Proportional valve control calculation. Part 2.

Part 2. Proportional control in the regeneration mode.

Regen mode - proportional control

Fig.1. Regeneration mode schematic

See also:
Part 1. Proportional control of double acting hydraulic cylinder.
Online calculator for cylinder proportional control.

The goal of the calculation is definition of a cylinder velocity at the current operating voltage at the valve’s solenoid.
For calculations we need to know: cylinder and valve parameters (take them from manufacturer catalog), values of forces at cylinder rod (load) and pressure drop at return line (can be assumed as zero if unknown).

2.1. Regeneration mode schematic
2.2. Regeneration Mode: Cylinder extension calculation
2.2.1. Balance of forces at the cylinder
2.2.2. Spool lands ratio and pressure drops at the spool lands
2.2.3. Hydraulic cylinder extension. System pressure calculation
2.2.4. Flow through spool lands: max. and at current voltage
2.3. Hydraulic cylinder retraction calculation
2.4. Final words….

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2.1. Regeneration mode schematic

So, first let see the schematic for proportional control of the cylinder in regeneration mode (Fig.1.). The first question you can ask me: why I connected the pressure line to the port “B” of the proportional valve? Why it is not a port “P”?

And this is a good question!

In real practice you have a lot options/ways how to connect proportional valve in regen mode. But general rule is: flow in ports in both valve work positions has to match to flow ratio of the spool (what you can find in valve’s catalog). For example, lets use Bosch Rexroth valve 4WRA10 for our calculations. The values of flow trough the valve’s spool we can find in the catalog:

Rexroth 4WRA valve

Fig.2. Rexroth 4WRA valve

It means, for this valve the the spool lands work area ratio (what I discussed in Part 1 of this article) is r_s=2, or 2:1.

Lets see will this rule work for my schematic or not. For cylinder retraction (Fig.3.) flow in ports A->T is much bigger than flow in ports B->P:

Cylinder retraction

Fig.3. Cylinder retraction

Also, please pay attention to next moments:

  1. During retraction the 2-way/2-pos. valve to be triggered to connect our system with the tank line.
  2.  The check valve I use in the schematic prevents bypassing flow and directs flow trough the valve. It helps to create equal pressure drops on both lands (otherwise, without this check valve pressure drop at lans A->T will be close to 0).

So, in this spool position, the flow through spool lands matches to requirements in the valve catalog.

For cylinder extension (Fig.4.) flow in ports P->A is much bigger than flow in ports B->T because the quantity of the flow from pump to cylinder blind port displaces more flow from rod port (do not forget, the displaced quantity of the flow from the rod port comes to the blind port as well and pushes the piston until forces from both sides of the piston will be equal):

Cylinder Extension (regen mode)

Fig.4. Cylinder Extension (Regeneration mode)

During cylinder extension the 2-way/2-pos. valve has to be triggered to block the system from the tank line.

So, in this spool position the flow through spool lands also matches to requirements in the valve catalog.

2.2. Regeneration Mode: Cylinder extension calculation

Let calculate the cylinder extension only (regeneration mode). There no difference in calculation of cylinder retraction, so see Part 1 of this article for info about that.

2.2.1. Balance of forces at the cylinder

Let’s figure out all forces what are at the cylinder at the constant velocity, acceleration and deceleration of the piston.

At the left image (Fig.5.) the load pushes the rod, at the right image the load pulls the rod:

Fig.5. Forces at the cylinder during extension at regen mode

Fig.5. Forces at the cylinder during extension at regen mode

F_{load} – Load force from mechanism or/and from tool weight.
F_f – Friction force, what includes piston side breakaway force and friction in the mechanism.
F_p – Force created by hydraulic pressure in the piston area (from port “A” of proportional valve): F_p=p_A\cdot A_p
F_v – Force created by backpressure in the annular area (from port “P” of proportional valve, because it works as an orifice):F_v=p_P\cdot A_a

So, at the constant velocity, when external load pushes the rod, the force balance is:


and, when external load pulles the rod, the force balance is:


So, the generic formula for cylinder extension is:

F_p=F_v+F_f\pm F_{load}

F_{load} > 0 (positive value) when the load pushes the cylinder rod;
F_{load} < 0 (negative value) when the load pulls the cylinder rod.

