# Why does DMM show AC and DC current when source is a battery? (How to calculate power in this case?)

The circuit is an altered Joule Thief of my design that is driven by an AA or AAA battery. I’m observing the current at the battery anode with my DMM in series. (Extech EX330)
This circuit is extremely efficient (In relation to other Joule Thief I've build.) and does some weird stuff - like, it needs a pushbutton sometimes.

Measurements:
It is drawing 9.21 mA DC and 6.48 mA AC in series with the battery.
It shows 1.364 VDC and 0.010 VAC in parallel with the battery.
Output on the LED is 2.572 VDC and 0.011 VAC with no measurable frequency. (with my DMM at least)
All measurements are taken with the same configuration and voltage.

But how is it drawing AC and DC and how do I calculate power consumption?
--> Could it be because the emitter pin of the NPN is directly connected to ground?
--> Or is the circuit just "switching on and off"?

(I tried to simulate the circuit in LTspice. Thanks for making me aware of this amazing program, Sredni Vashtar!)

What the pushbutton is used for:
The pushbutton is used to "kickstart" the oscillation when it wont start by itself.
For example: I have to manually "oscillate" using this button to start the circuit if the capacitor is too big and battery voltage is low.
Pushing the button releases the energy in the transformer into the capacitor. If it reaches LED forward voltage the circuit keeps oscillating on it's own.
It does work without manual intervention when then capacitor in parallel with the LED is not too big, like <5000 µF, and it's a fresh battery.

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• The meter isn't necessarily going to give you a valid reading on the wrong setting, but in fact a circuit like this will have both AC and DC currents in it. Because it is pulsed there is an AC element to it, and because those pulses are overwhelmingly of the polarity of the battery, there is a DC offset. Commented Mar 26, 2018 at 5:12
• I wish I had an oscilloscope... How do I calculate the overall power consumption then? Actually... the battery cathode is only connected through diodes. (including the NPN emitter) How is there AC? Commented Mar 26, 2018 at 5:13
• You are switching a load on and off. That is AC. Commented Mar 26, 2018 at 5:24
• I thought it has to go negative to be AC. But it makes sense. Ok how do I calculate power consumption then? Add them? Commented Mar 26, 2018 at 5:35
• The Joule Theif works by creating a step-up pulse through the transformer. There better be an AC component or else your circuit wouldn't work. AC doesn't mean it goes negative, just that it alternates. Commented Mar 26, 2018 at 5:37

Battery current and voltage can vary

The gist of your question is "how can a battery, that is supposed to supply a fixed, constant voltage give rise to an 'alternating' current measurement?". I believe that what you find strange is the fact that being the battery voltage constant, the current drawn from it should also be constant.

You might be surprised by the fact that not even the voltage supplied by the battery is constant, but varies instead with the amount of current drawn. To see how this can happen we just need to model the battery with an ideal constant voltage source (with zero internal resistance) and a series resistance. Something like this - where I intentionally exaggerated the series resistance value

will give you a voltage Vbatt that is a function of the current supplied to the load (not shown above). In fact, without any load attached the voltage you measure at the 'external' terminals is the nominal voltage of your battery (let's say it's 1.5V, or whatever value you happen to have). When there is a 'load' (i.e. a circuit that is attached to Vbatt and GND) that draws current, that same current has to go through Rs, and in doing so it will cause a voltage drop that will subtract to the nominal voltage of the battery.

Computing power

So both the voltage across and the current supplied by the battery can vary, and depending on the powered circuit they do it with a time dependance that can be very complex, not necessarily symmetrical, nor periodic. Why is that important? Because to compute the instantaneous power p(t) = v(t) i(t) absorbed by your circuit you need to get both i(t) and v(t) right.

To illustrate all this, let's consider this simulation of a Joule Thief circuit powered by such non-ideal battery:

As you can see, a pulsed current drawn from the circuit results in a pulsed voltage variation at the battery terminals:

The voltage swing at the battery here is grossly exaggerated by the unusual high internal resistance chosen. Two notes on the signs: first, I had to put a minus sign before I(Vnom) because LTSpice adopts the user convention for the current in the battery - since I want to call i(t) the current exiting the positive pole of the generator, I have to use i(t) = -I(Vnom). The product v(t) i(t) = -V(Vbatt)*I(Vnom) gives you the instantaneous power delivered by the battery and drawn by the attached circuit. Second, the circuit is higly reactive and the swings in current are so wide that they can also lead to a sign inversion. This means that at certain times the current is entering the battery from the positive pole, resulting in a negative power, i.e. power wasted by the battery instead that used up by the load. This is an extreme effect of the humungous values for Rs, associated with the back-kick coming from the coils.

At any rate, once we have v(t) and i(t) at every t, we know all we need to know about power. Moreover, since in this case the waveforms tend to a periodic steady state (or so it seems on the selected timescale), we can even compute a meaningful average power by integrating over an integer number of periods and by dividing by the time interval.

LTSpice can do that for you: hold down the CTRL key and left click on a power trace name. A small window will pop up showing the average power and the average energy over the selected time interval:

This Joule Thief appears to require some 7.5 mW to run, when the power delivered to the LED only is 4.5 mW.

A more realistic example

As already mentioned, the battery internal resistance has been grossly exaggerated in order to highlight the variability of battery voltage. If we reduce its value to a more believable 0.8 ohm (still some 4 times the average series resistance of an AA cell) the effect on the overall battery voltage is greatly diminished. The circuit will behave so differently that other modifications are needed in order to avoid burning the LED. By changing base resistance and using a different transformer with much higher inductance and a 2:1 ratio, we get this circuit

that appears to oscillate at a frequency of 56 kHz with a peak current in the LED of nearly 60 mA (average current is only 10 mA). Note how small is the voltage swing at the battery terminals, now.

