What is the basic idea behind the so-called "improved Howland current source"?

Question

I dedicate my question and answer to Gordon Deboo - the inventor in the shadow of Howland.

The motive

The idea for this Q&A came about yesterday after a related question was closed.

Original Howland current source: There are circuit solutions that are legends and they usually bear the names of their creators. Such a legendary circuit is the current source for a grounded load. It is believed to have been invented by Howland and is therefore named after him. But in 2007, I had the good fortune to correspond with Gordon Deboo and he shared with me that at the same time he invented his Deboo integrator based on the same circuit. Apparently, as is often the case, they both independently arrived at the same circuit solution.

^{simulate this circuit – Schematic created using CircuitLab}

Improved Howland current source: Later, the "improved" solution appeared. It seems it is not Howland's doing; rather, it is named after him as a sign of appreciation and respect.

^{simulate this circuit}

Simple analysis

The voltage across R5 is VR5 = -V.R3/R1. So, the load current IL flowing through R5 and accordingly, through RL, is IL = VR5/R5 = -V.R3/R1/R5. If R2/R1 = R4/R3 or, in the simplest case, R1 = R2 = R3 = R4, and the R3+R4 voltage divider has high resistance enough, IL = -V/R5 (Ohm's law). Brilliant simplicity...

Looking for the idea

Much more important than a formal analysis, however, is to reveal the idea on which the circuit is built. This is the purpose of my question. To help find an answer, I have outlined my view below. I hope you will add to it with more explanations.

Answer 1

Basic idea

The "improved" solution is not Howland's doing. Rather, it is named later after him as a sign of appreciation and respect. However, it is interesting that the two circuit solutions are connected not only by his name but also by a common idea. What is it?

The general idea that connects the two circuits is to make an imperfect current source perfect by helping it with an additional source:

• voltage source in series (improved Howland pump)

• current source in parallel (original Howland pump)

Building a conceptual circuit

Let's follow how this is done with the first solution by building and exploring its conceptual circuit step by step.

Imperfect current source shorted

Since there are no (convenient) current sources in nature, we have to make them ourselves. The simplest way is to connect a resistor in series with a voltage source. If we short-circuit it (connect a load with zero resistance), it works under ideal conditions, and the current is exactly I = V/R = 1 mA (Ohm's law).

^{simulate this circuit – Schematic created using CircuitLab}

Imperfect current source visualized

Let's now visualize the main quantities by connecting a voltmeter and an ammeter. To simplify the schematic, we can combine them with the resistors. For this purpose, we can set their respective internal resistance in the parameters window. For example, I have chosen to visualize in this way the voltage across the current-setting resistor R and the current through the load RL.

^{simulate this circuit}

Imperfect current source loaded

The biggest problem with the simple "resistor current source" is that the load usually changes its resistance (voltage).

RL = 1 kΩ: For example, if we increase the ammeter's resistance to 1 kΩ, the current decreases to 0.5 mA.

^{simulate this circuit}

We can see it graphically by sweeping RL from zero to 1 kΩ.

Imperfect current source compensated

The problem with the above circuit is that the voltage drop VL across the load is subtracted from the input voltage and, as a result, the current decreases.

RL = 1 kΩ: The remedy is obvious - to add as much voltage as is lost across the load. So, we insert another ("helping") voltage source in series and in the same direction as the main voltage source V, and adjust its voltage equal to VL. As a result, the voltage across R is 1 V and the current through RL is 1 mA as we want.

^{simulate this circuit}

RL = 2 kΩ: It is a following voltage source; so when RL increases to 2 kΩ, the "helping" voltage increases to 2 V. As a result, the voltage across R is 1 V and the current is 1 mA as above.

^{simulate this circuit}

So, as a result of this clever trick, the voltage at the upper R terminal follows the voltage of its lower terminal; the voltage drop across R and the current do not change.

Implementation

This was only a conceptual circuit implemented by man-controlled basic electrical devices. Let's now make its electronic version.

With voltage follower

If the input voltage source V can be "floating" (not connected to ground), we can connect it between the load and the input of a voltage (op-amp) follower OA. Thus its voltage will be added to the load voltage and the sum will appear at the upper R terminal.

