as D2 decouples the negative drop on C1 from Vload. However, we do have R2 connected to Vload during this time and it's loading creates the negative going voltage ramp we see on CH3 during the time the SG pulse is 0v.
SP4 gives us a view of the alternating voltage changes on C1 with CH3 connected to Vc1.
So, during this first complete cycle, we have supplied energy to the Vload supply and to our load resistor R2. This is our goal.
This same action repeats over the next 6 cycles.
The reason that the current in CH4 is seen to go negative during the time C1 is decoupled from Vload is because Vload is supplying energy to R2 during this time in the cycle. During the rest of the cycle, C1 is supplying energy both to Vload and R2.
I hope this makes sense and helps.
Pm
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Topic: partzmans board ATL (Read 36189 times)