By selecting a time interval where the waveform are periodic, and by using the CTRL + right click trick, we find that in this case the average power drawn from the battery is 47.3 mW, while that used up by the LED is 42.5 mW. This amount to nearly 90% efficiency.

Incidentally, in LTSpice you don't have to simulate imperfect batteries by manually adding an external resistance. You can specify Rser's value as one of the parameters of any voltage generator. In the above circuits I made it external in order to better see the origin of the 'non constant' value of a DC battery.

Real circuits and measurements

Now, this is all in the realm of simulation. What happens in the real world can be different: qualitative and quantitative aspects can play a determinant role. First, your battery will have a much smaller internal resistance, so the voltage drop will be much lower. Nonetheless, you can still have variable current draw - this means you might see appreciable AC current content and negligible AC voltage content.

Second, you will have a hard time in simulating the transformer, that is in determining what value to give to the required parameters. Stray capacitances can also alter the behavior of this circuit. As another poster has noticed, even the multimeter cables - or the oscilloscope's probes - can change the way the circuit will work.

The results of Joule Thief simulation can be significantly different from what you get in the real world. I would not rely on a simulation to actually compute the power drawn by such a circuit. The simulation's value is that it makes you see what kind of waveforms you are dealing with: rapidly varying voltage and current at the battery terminals.

The role of the instrument

How this time changing current value will be interpreted by your multimeter depends on how your instrument is designed to measure AC current. Some multimeter expect zero DC content to give a meaningful AC reading, and even then, they might give a meaningful reading only if the varying current is sinusoidal (this is especially true when 'not-exactly-true' RMS measures are involved). Also (I almost forgot), the frequency of the variable signal is important: Joule Thieves tend to oscillate at several tens or hundreds of kHz, and your multimeter might not be able to understand what's going on.

In the end, as usual, what you read depends on how you read it, and with what instrument you read it.

To measure power in a circuit like yours, I'd use an oscilloscope to get voltage and current waveforms and then use the math function to multiply them together. This will give you the instantaneous power waveform. From there to average power, it's just a matter of integration (and division by time interval).

• @Distelzombie For the negative power: it's a matter of convention. The power supplied by the battery is Vbatt * Ioutofbatt; Since LTSpice gives I positive when entering the battery, you have to add a negative sign. Now, the power computed with the right sign for I can still assume negative values; what does that mean? That the circuit is pushing current into the battery and its internal resistance is turning that into heat, hence power is wasted inside the battery. (I will later update my answer with the conventional signs for current and power). Commented Apr 2, 2018 at 23:55
• In your new circuit you label a node "VLED", but that's not the voltage across the LED, it's the voltage of that node with respect to ground (click and drag to measure differential voltage). And as for the efficiency, well, 22.65% efficiency looks like a legit value. But before computing that, you should isolate a time interval where the waveform are periodic. The Joule Thief I simulated takes an average of 7mW from the battery and gives an average of 4.5 mW to the LED. Commented Apr 2, 2018 at 23:59
• Didn't you make the same "mistake" in your pictures? It's also vled*i Commented Apr 4, 2018 at 20:27
• my LED has the cathode at GND. Yours doesn't. Click and drag the probe: you will see the red probe icon fixed and a new black probe icon that you will place wherever you want your measured voltage referred to. See electronicspoint.com/resources/… (last part of fig. 3) Commented Apr 4, 2018 at 20:31
• Oh. Ok, so it is exactly like alt+clicking on the LED. I thought it didn't change anything, but it does. Thanks again :) Commented Apr 4, 2018 at 20:42

Sredni's response is excellent, so I'm only going to write about measurement.

Since the Joule Thief draws pulsed current, an ammeter will give both AC and DC values. Additionally the ammeter may add impedance in the supply due to wiring, which may change the joule thief's frequency, efficiency, etc.

The simplest is to decouple the supply with a capacitor as close as possible to the circuit. Use a low-ESR cap, in the "low tech" spirit of the Joule Thief I'd go with a large electrolytic cap from the junk parts bin, like 1000µF or more, as ESR of general purpose caps goes down with capacitance. Or a ceramic of 10µF or more. Or any cap with ESR lower than the battery.

Anyway. The cap at the input of the joule thief will supply the AC part of the current, so the ammeter in series with the battery will measure something closer to DC only.

• Oooh... I just learned so much. It all makes sense now! I cannot get rid of the added impedance, can I? I got three nice Rubycons with 9,900µF combined. They'll get rid of most of it. I can't get it lower than 0.02mA AC, though. And if I touch anything or blow on it, it fluctuates horribly. Damn breadboards... I didn't know there are ceramics with such high values. Might be useful to get some of these. Commented Apr 4, 2018 at 16:50

The power consumption from a DC power supply is $V_{DC}\times I_{DC}$. Taking into account the AC part of the current is nonesense if the supply is DC.

Why is this so you might ask.

Power is volts x amps at its most basic and if the current taken from a supply is a sinewave wholly superimposed on a DC value then multiplying $V_{DC}$ by the AC content of the sine wave produces a power waveform that is positive in one half cycle and negative in the other. The average power would therefore be zero.

• What happens if the Idc is not DC but has spikes? Not a symmetrical sinusoidal variation superposed to the DC value, but one direction only spikes due to different current draw at definite time intervals? Commented Mar 30, 2018 at 22:29
• @SredniVashtar well, with the scenario you paint, my answer remains perfectly intact - do you understand why? Commented Apr 2, 2018 at 11:12
• The problem with your answer is that it is based on the existence of a "DC value" for either voltage or current. That is generally not the case. I can easily find a simple circuit where there are not even well defined average values for "external" battery voltage and current (you'll have to resort to a moving average) Commented Apr 2, 2018 at 17:48