# Help Calculating Current Draw

#### ben sorenson

Joined Feb 28, 2022
180
Hello, I am running a DC Solinoid with a resistance of 45 ohms through a 5 ohm resistor. I am using pulsed DC to operate the device. The Power is being delivered by a 20V Battery.

I am just trying to figure out how much current the device is drawing as my AMP metered is currently non functional.

The last test run I had, I beleive i started with a battery voltage of 19.2, I did not check the voltage when I was finished, as I should have. I ran the demonstration for a period of 60 min and when i was done I put the battery on the charger.

When the charging was done, it had been charging the battery for 13m30s and said it charged or "used" 130mAh.

is there any way to use the the amount of time I ran the demonstration compared to the charge time to calculate some type of average current usage?

#### WBahn

Joined Mar 31, 2012
29,885
Assuming the numbers are sufficiently accurate, you can simply use the definition of current, which is charge flow per unit time.

I_avg = 130 mAh / 60 min
I_avg = 130 mAh / 1 h
I_avg = 130 mA

Now, let's see if that seems at all reasonable.

In steady state, the solenoid should draw

I_ss = 20 V / (45 Ω + 5 Ω) = 400 mA

This is assuming that the full battery voltage was being applied to the coil/resistor. How was the switching being done.

It also neglect the falling battery voltage, but we can us an average of 19.6 V to probably get a better estimate.

If we ignore the time constant due to the solenoid's inductance, that would indicate a duty cycle of about

duty cycle = 130 mA / 400 mA = 32.5%

You said this was pulsed DC, but gave no indication of what the duty cycle was. Is being on 1/3 of the time seem about right?

Do do better than this, we need to know the inductance of the coil and the frequency and duty cycle of the pulses.

#### MisterBill2

Joined Jan 23, 2018
17,885
If we can assume that the solenoid was connected for a time long enough for the inductance effect to stabilize, then the steady state current would be determined by ohms law. But given the small capacity of the battery that would probably be correct for the first instant after the current became stable. Batter voltage drops as current flows out of the battery. Recharging replaces not only the energy flowing out as current, but also the energy that created internal heat by flowing through internal resistance.
So you can measure the voltage at the instant of connection and then observe how quickly it drops and guess at an average voltage. Then, as resistance is known and a voltage is known, I=V/R, which is probably close to the value determined by recharging.

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#### ben sorenson

Joined Feb 28, 2022
180
Assuming the numbers are sufficiently accurate, you can simply use the definition of current, which is charge flow per unit time.

I_avg = 130 mAh / 60 min
I_avg = 130 mAh / 1 h
I_avg = 130 mA

Now, let's see if that seems at all reasonable.

In steady state, the solenoid should draw

I_ss = 20 V / (45 Ω + 5 Ω) = 400 mA

This is assuming that the full battery voltage was being applied to the coil/resistor. How was the switching being done.

It also neglect the falling battery voltage, but we can us an average of 19.6 V to probably get a better estimate.

If we ignore the time constant due to the solenoid's inductance, that would indicate a duty cycle of about

duty cycle = 130 mA / 400 mA = 32.5%

You said this was pulsed DC, but gave no indication of what the duty cycle was. Is being on 1/3 of the time seem about right?

Do do better than this, we need to know the inductance of the coil and the frequency and duty cycle of the pulses.
Ok. Sounds good and thank you for spending time to help me. I will try again tommarrow and check what the duty cycle is on the oscilliscope. Thanks again.

#### WBahn

Joined Mar 31, 2012
29,885
Ok. Sounds good and thank you for spending time to help me. I will try again tommarrow and check what the duty cycle is on the oscilliscope. Thanks again.
Also measure the rise and fall times of the current (measure the voltage across the 5 Ω resistor). If you don't know how to do that, just measure the time it takes the waveform to go from 10% to 90% of the way from one value to the other. We can help you back out the time constant from that (posting a screen shot or picture of the scope image would be useful, too).

#### MrAl

Joined Jun 17, 2014
11,280
Hello, I am running a DC Solinoid with a resistance of 45 ohms through a 5 ohm resistor. I am using pulsed DC to operate the device. The Power is being delivered by a 20V Battery.

