New Siglent SPD4000X series power supply

[Furna]

They're on the water so waiting impatiently.......
Top and bottom models are coming, not the middle one at this time.

Are you also going to teardown them?
I have the middle one so we are covered ... if I get courious enough and have something to compare to, I could teardown mine.
I do not have a pro setup for photos/videos tough.

[tautech]

They're on the water so waiting impatiently.......
Top and bottom models are coming, not the middle one at this time.

Are you also going to teardown them?

Defpom will get one to have a good look at. 

[jayk]

Will be interested to hear about the web interface on these.  The GPP-4323 is great but has a crap web interface.  Hopefully Siglent did a better job with the SPD4000X.

[Furna]

Will be interested to hear about the web interface on these.  The GPP-4323 is great but has a crap web interface.  Hopefully Siglent did a better job with the SPD4000X.

An image is better than 1000 words?

[mawyatt]

Could you  do some of the tests discussed here:

https://www.eevblog.com/forum/testgear/power-supply-for-home-lab-do-i-really-need-a-rs/

Specifically with a Voltage and Current Limit of 20V and 10mA respectively using a Blue LED as load DUT. Record Turn On Characteristic of Voltage (or Current) at DUT.

Also do the above with Voltage Limit of 8V and 50Ω DUT.

The GPP-4323 seems to be the best of the bunch.

Thanks in advance if you have the time

Best,

[folays]

SPD4306X ; set to 0.020 A.
3 screenshoots : 5V / 10V / 20V.

Green led, vendor says Vdrop=2.2V 20mA.

The green leds didn't like that. 3 are dead, some are more resilient. They flash yellow at 20V, but some still continue to work.

Sorry it's only a scope measurement of Voltage at the led terminals.
3V per horizon div, 2ms per horizontal div.

At 20V, the SPD in CC=20mA settles at 3.2V.

I will insert a resistor in series to measure current.

[folays]

Here is fresh green led (not one I martirized before) at 3V / 5V / 10V / 20V.
The resistor is a really cheap one marketted as 100 Ohm but real BM869's measured resistance was ~ 90.00 Ohm.

For what is worth, wiring is : Vcc -> pushbutton -> Led -> Resistor -> Ground.
I didn't connected the SPD "sense" leads, but I had the BM869s measuring Vcc at the breadboard, got .00 decimals at each of the 4 voltage ranges.

Here are 4 traces at 50ms / div.

The yellow C1 trace is the SPD4306X Voltage.
C2 (not displayed) is the resistor Voltage drop.

F2 (math) is C1/90 ; so that's the current. At 30 mA / div.

At 20V, the current peaked at 169mA.

Bonus 5th picture at 20V and 10x more zoom (5ms / div). Beware the vertical division of Amperage has changed, it is now 50mA / div.
Peak start at 170mA, after 45ms it seems to be still at ~120 mA.

[mawyatt]

You want to be using the ON/OFF Enable Button on the Supply, not an external Push Button. The Push Button places the PS output capacitance fully charged to set voltage across the DUT. At higher set voltages like 10 and 20V this will damage/destroy the LED as too much energy gets dumped into the DUT LED.

Best

[folays]

Here using the button of the SPD.
I missmeasured the resistance (for calculating the current) it is 99 Ohm, also checked with Shannon's tweezers.

F2 = Vresistance / 0.099 Ohm (scope wouldn't let me graph F2 below 5mV/div at 3mV/div so I divided by 0.099 to graph in V/div).

F2 Rise slowly to 19+ V, it means 19mA, as set on CC supply (20mA).

I like the fact that we can see the current rising once we approach the Vdrop of the LED.

[tautech]

Here using the button of the SPD.
I missmeasured the resistance (for calculating the current) it is 99 Ohm, also checked with Shannon's tweezers.

F2 = Vresistance / 0.099 Ohm (scope wouldn't let me graph F2 below 5mV/div at 3mV/div so I divided by 0.099 to graph in V/div).

F2 Rise slowly to 19+ V, it means 19mA, as set on CC supply (20mA).

I like the fact that we can see the current rising once we approach the Vdrop of the LED.

Q
Why have you not set the scope to show A with the correct scaling ?
Hint, look in the channel menu. 

[folays]

Why have you not set the scope to show A [...]

The Math channel can't let me set the unit on it, Ohm's law doesn't exist, and if Math channel divides Volts by Resistance, then Scope law's which is above Ohm's law mandates that the result is still Volts

I could force Math channel to be in Amp, if I were to set the C2 (Vdrop at the resistance) to be Amps.

