# Improving Bandwidth of an Enhanced Howland current source

Joined Sep 9, 2019
20
Hello All!

I am currently designing an improved Howland current source to sink current from a TIA. The specifications of the circuit is as follows:

Bandwidth = 200MHz
Maximum current it should sink = 2mA
Load(TIA input impedance) = 100 to 500 ohms

I have chosen TI opamp LMH6629, suppling it with +2.5V and -2.5V and implemented it as follows:

The circuit is working perfect, just for the matter that the bandwidth of this circuit is below 100MHz - both in simulations and hardware implementation. I am been trying to improve the circuit bandwidth using few techniques. But nothing works better. I know I have to use a better opamp to be able to sink in current upto 2mA. But for the starters, I am looking for any bandwidth improvement techniques. If anybody could suggest any, it would be great.

#### crutschow

Joined Mar 14, 2008
26,994
Two things may help:
Higher frequency op amp.
Lower circuit feedback resistor values.

#### crutschow

Joined Mar 14, 2008
26,994
Why is a TIA being used as a load?
It sounds like you are going from voltage to current to voltage.
Is this for a test?

Joined Sep 9, 2019
20
Why is a TIA being used as a load?
It sounds like you are going from voltage to current to voltage.
Is this for a test?

Yes it is for a test circuit in the lab.

Could you also suggest any external circuit compensation network that could be used?

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#### crutschow

Joined Mar 14, 2008
26,994
Why do you need the complexity of a voltage to current converter for the test?
Just a resistor in series with the signal generator will provide a current into the transimpedance amp, whose input is a virtual ground.
That will give a signal bandwidth limited mainly by the signal generator.

Last edited:

Joined Sep 9, 2019
20
Why do you need the complexity of a voltage to current converter for the test?
Just a resistor in series with the signal generator will provide a current into the transimpedance amp, whose input is a virtual ground.
That will give a signal bandwidth limited mainly by the signal generator.
True! This solution already exists in our lab. Howland circuit is to replace it with a standard active circuit(capable of handling fast pulses) to sink/source current in the range of 2uA-2mA.

#### OBW0549

Joined Mar 2, 2015
3,566
Why do you need the complexity of a voltage to current converter for the test?
Just a resistor in series with the signal generator will provide a current into the transimpedance amp, whose input is a virtual ground.
That will give a signal bandwidth limited mainly by the signal generator.
True! This solution already exists in our lab. Howland circuit is to replace it with a standard active circuit (capable of handling fast pulses) to sink/source current in the range of 2uA-2mA.
What makes you think a Howland circuit is going to be any faster than an ordinary resistor? To me, the statement highlighted above makes no sense at all. Any Howland circuit you can build is going to be slower than a simple series resistor, not faster.

@crutschow was spot-on in post #2: you need a MUCH faster op amp, and you also need to reduce the values of the resistors in your circuit lest the dynamics of your circuit get hopelessly contaminated by the effects of stray capacitance.

Joined Sep 9, 2019
20
What makes you think a Howland circuit is going to be any faster than an ordinary resistor? To me, the statement highlighted above makes no sense at all. Any Howland circuit you can build is going to be slower than a simple series resistor, not faster.

@crutschow was spot-on in post #2: you need a MUCH faster op amp, and you also need to reduce the values of the resistors in your circuit lest the dynamics of your circuit get hopelessly contaminated by the effects of stray capacitance.
Sorry the main highlight in my previous reply should have been 'to sink/source current in the range of 2uA-2mA'. In case of a passive circuit, I need to re-solder different resistors to obtain different currents. Hence to save all the trouble.

Also to lower the values of resistors in my circuit is to reduce the gain? which in turn increases the bandwidth? If yes, I am a bit confused because this circuit has both positive and negative feedback paths(which is a bit complicate) and the transconductance gain is determined by R3. Thanks!

#### OBW0549

Joined Mar 2, 2015
3,566
In case of a passive circuit, I need to re-solder different resistors to obtain different currents. Hence to save all the trouble.
Why wouldn't you need to change resistors in the case of a Howland circuit? Seems to me that what's true for one approach is also true for the other, in that regard.

Also to lower the values of resistors in my circuit is to reduce the gain?
No. The gain is determined by the ratios of the resistors (other than R3), not their absolute values.

At this point, having given my advice (which is the same as that given by @crutschow), I'm done here. If you choose to follow that advice, fine; if you choose not to, that's OK too.

Joined Sep 9, 2019
20
Why wouldn't you need to change resistors in the case of a Howland circuit? Seems to me that what's true for one approach is also true for the other, in that regard.

No. The gain is determined by the ratios of the resistors (other than R3), not their absolute values.

At this point, having given my advice (which is the same as that given by @crutschow), I'm done here. If you choose to follow that advice, fine; if you choose not to, that's OK too.