I am just trying to figure out how much current the device is drawing as my AMP metered is currently non functional.

The last test run I had, I beleive i started with a battery voltage of 19.2, I did not check the voltage when I was finished, as I should have. I ran the demonstration for a period of 60 min and when i was done I put the battery on the charger.

When the charging was done, it had been charging the battery for 13m30s and said it charged or "used" 130mAh.

is there any way to use the the amount of time I ran the demonstration compared to the charge time to calculate some type of average current usage?
Hi,

I am sorry to say I cannot see a clear path to any reasonably accurate calculation from this information, and even hard to be sure about any reasonable approximation. That's because there are far too many variables involved. Let's consider some of them here.

First, we dont know how much the battery voltage changed throughout the experiment.
Second, we don't know what the inductance of the coil is nor the frequency and pulse width of the pulses.
The measurement of the 130mAh is nice, but we don't know if that is with a source of 20v or it went right down to 10v as soon as the experiment started. This ties in with the type of battery and the "P factor" of the battery, which changes wildly for different batteries and with different battery ages and some other possible changes.
We also do not know if the solenoid is being turned on and off, or just on and remains on.

If we can use that 130mAh figure, then with 100 percent charge acceptance the current would be 130ma, but with 50 percent charge acceptance the current would be 65ma. The other problem though is that we don't know if the battery was fully charged to begin with. If it was only half charged that could change everything quite a bit.

Do you have a voltmeter on hand? If so, you could measure the voltage across the 5 Ohm resistor and use Ohm's Law to calculate the current.

#### MisterBill2

Joined Jan 23, 2018
17,885
For determining the current, I used the assumption that the on time was beyond what was required for the inductor to saturate, and so it would be limited by resistance for most of the time. that left the battery voltage droop as the unknown, as well as any resistance increase due to heating.

Joined Jul 18, 2013
28,533
Hello, I am running a DC Solinoid with a resistance of 45 ohms through a 5 ohm resistor. I am using pulsed DC to operate the device. The Power is being delivered by a 20V Battery.
What is the reason for pulsed DC? is this constantly pulsed?

Joined Jul 18, 2013
28,533
An answer to my previous question would help, also with DC operated inductive devices, it is possible to drop the voltage a significant amount without loosing retention on a DC operated relay/solenoid device etc.

#### ben sorenson

Joined Feb 28, 2022
180

#### ben sorenson

Joined Feb 28, 2022
180
Hi,

I am sorry to say I cannot see a clear path to any reasonably accurate calculation from this information, and even hard to be sure about any reasonable approximation. That's because there are far too many variables involved. Let's consider some of them here.

First, we dont know how much the battery voltage changed throughout the experiment.
Second, we don't know what the inductance of the coil is nor the frequency and pulse width of the pulses.
The measurement of the 130mAh is nice, but we don't know if that is with a source of 20v or it went right down to 10v as soon as the experiment started. This ties in with the type of battery and the "P factor" of the battery, which changes wildly for different batteries and with different battery ages and some other possible changes.
We also do not know if the solenoid is being turned on and off, or just on and remains on.

If we can use that 130mAh figure, then with 100 percent charge acceptance the current would be 130ma, but with 50 percent charge acceptance the current would be 65ma. The other problem though is that we don't know if the battery was fully charged to begin with. If it was only half charged that could change everything quite a bit.

Do you have a voltmeter on hand? If so, you could measure the voltage across the 5 Ohm resistor and use Ohm's Law to calculate the current.
I am having an extremely hard time pinning down a frequency. My handheld is only good to 10MHz.

Joined Jul 18, 2013
28,533
Appears to be a great deal of pulsing going on??

#### ben sorenson

Joined Feb 28, 2022
180
Appears to be a great deal of pulsing going on??
[/QUOTE
Just a tiny bit.

#### MisterBill2

Joined Jan 23, 2018
17,885
If there is a consistent duty cycle for the pulsing, then a reasonable guess could use the percentage multiplied by the total time spent pulsing. Pulsing will allow less heating then a constant draw, so that will have some effect.
But really, there are too many unknowns, as already stated, to have an accurate number. Possibly a "good guess", though. Maybe.