I very would like to be able to change the Math channel independently, or if the scope would let just enter a suffix string.
I've entered the Steinhart-hart equations in a math channel for the fun, and I had to display the temperature in V.

[...] with the correct scaling ?

The C2 channel was set a 1V/div.
F2 was set at F2=C2/99 (99 Ohm).
So I indeed had the target LED's 20mA, but displayed as 20mV, but if I changed the C2 unit from V->A, then I could have F2 change from mV->mA.

But I wanted the current (~20 mA, or 20mV badly displayed) to be graphed with the 7 full spans on division available in the positive range.
so 20m[A or V] divided by 7 = roughly 3 wheen rounded to ceil.

So I wanted to set F2 to 3 mV vertical div, but the scope wouldn't let me go below 5mV. Without any message, even if I had the "coarse" unchecked, when I entered something below 5milli, it raised it up to 5milli.

Attached is a picture when I set back F2=C2/99, then I can't set F2 vertical div to below 5mV. When I lower it, it ignores it and briefly writes a big 5mV in White in the bottom center of the screen.

So I changed F2 from F2=C2/99 to F2=C2/0.099 and set the vertical div of F2 from 3mV(impossible minimum 5mV) to 3V.

Ah and in the oddities, the Steinhart-hart equation is something along :

Code: [Select]

// let a, b, and c be three coefficients which to me, may as well have been determined by black magic
F3=ln(voltage)// natural log, not log10 // F4 and math channels have a string length limit, so I have to use a temporary channel to reduce size
F4=a + b*F3 + c*F3*F3*F3
The scope wouldn't allow have F3 be cube'd to the power of 3.
So It worked with the mathematically-equivalent of factorized :

Code: [Select]

F4=a + F3*(b + c*F3*F3)
And somehow it accepted it, like if we weren't anymore be ^3 a chan source.
I guess that it comes from limit in the math operations accepted by whatever chips calculate the maths (FPGA ?).
Probably the FPGA cannot do F3*F3*F3, probably it can do F3*F3*(F3+0). Or something along those lines.

I guess square/square root are often used in the electrical domains, cube/cube root less often, and I wanted to ^3 but it was for something probably rarely usefull on a scope.

[mawyatt]

Can't you just set the CH2 Units to Amps, the Probe to proper V/A? The DSO will then display in mA

100Ω is large for current measuring, as 100Ω * 20mA = 2V.  Try 1 or 10Ω for less voltage drop, your nice DSO should have plenty of low noise sensitive to handle this.

Best

[KungFuJosh]

Probably should put the 20M channel filter on too.

[folays]

C2 set to A, with 20M filter.
C2 measured Vdrop changed to ~ 1.53 Ohm.

Still at 5mA div on F2, cannot set it below to 3mA per div.

[Furna]

Could you  do some of the tests discussed here:

https://www.eevblog.com/forum/testgear/power-supply-for-home-lab-do-i-really-need-a-rs/

Specifically with a Voltage and Current Limit of 20V and 10mA respectively using a Blue LED as load DUT. Record Turn On Characteristic of Voltage (or Current) at DUT.

Also do the above with Voltage Limit of 8V and 50Ω DUT.

The GPP-4323 seems to be the best of the bunch.

Thanks in advance if you have the time

Best,

Hi,
this what I have been able to do so far.

Setup: SPD4121X
- 2*100Ω wire wound cement resistor in parallel => 50Ω (see IMG_20241006_125209.png)
- CH1 8V - 0.010A
- CH2 8V - 0.010A

Results:
- CH1 overshot to 3.48V => 0.0696Ω (see CH1-8V-10mA-50R-20msec.png)
- CH1 overshot to 3.48V => 0.0696A (see CH1-8V-10mA-50R-20msec.png)
- CH1 recover to to 0.50V in about 30msec (see CH1-8V-10mA-50R-50msec.png)
- CH2 overshot to 1.32V => 0.0264Ω (see CH2-8V-10mA-50R-20msec.png)
- CH2 overshot to 1.32V => 0.0264A (see CH2-8V-10mA-50R-20msec.png)
- CH2 recover to 0.50V in about 90msec (see CH2-8V-10mA-50R-50msec.png and CH2-8V-10mA-50R-100msec.png)

So the "issue" is present on SPD4000 series too, anyway with lower magnitude if I read correctly the other posts.
CH2 overshots less than half than CH1 but recovers much slower, anyway much faster than 12 sec I read in other posts.

