# Estimating Ulka EP5 Pump Inductance

As part of my work to modulate the Ulka EP5 pump pressure in my Espresso machine, I decided to try and simulate the pump in SPICE. This would necessarily be a very simplified model, but might be useful nonetheless to understand the behaviour of snubber circuits for the pump PWM controller. To create the model, I needed an estimate of the pump inductance.

## Previous attempts

I could only find one previous reference to the pump inductance online. It’s not clear which Ulka pump was used, but it looks like an E5 type (EP5 or EX5) and a circuit diagram refers to 41W rating (consistent with the 120V 60Hz EP5). They took the approach of measuring the current drawn by the pump at 0.57A and, assuming a line voltage of 120V, calculated the reactance as follows:

XL = V/I = 120V / 0.57A (AC) = 211

With the formula XL = 2πfL = 2π * 60Hz = 377L, the inductance L was estimated as follows (for a 120V 60Hz pump):

L = 211 / 377L = 0.56H = 560mH

However, XL = VPEAK/IPEAK and the above voltage current values may be average or RMS values rather than peak values. For 120V AC mains the peak voltage would be VPEAK =12√2 = 170V. Similarly, when measuring AC current, some multimeters only give accurate results for a sinusoidal AC waveform, and may not be accurate for AC chopped by the series diode in the Ulka pump. This doesn’t necessarily mean the result is wrong of course, just difficult to be certain.

## Manufacturer data on the pump

The manufacturer data states that:

• Ulka EP5 240V~50Hz is rated at 48W
• Ulka EP5 120V~60Hz is rated at 41W

## Estimating pump electrical properties

There’s an integrated series diode in the pump, but the type is not indicated. However, based on similar Ulka pump models described in Ulka/CEMA specifications, the diode is assumed to be 1N4007 which according to manufacturer datasheets have a typical forward voltage VF = 1.0V when IF = 1.0A (noting that there are some small variations in the specifications for 1N4007 between manufacturers).

Attempting to measure the pump coil resistance directly with a resistance meter could give misleading results, due to the internal diode. To overcome this, a known DC voltage and current can be used to estimate the resistance.

With a ~12V DC power supply connected across the Ulka EP5 240V~50Hz pump, the measured voltage was 12.33V and the measured current was 70.1mA. At this relatively low current, it is likely that the diode forward voltage will be less than 1.0V. To estimate the diode VF, an individual 1N4007 diode was connected in series with a 220R resistor and ~12V power supply, giving IF = 51.9mA and VF = 0.775mV.

Assuming that VF = 775mV at IF = 70.1mA, the actual voltage across the coil will be reduced to (12.33 – 0.775) = 11.56V and therefore a first estimate of the pump coil resistance is as follows:

R = 11.56 / 70.1×10-3 = 165Ω

To calculate the inductance, the manufacturer voltage and power ratings are used to estimate the reactance as follows:

Although the Ulka EP5 240V~50Hz model is rated 48W, the internal diode means that the pump is only active for half the cycle. Considering the coil without the series diode, we therefore assume that the AC power consumption would be 96W and use this figure to calculate the AC current:

P=VI therefore I=P/V

IRMS = PRMS/VRMS = 96/240 = 0.4A

Note that the mains line voltage and current will of course vary in practice due to local mains supply quality.

The peak voltage and current are then estimated as follows:

VPEAK   = VRMS√2 = 24√2 = 339V

IPEAK      = IRMS√2 = 0.4√2 = 0.57A

The inductive reactance for an inductor energised by a sine wave is calculated as follows:

XL = VPEAK/IPEAK =  339 / 0.57 = 599.27

The inductance can be calculated based on the reactance and frequency as follows:

L = XL / 2πf

Where f is the mains line frequency 50Hz:

L = 599.27 / (2π × 50)  = 599.27 / 314.159 = 1.91H

However, on closer inspection, I noticed that the exact pump model I tested is labelled 230V~50Hz on the side. Repeating these calculations with 230V, the following results are obtained:

IRMS = 96/230 = 0.42A

VPEAK = VRMS√2 = 23√2 = 325V

IPEAK  = IRMS√2 = 0.4√2 = 0.59A

XL = VPEAK/IPEAK = 325 / 0.59 = 550.59

L = 550.59 / 314.159 = 1.75H

So this estimate puts the pump inductance for the 230V~50Hz EP5 pump at somewhere between 1.75H and 1.91H and the resistance at 165Ω.

