Electronics > Beginners
Two PSUs in Series – Poor Thinking = Poor Results?
sleemanj:
--- Quote from: t1d on October 20, 2019, 11:14:02 pm ---The load has not changed = 1R.
--- End quote ---
Ok, so both supplies are 3A max, and they are in series, so you can not draw (sensibly) more than 3A, but you have both supplies set to 2.2v, giving you 4.4v, and a 1R load...
Reduce the load (increase resistance, you know what I mean, 1.5R at least) to draw less than 3A and see if that changes things.
Brumby:
--- Quote from: sleemanj on October 20, 2019, 11:25:33 pm ---
--- Quote from: t1d on October 20, 2019, 11:14:02 pm ---The load has not changed = 1R.
--- End quote ---
Ok, so both supplies are 3A max, and they are in series, so you can not draw (sensibly) more than 3A, but you have both supplies set to 2.2v, giving you 4.4v, and a 1R load...
Reduce the load (increase resistance, you know what I mean, 1.5R at least) to draw less than 3A and see if that changes things.
--- End quote ---
Yes. A very simple case of overloading. Ohm's law shows this clearly.
Your supplies would seem to be behaving as I would expect.
t1d:
--- Quote from: Brumby on October 20, 2019, 11:39:08 pm ---Ok, so both supplies are 3A max, and they are in series, so you can not draw (sensibly) more than 3A, but you have both supplies set to 2.2v, giving you 4.4v, and a 1R load...
--- End quote ---
Okay, I was thinking backwards, when I did the math... I was thinking as if they were in parallel.
I will change the load and voltage to see if the amps will stabilize. Thank you for your continued help.
ArthurDent:
There is a problem with what you’re saying about your measurements. With your drawing showing only 1 current path there can only be 1 current so it is impossible to measure one supply current as 2.15A and the second at 2.38A, they have to be exactly the same. If you switched the positions of the two supplies the current measured will still be the same one value. The question is how are you measuring those two currents? If you are relying on the meters on the supplies then they may have a calibration or accuracy problem or just differences between the 2 supplies.
Where the supplies are floating you don’t need the center connection to ground between the two sets of diodes because no current will flow through that ground connection because there is no return path from that ground connection to the voltage output terminals of the supplies. As others have pointed out, you need the diodes across the supplies, not in series with them. The way you have your circuit drawn you have a single loop of 2 supplies in series with 2 sets of dual diodes in series with a resistor load. If your ground (or common) is connected to the center of the 2 sets of resistors then each supply is supplying power separately to half the load and now there can be 2 different currents.
As you now know the load as you have drawn it is not a split rail power source but a single rail. With the load resistor as drawn being 1 ohm the maximum combined voltage from the 2 supplies for 2A output could be about 3V minus the .6V to.8V drop through the diodes plus wiring drops to give about 2V at 2A across the 1 ohm load. That makes sense because you had 1.1V + 2.1V or 3.2V total which is pretty close to my figures. I have re-drawn the circuit to show what I think you meant to do showing values you’d get with the resistors you originally used.
t1d:
--- Quote from: ArthurDent on October 21, 2019, 02:54:24 am ---There is a problem with what you’re saying about your measurements. With your drawing showing only 1 current path there can only be 1 current so it is impossible to measure one supply current as 2.15A and the second at 2.38A, they have to be exactly the same. If you switched the positions of the two supplies the current measured will still be the same one value. The question is how are you measuring those two currents? If you are relying on the meters on the supplies then they may have a calibration or accuracy problem or just differences between the 2 supplies.
--- End quote ---
Please see my post following.
--- Quote from: ArthurDent on October 21, 2019, 02:54:24 am ---Where the supplies are floating you don’t need the center connection to ground between the two sets of diodes because no current will flow through that ground connection because there is no return path from that ground connection to the voltage output terminals of the supplies. As others have pointed out, you need the diodes across the supplies, not in series with them. The way you have your circuit drawn you have a single loop of 2 supplies in series with 2 sets of dual diodes in series with a resistor load. If your ground (or common) is connected to the center of the 2 sets of resistors then each supply is supplying power separately to half the load and now there can be 2 different currents.
As you now know the load as you have drawn it is not a split rail power source but a single rail. With the load resistor as drawn being 1 ohm the maximum combined voltage from the 2 supplies for 2A output could be about 3V minus the .6V to.8V drop through the diodes plus wiring drops to give about 2V at 2A across the 1 ohm load. That makes sense because you had 1.1V + 2.1V or 3.2V total which is pretty close to my figures. I have re-drawn the circuit to show what I think you meant to do showing values you’d get with the resistors you originally used.
--- End quote ---
Your solution is similar to Zero's at Post #4. I am presently using Zero's schematic. I had wondered about his connection from between the diodes to circuit ground. I am still scratching my head about that. IIUC, you do not include it, because there is no return path; correct?
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