A Comprehensive Study of the Howland Current Pump
National Semiconductor Application Note 1515Robert A. Pease January 29, 2008
A Comprehensive Study
It is well known to analog experts that you can use the positive and negative inputs of an operational amplifier to make a high-impedance current source (current pump) using a conven-tional operational amplifier (op amp). This basic circuit can put out both + and - output current (or zero current) into various loads. The theory is simple. But the practical problems in-volved are not so simple or obvious.
There are two basic circuits -- the Basic Howland Current Pump, Figure 1, and the "Improved" Howland Current Pump.The Basic circuit does good service for simple applications,but if its weaknesses are unacceptable, the "Improved" circuit may do much better for critical tasks. See Figure 5.
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FIGURE 1. The Basic Howland Current Pump
Applications for the Howland Current Pump
Sometimes a unidirectional current source (or sink) is just right. It is easy to make them with high output impedance and wide range, using an op-amp and some Darlington-connect-ed transistors. But sometimes you need a current pump that can put out a current in either direction – or even AC currents.
The Howland current pump is usually excellent for that. Cur-rent sources are often used for testing other devices. They can be used to force currents into sensors or other materials.They can be used in experiments, or in production test. They can bias up diodes or transistors, or set test conditions. When you need them, they are useful — even if you only need them once or twice a year.
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A Comprehensive Study of the Howland Current Pump
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FIGURE 2. Basic Howland Current Pump with Trim of Z OUT
The "Basic Howland Current Pump" was invented by Prof.Bradford Howland of MIT, about 1962, and the invention was disclosed to his colleague George A. Philbrick (the analog computer pioneer who was head of Philbrick Researches,Boston MA at that time). This circuit was not patented. The Howland Current Pump was first published in the January 1964 "Lightning Empiricist", Volume 12, Number 1. It is Figure 5A on page 7 of an article by D. H. Sheingold, "Impedance &Admittance Transformations using Operational Amplifiers".This can be found at /1964-1_v12_no1_the_lightning_empiricist.htm. It was also included in the Philbrick Researches Applications Manual, in 1965. Its elegance arises because the feedback from the out-put to both the + and - inputs is at equal strength -- the ratios of R1/R2 and R3/R4 are the same. While it is possible to an-alyze this circuit mathematically, it is easiest to just analyze it by inspection:
If the "output" node Vx -- which is the + input of the op amp --is grounded, it is easy to see that the "gain" is 1/R1, that is,the output current per change of the input voltage is equal to 1/R1. So you don't need a fancy set of equations for that. The resistors R2, R3, and R4 have no effect when the output is grounded, and only the + input voltage is active.
When you move the - input upward, the gain to the grounded output node is – R4/R3 x 1/R2. Since t
he ratio of the resistors is defined to be R1/R2 = R3/R4, then that gain is also equal to - 1/R1. That is easy to remember! Note that the gain is reversed for the - input.
Thus it is easy to see that if both Vin+ and Vin- are moved together, then there is no change of Iout. When Vin + rises,the "gain" to the output node is "1/R1". Then it follows that the gain for the - input is also "1/R1" but with a negative sign. So this current pump can accept positive or negative inputs. It has true differential inputs. Now all that we need to show, is
that the output impedance is high, so that the gain is correct for all output voltages and impedances, and for all inputs.It is easy to see that the output impedance is very high, using this analysis: If both signal inputs are grounded, and if the "output" node Vx is lifted up, somebody has to drive the re-sistance "R1". But as the op-amp's + input is lifted, the - input must also rise up, and the output also rises, providing just enough current through R2 to cancel the current flowing through R1, thus making the output impedance very high in-deed. The principle of linear superposition says that no matter what is Vin+ or Vin-, and no matter what is Zload, and no matter what is Vout (within the limitations that you shouldn't ask the op-amp output to put out more than it can do, in volt-age or current), the Iout will be (Vin+ - Vin-) x 1/R1. If you like to see a lot of fancy equations, see at Appendix A.
Most applications notes just indicate the circuit and the ratio,that R1/R2 must be equal to R3/R4. However they do not in-dicate how important it is to have precise matched or trimmed resistors. If all 4 resistors were 10 k ohms with a 1% tolerance,the worst-case output impedance might be as bad as 250 k ohms -- and it might be plus 250k, or it might be minus 250k!For some applications, this might be acceptable, but for full precision, you might want to use precision resistors such as 0.1% or even 0.01%. These are not inexpensive! But it may be preferable to use precision resistors rather than to use a trim pot, which has to be trimmed (and which may get mis-trimmed).
Note that if you use adjacent resistors from a tape of 1% re-sistors, the odds are that they will match better than 1/2%. But that is not guaranteed!
Figure 2, Figure 3, and Figure 4 show ways to use a trim-pot to make the output impedance very high. Typically, using 1/2% resistors and one trim pot, you can trim the output impedance to be 5 ppm of I (full scale) per volt.
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FIGURE 3. Basic Howland Current Pump with Trim of Z
OUT
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FIGURE 4. Basic Howland Current Pump with Trim of Z
OUT
However the resistor tolerance is not the only thing that needs to be trimmed out. The CMRR of the amplifier needs to be accommodated. F ortunately, an amplifier CMRR of 60 dB would cause the output impedance to degrade only to 10
megohms, not even as bad as 0.1% resistors would cause,
in the example above. And many amplifiers have CMRR bet-
ter than 80 dB. However, the CMRR of an op-amp is not
always linear -- it may be curvy or it may be otherwise non-
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degradelinear. Some amplifiers that have the advantage of rail-to-rail inputs may have a nonlinear Vos which may jump a millivolt or more as the CM signal gets within a couple volts of the +rail. Amplifiers with bipolar inputs often do have this kind of nonlinearity. Amplifiers such as LM6142 and LM6152 have nonlinearities of this type. See Appendix B. Some CMOS am-plifiers such as the LMC6482, LMC6462, etc. (See Appendix C) have a fairly linear curve of Vos, with no jumps, due to proprietary input process and circuit design.