But, if the piston moves with acceleration or deceleration, in accordance with Newton’s Second Law:

F_a=m\cdot a=F_p-(F_v+F_f\pm F_{load})


F_p=F_v+F_f\pm F_{load}\pm m\cdot a

F_a – is the acceleration/deceleration force;
m – is cylinder’s pulled or pushed mass;
a > 0 (positive value) – acceleration value;
a < 0 (negative value) – deceleration value;
a = 0 – for calculations of constant velocity.

To simplify future formulas I want to involve “Result external force” F_{res}:

F_{res}=F_f\pm F_{load}\pm m\cdot a

Now, just transfer this formula for future calculations:


p_A\cdot A_p=p_P\cdot A_a+F_{res}

p_A=p_P\cdot \frac{A_a}{A_p}+\frac{F_{res}}{A_p}


So, the generic force balance formula for cylinder extension in regen mode is:

p_A=\frac{p_P}{r_c}+\frac{F_{res}}{A_p}          (1)

Formula (1) will be used in future calculations.

2.2.2. Spool lands ratio and pressure drops at the spool lands

For the calculation we assume the pressure compensated pump with variable displacement is used as a source of pressure and flow. It means the constant pressure at port p of proportional valve and variable flow to keep the same pressure to compensate different contribution.

At the Fig.7. you can see simplified hydraulic schematic of proportional control the cylinder extension (Regeneration mode) and its equivalent where proportional valve shown as two separate orifices:

Fig.7. System pressure description during cylinder extension.

Fig.7. System pressure description during cylinder extension.

\triangle p_{PA} – pressure drop at the spool land P>A
\triangle p_{BT} – pressure drop at the spool land B>T
p_A – pressure at the piston/blind side of the cylinder (at the port “A” of the valve)
p_P – pressure at the annular/rod side of the cylinder (at the port “P” of the valve)
A_p – cylinder’s piston (bore) area
A_acylinder’s annular area

For future calculation let we assume r_c as a cylinder area ratio:

r_c=\frac{A_p}{A_a}          (2)

The the spool lands work area ratio r_s (see detailed description of this coefficient in Part 1):

r_s=\frac{A_{PA}}{A_{BT}}          (3)

A_{PA} – work area of the spool land P>A
A_{BT} – work area of the spool land B>T

All proportional valve spools can meter fluid in and out. Because of this orifice function, the equation for flow through an orifice applies:

Q=c_d\cdot A_l\sqrt{\frac{2\cdot\triangle p}\rho}

c_d – orifice discharge coefficient
A_l – spool land work area
\triangle p – pressure drop at spool land
\rho – fluid density

So, the flow flow at both spool lands will be:

Q_{PA}=c_d\cdot A_{PA}\sqrt{\frac{2\cdot\triangle p_{PA}}\rho}          (4)

Q_{BT}=c_d\cdot A_{BT}\sqrt{\frac{2\cdot\triangle p_{BT}}\rho}          (5)

The cylinder’s piston speed is:




Q_{PA}\cdot A_p=Q_{PA}\cdot A_a+Q_{BT}\cdot A_a

Q_{PA}\cdot A_p-Q_{PA}\cdot A_a=Q_{BT}\cdot A_a

Q_{PA}(A_p-A_a)=Q_{BT}\cdot A_a

\frac{Q_{PA}}{Q_{BT}}=\frac{A_a}{A_p-A_a}=\frac{A_a}{A_r}=r_r          (6)

At the formula (6) I have involved new ratio r_r (Annular area to rod Area) just to save space in future formulas.

Now, formulas (4) and (5) we put in formula (6) to determine the relation between cylinder area ratio and pressure drop at the lands:

r_r=\frac{A_a}{A_r}=\frac{Q_{PA}}{Q_{BT}}=\frac{c_d\cdot A_{PA}\sqrt{\frac{2\cdot\triangle p_{PA}}\rho}}{c_d\cdot A_{BT}\sqrt{\frac{2\cdot\triangle p_{BT}}\rho}}=\frac{A_{PA}\sqrt{\triangle p_{PA}}}{A_{BT}\sqrt{\triangle p_{BT}}}=r_s\frac{\sqrt{\triangle p_{PA}}}{\sqrt{\triangle p_{BT}}}

\Rightarrow r_r=r_s\frac{\sqrt{\triangle p_{PA}}}{\sqrt{\triangle p_{BT}}}

\Rightarrow \triangle p_{PA}=\triangle p_{BT}\left(\frac{r_r}{r_s}\right)^2          (7)

Formula (7) will be used in the future calculations.

