# Still messing around with zeners.

#### Hextejas

Joined Sep 29, 2017
187
I got my lamp circuit to work and the light is glowing brightly.
The lamp is rated at 5v .06A and measures at 12 ohms.
So, before I plug the lamp in, the voltage at the lamp measures 6.7v.
Then I plug the lamp in, measure again and it shows 4v.
I don't get it.
Where did the 2.7 v go ?
Which leads me to a more general question in
Will the load try and get as much current or volts, as it needs, or will it try as best it can with what's available ?
Something similar is my cell phone charger. It has the ability to charge 2 devices so I asked a buddy that if I were charging 2 devices would they take twice as long. He said no.
I didn't continue the discussion as to asking that there must be a limit somewhere, in that I could not have an infinite number of devices.

And I am hoping someone here could point me to something that I can read that might explain it.

Thanks

#### crutschow

Joined Mar 14, 2008
23,159
Where did the 2.7 v go ?
Post your circuit diagram and we can tell you.
I didn't continue the discussion as to asking that there must be a limit somewhere, in that I could not have an infinite number of devices.
Of course there's a limit -- when the current drawn by all the devices being charged exceeds the current capacity of the charger.

#### Hextejas

Joined Sep 29, 2017
187
Post your circuit diagram and we can tell you.
Of course there's a limit -- when the current drawn by all the devices being charged exceeds the current capacity of the charger.

The 1st problem that I was working on, and posted quite a bit about, had to do with a zener maintaining a steady voltage to a simple lamp. Well I have been struggling with it and I finally got it it work. This is a link to the circuit.
https://electrojets.blogspot.com/p/dirltr-styletext-align-left-trbidion.html

I was trying to figure out why the load voltage at the lamp changed.

Re the cell phone charger, thank you.

#### MrChips

Joined Oct 2, 2009
19,160
You are asking too many questions at once. Focus on one question at at time.

There are different types of power sources. There is a voltage source and there is a current source. You need to understand what type of source you are referring to.

No source is ideal. All sources have limitations. When you start drawing current from a voltage source, the voltage could start to sag. How much it sags depends on the source and the load.

You can put two loads on the same source. If the source is able to deliver the current demanded by the two loads combined, then the charge time will remain the same. If the two loads start to make the voltage sag, then yes, the charge time will be longer.

#### Hextejas

Joined Sep 29, 2017
187
You are asking too many questions at once. Focus on one question at at time.
Ok

There are different types of power sources. There is a voltage source and there is a current source. You need to understand what type of source you are referring to.
Really ? I had thought that they were the same, IE the same battery. Though, I guess I could have 2 or more.
Curious.

#### hobbyist

Joined Aug 10, 2008
887
Where did the 2.7 v go ?
It's used up in the power source internal impedance.
The lamp is not a constant resistive load.

#### neonstrobe

Joined May 15, 2009
73
To enlarge on Mr Chps's answer - most power supplies are voltage sources, but some battery chargers are constant current. Voltage sources have an internal impedance, or resistance, which drops the voltage when loaded. Current sources will drop the voltage depending on the load resistance. In each case it is called compliance. A voltage source has current compliance, or limit, and a current source has a voltage compliance, or limit. Your lamp resistance probably won't change much for a D.C. current, but it is likely that it is too great a load for your charger. You need to check the charger current ratings to see what it is capable of. Some simple power supplies use a diode rectifier with a capacitor to smooth the output. When not loaded, the voltage can reach 1.4 times (this is the square root of 2) higher than the RMS value of the AC voltage. When loaded the diode voltage drops and resistances of the transformer will cause the output voltage to drop.

#### Hextejas

Joined Sep 29, 2017
187
That is so interesting and then begs this question, at least to my untrained mind.
When I had what I thought was a working circuit, I measured the load voltage and it was close to what I was looking for. Kewl says I, now let's plug in the lamp. Hey ! What happened, and now I know.
However, what would I have to do to assure that the load voltage stayed at what i wanted ?