One of the weaknesses of the Basic Howland Current Pump is its output capability. Its output node does not normally swing very close to the rail. For example, the basic 10k/10k/10k/10k scheme can only swing its output node to + or - 5 or 6 volts , with ± 15-volt supplies. If the output node rises a lot,the op-amp's output would have to rise about twice as high.When that is no longer possible, the "Improved Howland"should be considered.
If you kept the gain resistor R1 as 10k, and change R2 and R4 to 1k, you could make a 10k/1k, 10k/1k circuit, that would
let the output node rise to 10 volts with a good amplifier. How-ever, this is a little less accurate, with more offset and noise.Another weakness of the Basic circuit is the inefficiency. If you want to have a
gain of (1/100 ohms), with R1 at 100 ohms,the amplifier has to put out a lot of drive, if the load voltage swings a lot. If the load is a low voltage, such as a diode, that may not be so bad. If the load only rises a half volt, only a few mA will be wasted. But if it had to rise 5 volts, that is a lot of power wasted!
If you had to drive a heavy load, you do not have to have equal resistances at R1 and R3. You could have 100 ohms, 100ohms for R1/R2, and 10k/10k for R3/R4. Then if you want to drive the - input, the input impedance will not be very heavy.However, when you have this imbalance, you must be careful that the amplifier's Ib does not cause a big error, which may be significant if a bipolar input op amp is used.
To avoid these weaknesses of the Basic Howland, the "Im-proved" Howland generally does solve many of these prob-lems, very well. See Figure 5.
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FIGURE 5. The “Improved Howland” Current Pump
The “Improved Howland” Current Pump
In this case, you still have to trim the R's to get good CMRR and high output impedance. But the gain is set by R13, mod-ified by the ratio of R14/R15 (which is typically 1/1). Conse-quently you can use low values for R13, and keep all the other resistors high in value, such as 100k or 1 Megohm.
In the "Improved" Howland, note that it is not just the ratio of R11/R12 that must match R14/R15; it is the ratio R11 / (R12+ R13) that must be equal to R14/R15. If you do the intuitive
analysis as mentioned above, you can see that if R14 = R15,R12 will normally be (R11 - R13). Conversely, you could make R11 a little higher, to get the gain to balance out. You could put a 2k pot in series with R11. This "improved" circuit can now force many milliamperes (or as low as microamperes, if you want) into voltages as large as 10 volts, with good effi-ciency. See Figure 6.
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FIGURE 6. The “Improved Howland” with Trim for Z
OUT
Dynamics
Most engineers know (if you remind them) that it's a good idea to add a feedback capacitor across the feedback resistor of an inverting amplifier. The Howland Current Pump does like a little bit of feedback capacitor there, across R4 (or R15). A small feedback cap of 3 to 5 to 10 pF is almost always a good idea. If you are putting in a really slow current, and if the rate of change of the output voltage at the output node is not high, you could make the Cf equal to 100 or 1000 pf, to cut down the bandwidth and the noise.
Most engineers have not analyzed the dynamics of this circuit. According to the detailed analysis (at Appendix D) the "output capacitance", as seen at the output node can be as large as 80 pF, for an ordinary 1 MHz op amp. However, there are many fast amplifiers available these days, so it is usually easy to select one with a lot more bandwidth than that, if you need it. But you have to remember to design for that.
The equation for the output capacitance of the Improved Howland is derived in the latter part of Appendix E. This may be slightly better than for the standard Howland. Choice of Amplifiers
Almost any op-amp can be used in a Howland current pump. However, if you need a wide output volt
age range, a high-voltage amplifier, running on ±15 volts (or more) may be needed. Conversely, if you only need a small Vout range, a low-voltage CMOS amplifier may work just fine. As with any amplifier application, choosing the amplifier may take some engineering, to choose the right type. F or high impedance applications (resistors higher than 0.1 Megohm), FET inputs may be a good choice. If you have one left-over section of LM324, it can even do an adequate job, for resistors below
100k. A list of amplifiers with Bipolar inputs (and generally
wider signal ranges) is found in Appendix B. A list of CMOS
amplifiers with very high Zin (but smaller output range) is in
Appendix C.
If you needed a ±6 volt output swing, you might not need a
±15-volt op-amp. The "Improved Howland" may be able to
swing that far, using a good CMOS op-amp running on ± 7
volts, such as LMC6482. The "Improved" Howland is much
more effective in terms of output swing.
Current-feedback amplifiers are not normally good choices
for the Howland current pump, as they mostly work best at
low impedances. And their CMRR is rarely as good as 58 dB.
But if you need a blindingly fast current pump at fairly low
impedance levels (100 ohms to 2k), current-feedback ampli-
fiers can do a good job. Be aware that they may need a good
bit of trimming to counteract their poor CMRR.
Special Applications
The Howland Integrator
One of the obscure applications for the Howland Current
pump is the "Howland Integrator", shown in Figure 7. This is
sometimes called a “DeBoo Integrator”. If a capacitor is used
as the load, the Amplifier's Vout can be easily seen to be: Vout
= 2 x 1/RC x ∫ Vin/dt . (This assumes that R1/R2 = R3/R4 =
1.) Of course, this integrator has to have some means to reset
it, just as every integrator does. However, it is fairly easy to
reset the integrator with a single FET or switch to ground, as
shown. The Howland Integrator only works using its + input,
as the - input must be grounded. It has a positive gain, as
opposed to the conventional inverting integrator.
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