Nice DC-DC converter. I didn't expect it to work down to such a low voltage.
Could the overshoot be caused by the electronic load taking its time to detect the voltage and apply a load.
I'm with Psi. I don't see how it makes sense to use an electronic load in constant power mode during turn down and switch on tests. At switch on it will appear like a short circuit, and how many actual loads behave that way? Most real loads have lower power demand at lower supply voltages.
This is relevant to my interests! I was just looking at using non-isolated modules as an easy way to power projects from the 10-15VDC of a lead acid battery backup (worst case discharge to maximum charging voltage). The isolated same-voltage ones like in the video are also neat for powering isolated RS-485 transceivers and whatnot.
Is there somewhere in the US to buy modules like this that might be cheaper/better than the usual distributors (Digi-Key etc.)? DK's prices are still a savings over rolling my own but I suspect there's even better out there.
Noob question, but what is the point of these things? You put 5v in, and get 5v out. Is it something to do with the isolation?
Isolation, and a totally separate ground for the secondary side. Helps with noise, and is needed if you want to power say a amplifier or whatever that is on a high voltage rail, like reading the current flowing in a high side switch and converting it to digital to send to a grounded MCU, or to power a high side switch that is fed a control signal via an optoisolator.
What was it near the end?
0.6V in, 0.8A in. 0.067A out and 0.00V out (says the load). It slowly crawls up.
Then you unplug the DC-DC and it goes back to full 5.15V...
Nice DC-DC converter. I didn't expect it to work down to such a low voltage.
Could the overshoot be caused by the electronic load taking its time to detect the voltage and apply a load.
I would be interested in knowing how the electronic load behaves in constant resistance mode instead of constant power or current. Since constant power/current requires the load to adjust it self to the observed input voltage, at the same time the DC-DC converter tries to adjust its output voltage.
Hence they are fighting each other (which can easily explain your transients during start-up or at the very end of the current range of the electronic load)
(How does the constant resistance mode work in an electronic load, is it still a feedback circuit that adjusts the resistance or is the resistance truly constant during start-up/shut down of a power supply?)
The thing blowing up.
I have figured that much, but how could the bench power supply drop down to 0.6V?
For my USB power supply I will use the DCP020505P which is also a 2W isolator DC/DC converter with an efficiency of 80%. But there’s no minimum load and several devices can be connected in parallel.
Data sheet:
http://www.ti.com/lit/ds/sbvs011k/sbvs011k.pdf
The thing blowing up.
I have figured that much, but how could the bench power supply drop down to 0.6V?
Because of its build in current limit or because all the voltage was dropping over the ammeter.
0.8A as current limit.. sounds about right then. Thanks.
It couldn't have been ammeter drop. It would have to be rather big resistor which on 10A range it is not. Also he had over 400mA with almost unnoticeable drop. 800mA is two unnoticeables only, not 4.2V. 3.36W in ammeter I don't think would be a good thing.
Just one consideration about the MTBF:
A MTBF of 350,000 hours means the component will have a reliability (working without any failure) of just 37% in the first 350,000 hours. If you do the math considering a reliability of 95% (probability of working without any failure), this value is reduced to just 2 years! [ R(t) = e^(-t / MTBF) ].
Thank you very much for all your effort developing the blog.
You do a really nice job.
Just one consideration about the MTBF:
A MTBF of 350,000 hours means the component will have a reliability (working without any failure) of just 37% in the first 350,000 hours. If you do the math considering a reliability of 95% (probability of working without any failure), this value is reduced to just 2 years! [ R(t) = e^(-t / MTBF) ].
Thank you very much for all your effort developing the blog.
You do a really nice job.
It's even a little bit more complex :-) Most people understand MTBF the wrong way. MTBF = MTTF (mean time to failure) + MTTR (mean time to repair). A MTBF of 1000h could mean that a device brakes just after 1 hour usage and the repair takes 999 hours on average. If we want to specify the average time a device works until it breaks we should use MTTF! MTBF also implies that a broken device can be repaired (not just swapping the broken device).
Sagan has grown up!
The day is not far when he will say "Don tuwn it on take it apawt"
In Australia you cannot even change a light fitting unless you are a sparkie, and you need a plumber to change a tap washer. Then of course you are getting $hafted by Al (live as I say not in the mansion I have, and I will sell you the credits I legislated to make you pay) as well.
I'm thinking the same thing as eV1Te. I would like to see how the DC/DC converter behaves on startup when connected to an actual resistor instead of the electronic load.
Why in Gods name doesn't Dave use his uCurrent thingy?
Why in Gods name doesn't Dave use his uCurrent thingy?
Why should he? And does it work up to 400 mA?
Why in Gods name doesn't Dave use his uCurrent thingy?
Why should he? And does it work up to 400 mA?
Unfortunately it only goes to 300 mA, but just change one resistor I believe and it could go higher. Then the burden voltage-drop would be significantly lower than that of the multimeter directly.
Dave have you reviewed the electronic loads performance under different conditions in any video? (e.g. how constant resistance vs. current/power modes behaves compared to a real resistor?)
Load will not behave as a resistor under any circumstance. It is designed to be non linear, either keeping a constant voltage, constanty current or constant power over a range of input. A resistor is very linear, increase voltage and current increases, and power squares.
Load will not behave as a resistor under any circumstance. It is designed to be non linear, either keeping a constant voltage, constanty current or constant power over a range of input. A resistor is very linear, increase voltage and current increases, and power squares.
I assumed that the constant-resistance-mode of the load actually simulates a linear resistor (mosfet with controlled gate voltage that gives specific on-resistance or similar).
If the constant-resistance-mode is some kind of feedback-loop that is non-linear during switch-on/off then I do not understand what the purpose if it is?!
Would very much want to see the scope measure the "real" current (through .1 ohm resistor or similar) and voltage drawn from the load during transient changes.