RL = 0: Initially, the op-amp output voltage of 1 V is entirely across R5.

^{simulate this circuit}

RL = 1 kΩ: Then it increases to 2 V distributed 1 V each across R and RL.

^{simulate this circuit}

RL = 2 kΩ Finally, the output voltage increases to 3 V (1 V across R and 2 V across RL). So the circuit perfectly supports, regardless of the load resistance, 1 V across R and 1 mA current through the load. The VR voltage drop "moves" ("raised" from below by VRL) but does not change.

^{simulate this circuit}

As you can see in the graphs below, the load voltage VL appears at the op-amp output "lifted" with V (1 V). The difference VOA - VL is constant (horizontal line).

Howland circuit

However, floating sources are inconvenient for us. How then do we make the circuit work with a grounded source (what Howland and Deboo figured out in the 60's)? We start thinking...

We have to sum two voltages - V and VL, like in biasing circuits. We can use the same idea as with the op-amp differential amplifier - connect voltage dividers to the op-amp inputs and choose the resistance values ​​so that the gain of the non-inverting input is 1. The simplest is that they are equal (R1 = R2 = R3 = R4). So VL is first attenuated 2 times by the R3-R4 voltage divider. Then, it is amplified 2 times by the non-inverting amplifier (R1-R2 voltage divider and OA), and appears unchanged at the op-amp output.

The four resistors (R1, R2, R3 and R4) and the op-amp actually form an op-amp differential amplifier acting as a summer (because V is negative).

It remains to understand how the input voltage V appears at the output.

RL = 0: To do this, we start as usual with zero load resistance (voltage). Then the non-inverting input is at zero voltage and the circuit is just an inverting amplifier. The output voltage (and the voltage across R) is VOA = VR = -Vin.R2/R1 = V.

^{simulate this circuit}

RL = 1 kΩ: Now the load voltage increases to 1 V but the op-amp output voltage also increases with 1 V. So the voltage drop across R remains the same.

^{simulate this circuit}

RL = 2 kΩ: Finally, let's increase RL by another 1 kΩ and see that the result is the same.

^{simulate this circuit}

The load voltage VL is "lifted" with V (1 V) so the difference VOA - VL is constant (horizontal line); the VR voltage drop "moves" but does not change.

Input resistance

Only negative feedback: If the circuit had only negative feedback (inverting amplifier), there would be a "fixed" virtual ground on the right R1 pin and the input resistance would simply be R1.

Both positive and negative feedback: But here there is also positive feedback introduced to the non-inverting input. Since the op-amp maintains (almost) zero voltage between its inputs, the virtual ground voltage follows the voltage of the non-inverting input thus "moving" in the same direction as the input voltage. As a result, the voltage drop across R1 and accordingly the input current Iin, decrease, and the circuit input resistance Rin increases.

To calculate Rin, you have to find the voltage V- (V+), then the difference Vin - V-, after the input current Iin = (Vin - V-)/R1, and finally the circuit input resistance Rin = Vin/Iin. Note that two voltage dividers (R-RL and R3-R4) are cascaded and connected between the op-amp output and the non-inverting input.

No positive feedback: If the load does not change significantly its voltage (e.g. rechargeable battery, Zener diode, LED, etc.), practically there is no positive feedback. So, the input resistance is as above Rin = R1.

The bottom line is that the input resistance varies depending on the load.

Conclusions

Common

• The two types of Howland current sources are connected not only by the name of Prof. Bradford Howland but also by a common idea - improving an imperfect current source by connecting an additional "helping" source.

• It can be a voltage source in series (in the case of the improved circuit solution),

• or a current source in parallel (in the case of the original circuit solution).

• These "helping" sources change their quantity when the load resistance varies; so they are dynamic sources.

"Improved" solution

• The input voltage referenced to ground appears as a floating voltage across the current-setting resistor (R5 in the question body).

• This greatly simplifies the analysis and effectively reduces it to Ohm's Law.

• Seen from the side of the load, this resistor has extremely high differential resistance (aka "bootstrapped" resistor).

• The circuit can be considered as a constant current source with dynamic voltage that follows the load voltage.

• R1 ÷ R4 and the op-amp form an op-amp differential amplifier acting as a summer (because V is negative).