Wall wart questions

[greco]

I have two electronics measuring devices that each use 4 X 1.5 volt batteries.

I putting this question in the Amp Tech section of the forum as I will be using these pieces of equipment to help me with my first build.

I bought a 6 volt (so it says on the outside cover) "wall wart" (900 milliamps) power supply thinking that I could use it and not have to always be replacing batteries that run down (because I often forget to turn off the device).

However, the wall wart is actually measuring 9 volts DC output. Do I need to put resistors in series to drop the voltage to 6 volts before I use it to accomplish the above?

I don't want to risk damaging the measuring devices.

Do these typically have an output that is higher than stated?

Thanks

Dave

[iaresee]

I have two electronics measuring devices that each use 4 X 1.5 volt batteries.
I have to ask: are you sure the batteries are being run in series in these devices? It's better to read the device spec for its power needs than to assume 4 1.5V == 6V demand from the device. If all 4 batteries are side by side the power configuration could be 1.5V, 3V or 6V.

I bought a 6 volt (so it says on the outside cover) "wall wart" (900 milliamps) power supply thinking that I could use it and not have to always be replacing batteries that run down (because I often forget to turn off the device).

However, the wall wart is actually measuring 9 volts DC output. Do I need to put resistors in series to drop the voltage to 6 volts before I use it to accomplish the above?
Are you measure the drop across a load? Or did you just stick your multimeter probes against the tip of the power supply jack? You need to measure the drop across a load.

Do these typically have an output that is higher than stated?
Expect some room for error on the current rating, but not the voltage. It would be wrong for it to be supplying 9V when it said it was a 6V supply. I'm assuming it's a standard transformer-based supply though. If it's a digital switching supply I'm not sure what happens if you measure it unloaded.

[greco]

I have to ask: are you sure the batteries are being run in series in these devices? It's better to read the device spec for its power needs than to assume 4 1.5V == 6V demand from the device. If all 4 batteries are side by side the power configuration could be 1.5V, 3V or 6V.


Are you measure the drop across a load? Or did you just stick your multimeter probes against the tip of the power supply jack? You need to measure the drop across a load.


Expect some room for error on the current rating, but not the voltage. It would be wrong for it to be supplying 9V when it said it was a 6V supply. I'm assuming it's a standard transformer-based supply though. If it's a digital switching supply I'm not sure what happens if you measure it unloaded.


I measured the voltage drop over a .9 Mohm resistor as the load. Maybe I need to start with/use a larger load? How does one determine the size of the load for this type of testing?

I'll check the configuration of the batteries/specs again.

Thanks for all the help.


Cheers

Dave

[iaresee]

I measured the voltage drop over a .9 Mohm resistor as the load. Maybe I need to start with/use a larger load? How does one determine the size of the load for this type of testing?
I think that's sufficient.

I'd take it back.

[Wild Bill]

I think that's sufficient.

I'd take it back.

.9 meg is essentially no load at all! It's just way too big a resistor to allow any sort of current draw at all.

Try 500-1000 ohms. You'll see that voltage drop much closer to 6 volts.

When you say "electronics measuring devices" are you talking about a multimeter? If so, don't run it on a wall wart.

Wall warts aren't supposed to give perfectly accurate voltage. Pretty well all devices that use them don't care if the voltage runs a couple of volts high.

Measuring devices are different. They actually use the stable battery voltage as a reference. Using a wall wart means you've thrown your accuracy away!:eek:

My main meter uses a 9v battery. It lasts for a year or more without replacement.

:food-smiley-004:

[greco]

Thanks for the help.

Wild Bill...Yes, I am talking about an old multimeter. Given your advice, I'll stay with the batteries. It might be an old DMM, but it "runs good"*(* I got that quote from several adds in Auto Trader ...thought it might appeal to you ....LOL)

No sense in risking the accuracy for the "convenience" of not having to replace the batteries every so often.

Thanks again

Cheers

Dave

[Wild Bill]

Thanks for the help.

Wild Bill...Yes, I am talking about an old multimeter. Given your advice, I'll stay with the batteries. It might be an old DMM, but it "runs good"*

Dave

Always welcome, Dave!

This might be a good time for a bit of a basic power supply lesson, if you're interested.:smile:

Wall warts are a good example of basic DC supplies. They are simple and basic because they're cheap! You have a transformer to step down the wall voltage , which is usually around 120 vac in these parts but can vary from 125 to as low as 115, or more. Then a couple of diodes to rectify the AC. Diodes don't generate smooth DC. They work by allowing only AC pulses of one polarity to get through and blocking the AC wave when it swings through zero to the opposite polarity. That's why they call it "alternating current", after all.