#### qrb14143

Joined Mar 6, 2017
112
That is so interesting and then begs this question, at least to my untrained mind.
When I had what I thought was a working circuit, I measured the load voltage and it was close to what I was looking for. Kewl says I, now let's plug in the lamp. Hey ! What happened, and now I know.
However, what would I have to do to assure that the load voltage stayed at what i wanted ?
The voltage drop you experience when you connect the load is due to the voltage drop across the internal resistance of the source. It is a simple voltage divider circuit. If the open-circuit or "nominal" voltage of the supply is Vs, the resistance of the source is Rs and the resistance of the load is Rl then the voltage across the load, Vl is given by:

Vl = (Rl / (Rl + Rs)) * Vs

It follows that if you want a higher load voltage you need either:
1) A source with a lower internal resistance
2) A higher source voltage to begin with

The other option is to have some sort of regulator between the source and the load but this adds another level of complexity which you may or may not be comfortable with.

#### Hextejas

Joined Sep 29, 2017
187
" When an ideal voltage source has zero internal resistance, it can drop all of its voltage perfectly across a load in a circuit "
This confuses me a bit as I thought that there would some voltage drop, even if minimal, across every component in the circuit, not just the load.
Although maybe the article is referring to the complete circuit as "The load".

And I most thankfully appreciate your patience.

#### qrb14143

Joined Mar 6, 2017
112
" When an ideal voltage source has zero internal resistance, it can drop all of its voltage perfectly across a load in a circuit "
This confuses me a bit as I thought that there would some voltage drop, even if minimal, across every component in the circuit, not just the load.
Although maybe the article is referring to the complete circuit as "The load".

And I most thankfully appreciate your patience.
If you refer to my above post, you will see that if the source has zero internal resistance, then Rs = 0. Substituting this into the equation will show you that if the source has no internal resistance then ALL of the source voltage appears across the load.

You are correct in saying that some voltage will be lost in other parts of the circuit. In practice though, the resistance of the wires is so small that for most purposes it is assumed to be zero.

Typically when drawing a circuit diagram, any wires you see are assumed to be perfect conductors. If the resistance of the wires is considered relevant for that case, it will be represented by drawing an equivalent resistor resistor in the circuit. Without delving too much into transmission line theory, you will find that the resistance of the wire increases as you increase its length and decreases as you increase its cross sectional area. In most cases, you are using "short and thick" wires so the resistance is negligible.

#### Hextejas

Joined Sep 29, 2017
187
The voltage drop you experience when you connect the load is due to the voltage drop across the internal resistance of the source. It is a simple voltage divider circuit. If the open-circuit or "nominal" voltage of the supply is Vs, the resistance of the source is Rs and the resistance of the load is Rl then the voltage across the load, Vl is given by:

Vl = (Rl / (Rl + Rs)) * Vs

It follows that if you want a higher load voltage you need either:
1) A source with a lower internal resistance
2) A higher source voltage to begin with

The other option is to have some sort of regulator between the source and the load but this adds another level of complexity which you may or may not be comfortable with.
I might try and do that once I get my understanding up a level or 5.
Something that I didn't try, which would probably be enlightening, would be to measure the resistance of the complete circuit before I plug the lamp in.
Hmmm, does measuring the resistance with the lamp in make any kind of sense ?
Groan, this is giving me a headache.

#### qrb14143

Joined Mar 6, 2017
112
I might try and do that once I get my understanding up a level or 5.
Something that I didn't try, which would probably be enlightening, would be to measure the resistance of the complete circuit before I plug the lamp in.
Hmmm, does measuring the resistance with the lamp in make any kind of sense ?
Groan, this is giving me a headache.
I do not believe it will be possible to simply measure the internal resistance before you plug in the load.

My best advice to you would be to start with Ohm's Law, arguably the most important rule in all of electrical engineering and it only has three letters in it! From there, have a look at Kirchoff's Voltage and Current laws which will help you to understand what is going on in this series circuit.

There is an indirect way of working out the source impedance by measuring the current through and voltage across varying resistance loads. This is a classic first year electronics lab assignment and I would recommend that you have a go at that!

#### Tonyr1084

Joined Sep 24, 2015
3,473
Your supply is like a water pitcher. It can deliver only so much. If you take out a teaspoon here and a teaspoon there your supply can easily handle it. But if you try to fill a swimming pool with your water pitcher you'll quickly discover that the supply is not big enough for the job. You'll need a much bigger supply.