2.2.3. System pressure calculation in regen mode

With pressure compensated pump during the work operation the pressure equilibrium try to be established in the system all the time. The load creates a force at the cylinder, and the pressure in all points of the system depends from this force. If the force is not changed during lifting or lowering, the pressure in all points is not changed as well. We assume and accept it, because otherwise the calculations will be more complicated and should be considered specifically for each case separately.

Let see one more time the Fig.7.:

Fig.7. System pressure description during cylinder extension.

Fig.7. System pressure description during cylinder extension.

Now we need to determine the pressure in all points of hydraulic circuit:

Pressure in the port “A” of the valve we have already found by formula (1), but in addition, it can be found:

p_A=p-\triangle p_{BT}

Pressure in the port “P” of the valve:

p_A=p_P-\triangle p_{PA}

\Rightarrow p_P=p_A+\triangle p_{PA}

Pressure in the port “T” of the valve:


Pressure in the port “B” of the valve:


So, the next what we need to do is solving the system of equations, what includes formulas (1) and (7):

\begin{cases}  p_A=p-\triangle p_{BT}\\  p_A=p_P-\triangle p_{PA}\\  p_A=\frac{p_P}{r_c}+\frac{F_{res}}{A_p}\\  \triangle p_{PA}=\triangle p_{BT}\left(\frac{r_r}{r_s}\right)^2  \end{cases}

Lets solve this system of equations. 

p-\triangle p_{BT}=p_P-\triangle p_{PA}

\Rightarrow p_P=p-\triangle p_{BT}+\triangle p_{PA}


p-\triangle p_{BT}=\frac{p_P}{r_c}+\frac{F_{res}}{A_p}

p-\triangle p_{BT}=\frac{p-\triangle p_{BT}+\triangle p_{PA}}{r_c}+\frac{F_{res}}{A_p}

Times r_c and using formula (2):

p\cdot r_c-\triangle p_{BT}\cdot r_c=p-\triangle p_{BT}+\triangle p_{PA}+\frac{F_{res}}{A_a}

Next, use formula (7):

p\cdot r_c-\triangle p_{BT}\cdot r_c=p-\triangle p_{BT}+\triangle p_{BT}\left(\frac{r_r}{r_s}\right)^2+\frac{F_{res}}{A_a}

\triangle p_{BT}-\triangle p_{BT}\cdot r_c-\triangle p_{BT}\left(\frac{r_r}{r_s}\right)^2=p-p\cdot r_c+\frac{F_{res}}{A_a}

\triangle p_{BT}\cdot \left(1-r_c-\left(\frac{r_r}{r_s}\right)^2\right) = p-p\cdot r_c+\frac{F_{res}}{A_a}

So now we know pressure drop at the land B>T:

\triangle p_{BT}=\frac{p-p\cdot r_c+\frac{F_{res}}{A_a}}{1-r_c-\left(\frac{r_r}{r_s}\right)^2}

Using formula (7) we can find pressure drop at the land P>A:

\triangle p_{PA}=\frac{p-p\cdot r_c+\frac{F_{res}}{A_a}}{1-r_c-\left(\frac{r_r}{r_s}\right)^2}\cdot \left(\frac{r_r}{r_s}\right)^2

Pressure in the port “A” of the valve:

p_A=p-\frac{p-p\cdot r_c+\frac{F_{res}}{A_a}}{1-r_c-\left(\frac{r_r}{r_s}\right)^2}

Pressure in the port “P” of the valve:

p_P=p-\frac{p-p\cdot r_c+\frac{F_{res}}{A_a}}{1-r_c-\left(\frac{r_r}{r_s}\right)^2}+\frac{p-p\cdot r_c+\frac{F_{res}}{A_a}}{1-r_c-\left(\frac{r_r}{r_s}\right)^2}\cdot \left(\frac{r_r}{r_s}\right)^2

p_P=p+\frac{p-p\cdot r_c+\frac{F_{res}}{A_a}}{1-r_c-\left(\frac{r_r}{r_s}\right)^2} \cdot \left( \left(\frac{r_r}{r_s}\right)^2-1 \right)

The next step is definition of the flow through spool lands and the cylinder velocity.