Give me some time to better read the posts and see if I have a spare blue LED around.

Edit: I clearly made a copy and paste error mixing Ω and A ... sorry for misleading

[mawyatt]

Thanks Furna, interesting results!!

Think starting a new thread on Power Supply Turn ON and OFF characteristics might be in order to consolidate the various efforts of folks on this issue.

Best,

[Furna]

Setup: SPD4121X
- Generic Blue LED (see IMG_20241006_151727.png)
- CH1 12V - 0.010A
- CH2 12V - 0.010A

Results:
- CH1 overshot to 3.48V => PSU shows 2.840V 0.0.10A in long run (see CH1-12V-10mA-BlueLED-10msec.png and CH1-12V-10mA-BlueLED-50msec.png)
- CH2 no overshot => PSU shows 2.840V 0.0.10A in long run (CH2-12V-10mA-BlueLED-10msec.png and CH2-12V-10mA-BlueLED-50msec.png and CH2-12V-10mA-BlueLED-500msec.png)

Awaiting comments

PS SPD4121X CH2 max voltage is 12V
For this reason I choose to use 12V for all tests with Blue LED

[mawyatt]

Thanks Furna

Interesting how the two channels differ in details!!

Edit: BTW we don't see this with the SPD3303X, see link below.

Note the thread started here.

https://www.eevblog.com/forum/testgear/lab-power-supply-turn-on-and-off-characteristics/

Best

[MildInductor]

Setup: SPD4121X
- Generic Blue LED (see IMG_20241006_151727.png)
- CH1 12V - 0.010A
- CH2 12V - 0.010A

Results:
- CH1 overshot to 3.48V => PSU shows 2.840V 0.0.10A in long run (see CH1-12V-10mA-BlueLED-10msec.png and CH1-12V-10mA-BlueLED-50msec.png)
- CH2 no overshot => PSU shows 2.840V 0.0.10A in long run (CH2-12V-10mA-BlueLED-10msec.png and CH2-12V-10mA-BlueLED-50msec.png and CH2-12V-10mA-BlueLED-500msec.png)

Awaiting comments

PS SPD4121X CH2 max voltage is 12V
For this reason I choose to use 12V for all tests with Blue LED

Is this overshoot likely to damage any general IC circuitry? Say if I was supplying 3.3V to a general purpose IC that was only rated to operate up to 3.3V. Would the spill over be enough to damage it generally speaking?

[Furna]

Setup: SPD4121X
- Generic Blue LED (see IMG_20241006_151727.png)
- CH1 12V - 0.010A
- CH2 12V - 0.010A

Results:
- CH1 overshot to 3.48V => PSU shows 2.840V 0.0.10A in long run (see CH1-12V-10mA-BlueLED-10msec.png and CH1-12V-10mA-BlueLED-50msec.png)
- CH2 no overshot => PSU shows 2.840V 0.0.10A in long run (CH2-12V-10mA-BlueLED-10msec.png and CH2-12V-10mA-BlueLED-50msec.png and CH2-12V-10mA-BlueLED-500msec.png)

Awaiting comments

PS SPD4121X CH2 max voltage is 12V
For this reason I choose to use 12V for all tests with Blue LED

Is this overshoot likely to damage any general IC circuitry? Say if I was supplying 3.3V to a general purpose IC that was only rated to operate up to 3.3V. Would the spill over be enough to damage it generally speaking?

It depends on the circuitry; some components are more tollerant than others.
Also it is not the voltage to damage a circuit; it's the current. And SPD4000X is not the worst in the class (note CH2 didn't overshoot).
Anyway here we are in a very corner case where we are testing different PSU (see https://www.eevblog.com/forum/testgear/lab-power-supply-turn-on-and-off-characteristics/) deliberating pushing a high voltage and limiting the current to check how fast the over current protection kicks in.
IMHO this is not a factor to keep in consideration while choosing a PSU; it's a fact to be aware while using the PSU.
As a general rule: you should be aware of what you are going to supply current to and how tollerant it is and set up your PSU accordingly.
Sure we are human and here the over current protection ... but if you are powering a super expensive prototype circuit you will double take any precaution.

