Power adapter voltage drop/load

[TTL]

I've just completed a small oscilloscope kit and need a power adapter for it which should output no more than 10.5V (or it'll get damaged), so I'm thinking a 9V DC power adapter will do.

I thought I had one already, and actually did find a few 9V ones, but checked their output with a multimeter which indicated far higher voltages than 9V (one even reported 19V!). I was once told by someone that they're designed with a voltage drop/load in mind, but I'm confused. Obviously I don't want to plug in (and possibly fry) my max 10.5V oscilloscope, so are my existing power adapters just cheap, inaccurate ones or do they all output more than their "official" voltages, and once plugged in I should be fine?

 

[Harald Kapp]

Yes, those cheap wall warts often have no regulation at all. They rely in the resistance of the transformer's coil (and the limited current transfer ratio of the transformer) and the resulting voltage drop under load.
To protect your circuit I suggest you insert a simple 3 pin 9 V regulator between wall wart and oscilloscope.

A sweet little kit you have there. Although next time you build or buy an oscilloscope I recommen you go for at least 2 channels. It is always good to be able to compare two signals. Mind reporting some of your experience with this kit?

 

[(*steve*)]

Connect up a 7809 regulator. If it works, great. It is possible that under load the output voltage will drop to 7V under load, and this may be insufficient for the scope.

 

[TTL]

Yes, it seems like a nice oscilloscope -actually it's my very first one and I have no experience with them but need one to trace some signal faults in a (music) synthesizer. I couldn't find any low cost used ones and as far as I know there aren't any cheap 2-channel kits/low cost mini scopes like this around, so this seemed like the best option (a step up from the more popular and cheaper DSO 138 from the same company. I think it should be OK for my use, I hope...
There's a very nice review/step-by-step assembly walkthrough by someone who I think is an electronics engineer here. And you can also download the user-manual and schematics.

So the way these wall warts are supposed to work is that (say for a 9V rated one) it should never output more than 9V (either by measuring it directly with a meter or when it's under load), but a lower voltage under load is OK and normal?

If I can fit a voltage regulator inside one of those wall warts I'll probably do that as I don't think there's room inside the scope. I also read that this oscilloscope uses an ICL7660 for voltage regulation which has a max input voltage of 10V, but the LT1054 is a better performance part with input ratings of maximum 15V and should be pin to pin compatible with the ICL7660. But I'm not sure about desoldering an SMD component and putting a new one in.... Also I don't have any electronics component shops nearby so all has to be ordered online which takes a while.

Perhaps I should just go out and buy a wall wart which is said to be regulated, but I read in another thread about this oscilloscope that a switched power supply (i.e. worldwide 100-240V) should be avoided because it might generate the frequencies the scope is trying to measure. Any truth to this? I've just found such a wall wart with stabilized power, multiple voltages (3V, 4.5V, 5V, 6V, 9V and 12V) and short-circuit protection. Seems suitable apart from the worldwide input voltage...

 

[(*steve*)]

Switch mode wall warts will output the rated output voltage under no-load conditions, and it will remain constant up to the maximum load.

Wall warts with a transformer inside them often have a no-load voltage much higher than their nominal voltage. This will drop as the load increases, with the voltage not dropping to the nominal output voltage until near the maximum load.

Switch mode wall warts feel a lot lighter than the with a mains transformer

 

[Harald Kapp]

Any truth to this?

This is possible, but not necessarily so. It depends on the quality of the power supply. In any case you could add a filter in between power supply and scope to damp the noise coming from a switch mode supply.
You can build one (or Google for other instructions - there are lots of them).

 

[Audioguru]

You should look in Google or read the datasheets. An ICL7660 is not a voltage regulator. It and the LT1054 generate a negative voltage with an oscillator based circuit that creates a lot of noise frequency that might show on the 'scope.

 

[TTL]

Audioguru: it seemed from that posting to be a pin-pin compatible (but better) voltage regulator. Oh well. Strange that the designers of the scope chose a component that will negatively affect the readings, but it probably has something to do with cost.

Harald Kapp: good idea about the filter! Should fit a small box with a power socket on one side and power plug/cable on the other side. Maybe something ready made is available as well (eBay?).
I just purchased a multi-voltage wall-wart power supply today which should deliver regulated 9V/650 mA. Does it seem OK from the specs in that link? I haven't opened it yet (and haven't checked if the voltage is actually 9V), so it can be returned if it's unsuitable.

*steve*: does the amount of voltage drop depend on how much the device draws (i.e. more voltage drop if it draws say 600 mA while a device drawing say 100 mA will have less of a voltage drop)? So in other words, with an unregulated power supply it's important to know how much current the attached device draws.
How do you calculate this so as not to risk damaging the connected device with too high a voltage?

With a regulated supply, is the output voltage also automatically adjusted upwards when an attached device draws a lot of power so the actual voltage will be the same (i.e. 9V with a 9V regulated supply), regardless?

 

[Harald Kapp]

Specs look good. The quality of the output voltage you'll have to check when the wall wart arrives. With a bit of luck you don't need an additional filter.
As the scope is spec'd for 8 V .. 10 V input voltage there is bound to be an integrated voltage regulator on board which will have its part in filtering the voltage.