2.2.4. Flow through spool lands: max. and at current voltage

To help us to calculate the flow trough the valve spool lands (= find the cylinder velocity) manufacturers in all proportional valves catalogs provide with two important values: rated flow Q_N at rated pressure drop \triangle p_R. A 100% command signal (i.e. +10VDC = 100% valve opening) the valve flow at rated pressure drop \triangle p_R per metering land is the rated flow Q_N. Please do not mix flow rate and pressure drop trough the spool and trough the land, when take values from the valve data sheet!

As we already know, the rated flow trough land is:

Q_N=c_d\cdot A_{land}\sqrt{\frac{2\cdot\triangle p_N}\rho}

For other than rated pressure drop the valve flow changes at constant command signal according to the square root function for sharp edged orifices:

Q=c_d\cdot A_{land}\sqrt{\frac{2\cdot\triangle p}\rho}

If the electrical signal is other than max voltage (10 VDC) we need to proportionally apply area correction:

Q=c_d\cdot\frac y{y_{max}}\cdot A_{land}\sqrt{\frac{2\cdot\triangle p}\rho}


y – operation voltage
y_{max} – max. operation voltage (10 VDC)

The flow ratio can give us the relation between nominal and actual flow:

\frac Q{Q_N}=\frac{c_d\cdot\frac y{y_{max}}\cdot A_{land}\sqrt{\frac{2\cdot\triangle p}\rho}}{c_d\cdot A_{land}\sqrt{\frac{2\cdot\triangle p_N}\rho}}=\frac y{y_{max}}\sqrt{\frac{\triangle p}{\triangle p_N}}

\Rightarrow Q=Q_N\cdot\frac y{y_{max}}\cdot\sqrt{\frac{\triangle p}{\triangle p_N}}          (9)

Using formula (9) and value of pressure drop from formula (8), we can calculate the flow trough P>A land:

Q_{PA}=Q_N\cdot\frac y{y_{max}}\cdot\sqrt{\frac{\triangle p_{PA}}{\triangle p_N}}

Now, the cylinder velocity is:


The flow trough B>T land:

Q_{BT}=\frac{Q_N}{r_s}\cdot\frac y{y_{max}}\cdot\sqrt{\frac{\triangle p_{BT}}{\triangle p_N}}

Note: At this formula an addition area correction r_s has been applied to rated flow value (do not forget about spool lands ratio!).

As an option (or for double check), you can find the flow trough B>T land by easy way:


The max flow trough spool lands be at y=y_{max}:

Q_{PAmax}=Q_N\cdot\sqrt{\frac{\triangle p_{PA}}{\triangle p_N}}

Q_{BTmax}=\frac{Q_N}{r_s}\cdot\sqrt{\frac{\triangle p_{BT}}{\triangle p_N}}

If you have an opposite task (find operation voltage for target flow), you can easy reconfigure the formula:

y=Q_{PA}\cdot\frac{y_{max}}{Q_N}\cdot\sqrt{\frac{\triangle p_N}{\triangle p_{PA}}}=Q_{BT}\cdot\frac{y_{max}\cdot r_s}{Q_N}\cdot\sqrt{\frac{\triangle p_N}{\triangle p_{BT}}}


2.3. Hydraulic cylinder retraction calculation.

The calculation of cylinder retraction is in the same way how I described in the Section 1.2. at Part 1 of this article. Just carefully use ports designations, using diagram at the Fig.8.:

Fig.8. System pressure during the cylinder retraction

Fig.8. System pressure during the cylinder retraction

2.4. Final words…

These calculations are only basis and can give you very approximate values but they are good enough for most applications.

If you have any questions or find any mistakes, please let me know in comments.

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