[ArdWar]

Is this overshoot likely to damage any general IC circuitry? Say if I was supplying 3.3V to a general purpose IC that was only rated to operate up to 3.3V. Would the spill over be enough to damage it generally speaking?

With regards to this specific exercise, no. It should not damage your circuit IF you use your PSU in constant voltage mode.

The aforementioned overshoot only happen (so far) when PSU are configured (and expected) to startup directly into constant current mode. That isn't exactly normal nor usual mode of operation.
Even then, you normally shouldn't set the CV setpoint that far high (say, 8.0V here) when you expect your I*R product to never get that high (say, 40mA*50Ω = 2.0V here). In this situation a sensible user should set the CV setpoint at something like 3.0V.

[MildInductor]

I see, thank you both!

[mhsprang]

The aforementioned overshoot only happen (so far) when PSU are configured (and expected) to startup directly into constant current mode. That isn't exactly normal nor usual mode of operation.

Why isn't that a normal mode of operation? It is, after all, advertized as one of the modes a PSU can operate in. CV means "constant voltage" whereas CC means "constant current". What makes the meaning of constant different in both definitions?

Even then, you normally shouldn't set the CV setpoint that far high (say, 8.0V here) when you expect your I*R product to never get that high (say, 40mA*50Ω = 2.0V here). In this situation a sensible user should set the CV setpoint at something like 3.0V.

But what if I don't know the expected maximum voltage? I'll give you a real life example: I once had to repair a small heater that looked like a wood stove. This heater contained a string of orange LEDS, to light the fake flames. This string was connected to mains using a big ceramic resistor that got flaming hot and eventually burned. To find out the proper value of this resitor, I needed to push 20 mA through the string of LED's without knowing the actual V_FWD and the number of LED's in the potted piece of plastic. In that case, I could have taken a power supply, set to CC at 20 mA, connect the LED and hit the Output button in order to read the total V_FWD of the entire string. Is that such an odd application of a PSU?

And with some of the PSU's in this thread, I could have blown all the LED's, hypothetically.

[2N3055]

The aforementioned overshoot only happen (so far) when PSU are configured (and expected) to startup directly into constant current mode. That isn't exactly normal nor usual mode of operation.

Why isn't that a normal mode of operation? It is, after all, advertized as one of the modes a PSU can operate in. CV means "constant voltage" whereas CC means "constant current". What makes the meaning of constant different in both definitions?

Even then, you normally shouldn't set the CV setpoint that far high (say, 8.0V here) when you expect your I*R product to never get that high (say, 40mA*50Ω = 2.0V here). In this situation a sensible user should set the CV setpoint at something like 3.0V.

But what if I don't know the expected maximum voltage? I'll give you a real life example: I once had to repair a small heater that looked like a wood stove. This heater contained a string of orange LEDS, to light the fake flames. This string was connected to mains using a big ceramic resistor that got flaming hot and eventually burned. To find out the proper value of this resitor, I needed to push 20 mA through the string of LED's without knowing the actual V_FWD and the number of LED's in the potted piece of plastic. In that case, I could have taken a power supply, set to CC at 20 mA, connect the LED and hit the Output button in order to read the total V_FWD of the entire string. Is that such an odd application of a PSU?

And with some of the PSU's in this thread, I could have blown all the LED's, hypothetically.

You couldn't because you probably would not have enough voltage to light  them in a first place.

As to good practice, you would set voltage to some low value and set 20mA.
Connect DUT, enable output and slowly ramp up the voltage...

[mawyatt]

Is this overshoot likely to damage any general IC circuitry? Say if I was supplying 3.3V to a general purpose IC that was only rated to operate up to 3.3V. Would the spill over be enough to damage it generally speaking?

Overshoot Voltage can and will damage/destroy certain components, of coarse this depends on the component and amount of Over Voltage. CMOS failure is caused by gate oxide puncture, many of today's advanced chips have very thin and hence low breakdown oxides (some BJTs have thin base regions). 2 decades ago we were dealing with custom chips that ~3 volts would certainly destroy, they normally operated at ~1.2 volts!!

Another potential effect is when Current Limit is engaged such as the discussion mentioned here:

https://www.eevblog.com/forum/testgear/lab-power-supply-turn-on-and-off-characteristics/

Anyway, IMO ANY Overshoot Condition is a potential disaster in the waiting. A perfect Power Supply should not Overshoot during Turn ON or Turn OFF under any conditions. Actual supplies will deviate from perfect behavior is various ways, some better than others.

Best