So we have pulses after the diodes, all of the same polarity. They rise up from zero to a peak and back down to zero again. If you traced the up and down of the peaks you'd get another wave. This one is called "ripple" and that's what makes the hum you get from an unfiltered power supply or one with bad caps.

Lastly, they stick a filter capacitor after the diodes. It's normally an electrolytic of a big enough value to store the pulsed energy. That way the capacitor will act like a big reservoir for the current. With no load, it will charge up to the peak of the pulse. When a load starts to draw current the capacitor will supply stored energy between pulses, when the voltage is dropping down, and re-charge energy when the pulses are at maximum.

This tends to smooth out the ripple from the pulses into a more constant, or flat, voltage.

Since this is a basic circuit predicting the output voltage is not a simple thing to do. With AC it's easy! If we need 12 vac we just buy a 120:12 vac transformer. If the line voltage is a little low or high we still get close enough to our desired voltage. The input differences get scaled down by a factor of 10 on the low voltage side. What's more, that voltage tends to hold steady through the range of current drawn by the load, from minimum to maximum.

When we rectify it into pulses we don't have the peak voltage all the time. In fact, the peak of the pulse is higher than the RMS or average value of the AC wave. After the diodes those peaks charge that filter cap to the very highest peak of the AC wave. That's 1.41 times the RMS AC voltage you read with your meter as it comes out of the transformer! So a 10 vac output trannie will give you 14.1 volts DC across the filter cap.

But wait! As soon as you start to use that voltage to power your circuit it drops down! That's because the maximum voltage is only there for an instant at the top of the pulse's peak. You're drawing off current at a steady rate to power your circuit. The more you draw the more it drops, until the voltage gets down to the "fat" part of the pulse, where there's the most energy. Likely the voltage will steady at around 12 vdc. The cap has helped but it can never make things perfect! There's always a bit of ripple. Bigger caps and more stages of them will help by having a bigger reservoir of energy to smooth out the pulses but you have to make some hard decisions. How much are you willing to spend to make the power supply? How BIG a cap can you fit inside that wall wart?

Some circuits like digital or computer stuff demand a very flat, or well regulated power supply. That's because ripple peaks can get confused with a signal pulse. They will use fancy voltage regulator configurations. They cost more and take up a lot more room than a wall wart.

Still, that wall wart is just fine for things less critical, like guitar pedals. Analog transistor circuits don't draw much current and will operate just fine a few volts above or below their design voltage. So you don't need much voltage regulation.

So this is why you bought a 6 vdc wall wart and with no load got a higher voltage!

If you've followed me through all this, you've just learned about the power supply in your amp! It works exactly the same way as your wall wart!

The line voltage goes through your power transformer. It has windings to step the voltage down to 6.3 volts for the tubes and maybe 5v for a tube rectifier. For the plate voltage power it has a step up winding. When you choose your transformer you need to do a bit of "back of an envelope" arithmetic. Maybe you want to run those 6L6's at a plate voltage of 450 volts. You know that the RMS AC voltage of the winding will result in a peak voltage on the 1st filter cap of 1.41 times RMS. So you might choose a secondary voltage of 320 vac. If you're using a centre-tapped winding like with a tube rectifier you might see it spec'd as 320-0-320 vac. That's because each half of the winding will supply only the top or bottom pulse of the AC wave to be rectified.

Whoops! We forgot that this is only the peak value. It's a good thing to understand because when we choose the safe working voltage of a filter cap it's the peak voltage we have to worry about and not the average. The tubes however will draw current. We have the idle current when we bias the tubes and then more and more current as we crank the amp or drive it harder.

This means that the plate voltage will never be the exact same level no matter what's happening. Most schematics will show a reading when the amp is just idling but even then it may be a bit different from what's on the print. Maybe it's an older amp and the transformer was designed for the 115 or 110 vac that came from the wall in the 50's or 60's. The extra volts from the 120 vac we see today are stepped UP through the high voltage winding, magnifying the difference! If the print said 537 volts on that old Traynor schematic you might actually read 568 vdc, which is what I measured on the one I'm working on today!

The tubehead's rule of thumb formula is 1.3 x the RMS voltage of the HV winding to get the working voltage, at least with steady idle current being drawn. So to get 450 working volts DC for the plates we need a trannie that puts out 450/1.3=346 vac!