Is there a limit to how many devices you can put on a charger ? ? ? Yes. Suppose you have a 2 amp charger (one capable of pushing 5 volts at 2 amps (for instance)). You plug in a single 5 volt device such as a phone. Suppose your phone is only drawing 700 mA (0.7 amps). Suppose you plug a second phone in drawing 700 mA. Now you're drawing 1400 mA (1.4 amps). Both your phones will charge in the normal time. But suppose you plug a third phone into the charger, ALSO drawing 700 mA. Now you've exceeded the capacity of the charger and yes, things will start to slow down. The more you try to draw the slower the phones will charge. And that's also the point where you begin to see the voltage start to drop drastically. Because the voltage is a product of the current and the resistance; the resistance is the load of the phones and the current is the limited available current. If you can push 2 amps (without dropping the voltage) then you will see a reduction in the voltage when you overload the charger. 2 amps at 5 volts is 10 watts. You can't exceed the wattage.

Speaking theoretically, you're getting 10 watts. 2 amps at 5 volts also means you have a fixed resistance of 2 1/2 ohms (5v ÷ 2a = 2.5Ω). Two phones drawing 1400 mA means you're using 7 watts (5v x 1.4A = 7W). That means that each phone is contributing 5Ω load. Keep in mind I'm making up numbers for this example. Your actual numbers will most certainly be different. Ohms law on parallel resistance states that if you have two resistors of the same value in parallel then the total resistance is 1/2 of either resistor. Adding a third load means dropping that resistance even lower. A third resistor of the same value (or phone) will drop the resistance to 1.67Ω. With a limit of 2 amps your voltage will drop to (1.67Ω x 2A) 3.3 volts. THAT's why you see a big voltage drop when you add too much draw to your circuit.

Conversely, when your supply has NO load, then yes, you will see a higher than 5 volt reading. In theory, an infinite resistance (an open circuit) divided by 2 is still infinity. And your supply is NOT going to give you an infinite voltage. Theory is great for understanding how things work. But you also have to take into account the limitations of devices, the resistance caused by imperfections in those devices and the wires, line losses, magnetic resonance that escapes into the ether, and probably other things not mentioned. In other words, you lose something just because things are not perfect.

So there IS a limit to how many devices you can charge from a single device. Suppose you had a 12 volt car battery. How many chargers can you plug into that? Answer - a whole fricken lot! But how many chargers can you plug into a simple 9 volt transistor battery? Probably just one. And it's likely your charging time will be longer as well. It's all about supply and demand. If you demand too much then delivery slows down drastically.

Remember the water pitcher and the spoon (or glass if you prefer).

#### neonstrobe

Joined May 15, 2009
73
So if I can summarise ... your lamp regulator used a resistor and Zener diode to stabilize the voltage. I'm guessing the Zener was 6.8V. When you put a lamp across it the voltage dropped to4V. That is because your resistor dropped the voltage. You may possibly damage the resistor if it cannot handle the power, but a 6V 60mA lamp is 360mW so most resistors will be OK.
Now check what the power supply is that you are feeding the zener from. If that will handle the lamp power/current then the situation can be fixed by using a more powerful regulator. You can add an NPN power transistor with collector connected to your main supply (12V?), base to the Zener /resistor connection, and take the output from the emitter.
A couple of warnings: the output current will be limited by the transistor gain, but should be say 20x the current you can get from the resistor/zener alone. But the transistor might be damaged if you short the emitter to ground. A very common PSU design in the old days was to use just this with the addition of one power resistor in series with the transistor collector, to limit the current. This has to be high power and it and the transistor may get hot, so need to be put on a heatsink.

#### ebeowulf17

Joined Aug 12, 2014
2,934
There is an indirect way of working out the source impedance by measuring the current through and voltage across varying resistance loads. This is a classic first year electronics lab assignment and I would recommend that you have a go at that!
Love this idea. What better way to wrap your head around the not-so-obvious concept of source impedance than to do some hands on experiments and see it in action?!