Op-Amp Current Sources: The Howland
Current Pump
Dennis Feucht - May 06, 2014
The op-amp current source attributed to Brad Howland of MIT is used widely and is familiar in
circuit design. Yet it can be difficult to understand for those not close to it. The following explanation
begins with simpler, familiar circuits.
The first circuit to consider is the op-amp current mirror, shown below.
The inverting (–) and non-inverting (+) input voltages of the op-amp are
For correct op-amp operation, v– = v+. Then equating,
The op-amp input voltage is determined only by the ground reference of RL and it floats at vL relative
to ground;
The output current is independent of vL and of the load resistance, RL. What determines iL is the
voltage across Rf+ which must be the same as that across Rf–, regardless of what v– = v+ is, and
neither voltage across Rf depends on ground.
The current mirror is now modified for the second circuit by making the input source a voltage, as
shown below and returning Rf+ to vR.
For v– = v+,
First, let vR = 0 V. Then the one-op-amp differential amplifier (diff-amp) results. For equal input
voltages of the op-amp (v– = v+),
Next, let vR = VOS, a constant offset voltage. Then
The effect of connecting Rf+ to other than ground is to elevate the output voltage by the same
amount.
For the third variation of vR, we take a deep breath and set vR = vO. Then for the required op-amp
input condition (v– = v+), vI = 0 V. To set vI ≠ 0 V causes v+ to always be greater than v– and the opamp output will increase until it reaches the upper end of its range. To make the circuit work
correctly (v– = v+), v+ /vO must be less than 1. The excess voltage applied at vO to Rf+ is the
contribution to vO by vI, and this is
This is the amount by which the feedback voltage applied to Rf+ must be reduced.
To achieve this voltage reduction for the positive-feedback path, decompose Rf+ into two series
resistors, Rf – Rs and Rs. It is across Rs that the excess voltage will be dropped by causing an
additional current (iL) to flow through it in additional to the current in Rf – Rs of iB.
Separating the currents
To simplify the currents by separating iL and iB, an intermediate circuit has a ×1 voltage amplifier
that is inserted to buffer iB so that only iL flows through Rs. The circuit is shown below.
With Rs and its voltage, vS and current, iS, then
The op-amp input voltages are
where, as before, iI = iI– = iI+. Then substituting for vL and equating the input-voltage equations,
The resulting output current is
iL is independent of vL; vL appears nowhere in the formula for iL which is dependent only upon vI. If RL
increases causing vL to increase, it is amplified around the positive-feedback loop. An incremental
change of vl in vL is amplified and changes vO by the incremental amount of
H+ is the positive-path feedback divider from vo to v+. Then v+ is amplified by the non-inverting opamp gain, Av+. The result is that the gain from vl to vo is one; vo changes along with vl and bootstraps
the vL voltage so that no change in vS occurs. Thus, iS remains constant and vL does not affect iL
because of this bootstrapping behavior.
Finally, without the ×1 buffer, we have the Howland current source as shown below. Rf+ is
decomposed into two series resistances, Rf – Rs and Rs, so that the total resistance remains the same
as before, to satisfy current-source conditions.
The added complication over the buffered source is that now iL and iB are not separated but flow
together through Rs as
The two currents can nevertheless be separated by applying some circuit manipulation reminiscent
of the source-shifting theorem. Decompose Rs into two circuits, of iL and iB, as shown below. The left
circuit is part of the feedback circuit and the right circuit is part of the output circuit.
The feedback path still has a total series resistance of Rf, and vL is reduced from vO by vS, as before,
plus the voltage that is dropped across the Rs portion of Rf by iB. The current-source conditions
remain the same:
That is, the ratio of Rf /Ri for both positive and negative feedback paths must be equal. Then voltage
bootstrapping keeps iL independent of vL.