In the real world the catalog would offer something close, like 350-0-350 or 340-0-340. A few tens of volts either way won't matter on those tube plates. We DO have to recalculate the peak voltage! We have to worry about the rating of those filter caps! So if we went with a 340 volt trannie then the peak would be 340 x 1.41 = 479 VDC. This means we have to choose caps rated higher than the peak, like 500 or 525 vdc.

If we can't get ones this high or they're too expensive we can stack caps for a higher voltage rating. Caps in series just add their voltage ratings. So two 350 volt caps act like one rated for 700 vdc. More than safe enough. Their combined value is calculated the same as resistors only "backwards". When caps are in parallel just add them together. Two 40 mfd caps act like one 80 mfd caps. In series if they're the same value it's easy! Two 40 mfd caps look like only 20 mfd in series.

The power supply in an amp is a bit more complicated with extra stages of filter caps. The reason is that we need lower voltages in later stages of the amp, like with the preamp tubes. So we use dropping resistors. Each dropping resistor sees the current from all the tubes after that point. So the last dropping resistor might see only the input triode stages. The first resistor will see the current from ALL the preamp tubes and maybe the output tube screens, depending on if there's a screen choke or not.

If you want to raise or lower the plate voltage of those input 12AX7 stages you can guesstimate that each stage draws maybe a ma and a half. Take the current involved, the voltage drop you want and do a little OHM's Law to find the resistor value. Or measure the voltage drop across the resistor. Take that and the resistor value and you can calculate the current draw through the resistor. Once you know that you can choose a different resistor value for whatever voltage drop you want to put the right voltage on those preamp plates!

The reason we use more filter caps to tie each dropping resistor to ground is twofold. First, each stage smooths out more and more ripple. 98% of the ripple is filtered out by the 1st cap, and 98% of what's next by the next cap, and so on. Second, we have to do something called "decoupling". The filter cap is also a ground return path for the signal. Just like hooking up lights to batteries, you must always have a complete circuit with power leaving the source, flowing through the load and then returning. The return path in a tube circuit flows through that filter cap.

When several stages are connected to the same filter cap node the cap must be big enough to provide a "good enuff" low impedance path for all the signals on all the plates involved. If it's too small the signals might find it easier to flow over to plates in other stages instead of through the cap and ground. This can result in unwanted coupling between stages that results in squeals and oscillations. The rule of thumb here is at least 10 mfd for no more than 4 plates or stages involved. Usually in an amp you'll see no more than that at any given filter cap node. The dropping resistors must be at least a few hundred ohms to make each cap act like another decoupling stage and not just caps in parallel looking like one big cap.

Hope that's not too much to chew on at once, Dave! If you take the time to wrap your head around all this you'll have pretty well all you need to build, fix and tweak any amp power supply you run across.

And they all are just big versions of that wall wart!:smile:

:food-smiley-004:

[greco]

WOW

...Thank You Wild Bill:bow:

I appreciate you taking the time to write all this !!

I just finished reading it, took 2 tylenols, and will print it and read it again...several times (it will take my brain quite a few repetitions to assimilate all that you have written)

This could be part of one of the chapters for your future electronics textbook......you know...the book that I keep hoping you will write.

Can I cut and paste it into a new thread so others have the chance to read it? ....many will likely miss it as the "wall wart" title for my thread is not very "exciting" or "alluring". Please let me know.

Many thanks again for all your time and for sharing your vast wisdom. :food-smiley-004:

Dave

[Wild Bill]

Can I cut and paste it into a new thread so others have the chance to read it? ....many will likely miss it as the "wall wart" title for my thread is not very "exciting" or "alluring". Please let me know.

Dave

Go right ahead, Dave! I wrote it not just for you but for any other folk that might benefit.

I've been thinking about that book...:smile:

:food-smiley-004:

[greco]

I've been thinking about that book...:smile:

:food-smiley-004:

I am serious about the book!! You write very well and your great sense of humor would really add to the enjoyment of learning and understanding some of the tough stuff to wrap one's brain around.

If you don't want to go for a whole book at once, consider trying "booklets" (The cost could be kept reasonable by using standard 8.5" x 11" paper, cover stock and with plastic spiral bindings...remind you of other electronics books you have seen!!??) Consider starting with an electronics topic that you personally enjoy and one you feel will be helpful to others. If you wrote several booklets through time, you could put eventually put them together in one or more larger volumes.

Please understand, I'm really not trying to tell you what you shoud do, or how, I'm just trying to encourage and support you. My enthusiasm gets a bit carried away at times......Sorry.

Cheers

Dave