## Simulating the pump in SPICE

Running an LTSpice model with a 240V, 50Hz AC supply in circuit with a 1N4007 diode and an inductor modelled with L=1.75H, R=165Ω gives the following results:

Iavg=0.34A, Irms=0.46A, Vrms=240V, Pavg=35.67W

This doesn’t look too far off, however it is obviously lower than the rated power of 48W indicated by the manufacturer. Adjusting the LTSpice model inductance to L=1.37H gives results closer to the manufacturer ratings:

Iavg=0.39A, Irms=0.54A, Vrms=240V, Pavg=47.93W

## Measuring the pump inductance

More recently, another attempt was made to directly measure the inductance. This was achieved by drilling a small 1mm hole in the pump to access the coil terminal before the series diode. The measured inductance and resistance was L=854mH and R=165Ω. The series diode forward voltage was measured as 706mV.

## Conclusions

Modelling the pump as a simple inductor is obviously a gross simplification, as the pump is a spring loaded solenoid, whose inductance will vary as the internal plunger vibrates. Also, the hydraulic pressure in circuit is not considered. Finally, with high frequency PWM switching, the parameters estimated at 50Hz / 60Hz frequency may not be correct.

Nevertheless, the inductance values above when used in SPICE provide a first approximation which may be useful to estimate how the system will behave in practice, and do seem to give behaviour very similar to the real system.

If anyone has any corrections, comments, improvements or information to add, drop me a line in the comments below.

## 5 thoughts on “Estimating Ulka EP5 Pump Inductance”

1. Andres says:

Hey!

One simple remark. The pump, of course, converts electrical work to mechanical work.

This means that for higher mechanical power (pressure times flow rate), you would also be taking a higher electrical power.

In your model this means that your electrical resistance (not the one generating losses due to the wire resistance, and core losses in the inductor) simbolizes mechanical power, and is dependent on the operating point that you’re measuring.

One simple and very crude way to measure the inductance, if you have an oscilloscope, is to charge a capacitor with some voltage, then connect them in parallel and measure the frequency. It’s surprisingly accurate given its simplicity. (use f=1/(2pi sqrt(LC)) )

1. james says:

Hi Andres
Thanks for the comments above. I assumed that the series diode built into the pump would prevent a resonant/tank circuit with a parallel capacitor from working? (the diode is encased in the pump and there seems no easy way to take it out of circuit without damaging the pump)
Regards
James

2. Steve says:

Very interesting calculations, but i actually arrived here in a much simpler way.
I have measured a voltage difference between the metal framing of the pump and ground of approximately 35Vac when using the rubber mounts and connected to a 230Vac supply.
there is little technical info on the Ulka website, or indeed other sites of companies manufacturing similar pumps, and i wanted to know if that voltage reading would be normal?
I assume if it is then it would be because of the inductance from solenoid coil of an electrical reciprocating pump, but confess i don’t really know, would you?
Thanks and kind regards.

1. james says:

Hi Steve

For comparison I’ve just tried the same on my 240V Gaggia, and with the pump operating I measured roughly 30V AC between the metal frame of the pump and ground. So I would say that what you are seeing is “normal”.

The metal frame of the pump is not electrically connected to the pump terminals or to earth, but it is right next to a very strong alternating magnetic field from the coil in the pump. The coil is probably inducing this voltage in the metal frame and the meter leads.

Although the voltage may appear high, it’s only the potential difference between those two points and probably wouldn’t be able to deliver much current. e.g. it’s only measurable because the meter has a very high input impedance and there’s no continuity between the metal frame and ground that would keep them at the same potential.

So I think this is normal, and I do get a similar result to you.

Best regards
James