 

[Audioguru]

The ICL7660 and LT1054 are not voltage regulators. Instead they are voltage converters to produce a negative voltage from a positive voltage. Their voltage regulation is not good.

 

[davenn]

*steve*: does the amount of voltage drop depend on how much the device draws (i.e. more voltage drop if it draws say 600 mA while a device drawing say 100 mA will have less of a voltage drop)? So in other words, with an unregulated power supply it's important to know how much current the attached device draws.
How do you calculate this so as not to risk damaging the connected device with too high a voltage?

yes
the voltage stated on the unregulated plugpack is the voltage it will supply when the stated max current is drawn from it

eg 12V 600mA
so the output will be 12V when 600mA is drawn

With a regulated supply, is the output voltage also automatically adjusted upwards when an attached device draws a lot of power so the actual voltage will be the same (i.e. 9V with a 9V regulated supply), regardless?

NO .... the output is ALWAYS the stated voltage ( unless you try and exceed the stated current)


Dave

 

[TTL]

Sorry, by "upwards" I meant to ask if the (regulated) power supply automatically compensate for any load so that the correct voltage would be output anyway.

So if I understand you correctly: with a regulated power supply you will always get 9V from a 9V power supply, regardless of how much or little power the attached device draws (at least within the limits of what the power supply has to offer).
But with an unregulated power supply the supplied voltage depends on how much power the attached device draws (if the load is very low it receives a very high voltage while a larger load gives it a lower voltage).
If the last assumption is correct I assume that if you don't know how many mA the attached device draws you might risk frying it (with a power supply supplying a lot more Ma).
But if you know that the device draws the same amount of mA the (unregulated) power supply is rated at you're good to go (if the manufacturer has got it right, all the power supply components are OK and the input AC voltage hasn't fluctuated too much)?

Is there a safe way to check how much voltage it will supply a certain device WITHOUT plugging it in (and thereby avoiding to fry its circuitry if the voltage is too high)? I assume you need to know its current draw: say the oscilloscope I'm building is rated at 9V/250mA while an unregulated power I plan to use is labelled 9V/500mA -how do I calculate how many volts it will give the scope?

 

[(*steve*)]

Theoretically you load the power supply and measure the voltage.

The problem is that you don't know if the actual device represents a constant load. If it doesn't, the input voltage will vary (perhaps significantly).

 

[Audioguru]

A power supply with a low maximum output current is usually cheap and has such a high resistance that its voltage drops a lot when it has its rated load. So they simply make it with a higher voltage (when it has no load) so that its voltage is correct at its rated current.

A high current power supply gets too hot if it is made cheaply so it is made to have low resistance then its voltage regulation is much better than a cheap low current one.

My 9V/100mA wall wart produces 19VDC with no load. My 9V/500mA wall wart produces 13.4V with no load. They both produce 9V at their rated load current.

 

[davenn]

Sorry, by "upwards" I meant to ask if the (regulated) power supply automatically compensate for any load so that the correct voltage would be output anyway.

no, it's a fixed voltage regardless of it loaded or not .... see my previous comment

So if I understand you correctly: with a regulated power supply you will always get 9V from a 9V power supply, regardless of how much or little power the attached device draws (at least within the limits of what the power supply has to offer).

yes

But with an unregulated power supply the supplied voltage depends on how much power the attached device draws (if the load is very low it receives a very high voltage while a larger load gives it a lower voltage).

yes

If the last assumption is correct I assume that if you don't know how many mA the attached device draws you might risk frying it (with a power supply supplying a lot more mA).

yes, because the supply voltage will be higher, this allows a larger current to be drawn by the load

Is there a safe way to check how much voltage it will supply a certain device WITHOUT plugging it in (and thereby avoiding to fry its circuitry if the voltage is too high)? I assume you need to know its current draw: say the oscilloscope I'm building is rated at 9V/250mA while an unregulated power I plan to use is labelled 9V/500mA -how do I calculate how many volts it will give the scope?

measure the unloaded voltage from the supply ... it may be around 12 - 15V

not sure of the V and I relationship in this case .... @(*steve*) ?


Dave

 

[(*steve*)]

not sure of the V and I relationship in this case .... @(*steve*) ?

It's not something I'd like to try to calculate theoretically.

Empirical measurement is the best approach. In addition, we're only talking about voltage as measured on a voltmeter. There's also ripple which will vary with load.

 

[TTL]

Good news! I checked the switched/regulated power adapter I bought, set at 9V it supplied 9.135V so that should be well below the 10V limit of the oscilloscope.

Since I don't know how much current it draws (someone mentioned 150mA or was it 450mA? but I'm not sure if that's a fact or guessing) I don't want to risk anything with an unregulated supply.
Seems like these may be better suited for devices which you know the exact amount of mA it draw and you match that with a power supply with the exact same V and mA rating.

 

[Audioguru]

Voltage matching is very important but you do not need to match current. The starter motor in a car in winter draws 400A or more but the clock that is powered from the same battery uses only 0.01A. Get a power supply that provides the correct voltage and can supply the current that is needed or more current. It simply loafs along without sweating when it is not powered to its maximum current. The extra current that is available is simply not used.