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I am trying to do an analysis of the following modified howland current pump based on the circuit description in the datasheet of the LMP7701. This is a voltage controlled current pump that is driven by a sine generator with aroung 10 kHz.

schematic

simulate this circuit – Schematic created using CircuitLab

I want to derive the formula for the output current, that should only depend on the resistor \$R_s\$ and the input \$V_{in}\$. The formula is already given in the datasheet of the LMP7701 and in AN-1515:

\$ i_l = \frac{V_{in}}{R_s} \$

In order to derive this, I started with the followng nodal equations:

\$ i_1 = i_2 \Longleftrightarrow \frac{V}{R_1} = \frac{V_{out}-V}{R_2} \Longleftrightarrow V=\frac{R_1}{R_1 + R_2} V_{out} \ \ (1) \$

\$ i_3 = i_4 \Longleftrightarrow \frac{V-V_{in}}{R_3} = \frac{V_{L}-V}{R_4} \Longleftrightarrow V=\frac{R_4 V_{in} + R_3 V_L}{R_3 + R_4} \ \ (2) \$

\$ i_l = i_s \Longleftrightarrow \frac{V_{out}-V_L}{R_s} = \frac{V_L}{R_L} \Longleftrightarrow V_{out}=\frac{R_LV_L + R_sV_L}{R_L} \ \ (3) \$

Now I set \$ (1) = (2) \$ and derived after \$ V_{out} \$ to get

\$ V_{out} = \frac{(R_1 + R_2)(R_4 V_{in} + R_3 V_L)}{R_1 (R3 + R4)} \ \ (4) \$

Now I set \$ (3) = (4) \$ and solve after \$ V_in \$ to get

\$ V_{in} = \frac{R_3 R_s V_L}{R_4 R_L} \ \ (5) \$

Now, assuming that \$ R_1 = R_2 = R_3 = R_4 \$ and setting \$ V_L = \frac{R_L}{i_L} \$ I get that

\$ V_{in} = \frac{i_L * R_s}{R_L^2} \$

Which seems to be wrong, or am Imissing something? Are some of my assumptions wrong or are some things missing? I recalculated many times and tried different approaches but this is not my field and do not really now how to proceed further, would appreciate some help.

Thanks in advance.

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  • \$\begingroup\$ oh WOW, I already can see what I did wrong after posting this question....I assumed V_L = R_L/i_L, sometimes the simplest things can go wrong. SOLVED \$\endgroup\$ – gammaALpha Aug 26 '18 at 12:46
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Unfortunately, for some reason I made a mistake and assumed that \$ V_L = R_L/i_L\$, which is obviously wrong. Setting \$ V_L = R_L*i_L\$ and doing the calculations again I get now \$ i_L = V_{in}/R_S \$ which is what I wanted to derive.

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  • \$\begingroup\$ Now what are the practical limitations of this design? What happens say from currents of 1uA to 1A? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 26 '18 at 17:45

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