Improved howland AC current pump

Thread Starter

Neehaa

Joined Jan 20, 2017
7
Hi everyone i am using howland current pump to generate 600 micro Amphs current ,With 50KHz and load variable upto 100kilo ohms. I am practically designed attached circuit . But After 15 kilo ohms of load current is decreasing and negative peak of the output wave is clipping .
Please any one help me to solve the issue.

howland current pump_Neehaa.jpg
 

Thread Starter

Neehaa

Joined Jan 20, 2017
7
Hi i have a doubt
In one data sheet i have seen that
1. Bandwidth is 20 khz
2. Gain bandwidth product is 4 MHz
What is mean by bandwidth and gain bandwidth product
please explain me
 

kubeek

Joined Sep 20, 2005
5,656
What you have shown is not a howland current source. It is close, but not the same. Why is your load connected in the middle of Ra and Rb, instead of between R3 and Ra?
 

OBW0549

Joined Mar 2, 2015
3,135
What you have shown is not a howland current source. It is close, but not the same. Why is your load connected in the middle of Ra and Rb, instead of between R3 and Ra?
The circuit shown is a form of the Howland current source circuit commonly known as the "modified Howland current pump" or "improved Howland current pump." This application note by the late Bob Pease of National Semiconductor discusses this circuit; see Figure 5 on page 5.
 

Thread Starter

Neehaa

Joined Jan 20, 2017
7
Thanks for reply .
What should i do to increase load above 10 kilo ohms without any decrease in current.
If u have any idea please guide me.
 

OBW0549

Joined Mar 2, 2015
3,135
What should i do to increase load above 10 kilo ohms without any decrease in current.
Ohm's Law gives you the answer: if you are trying to drive ±600 μA through a 100 kΩ load, your circuit will need to be capable of applying at least ±60 volts to your load. An op amp will not be able to do that by itself; it would have to be augmented by some sort of high-voltage output stage. This application note from Linear Tech might help a bit; Figure 9 on page 7 might give you some ideas, although the circuit shown does not have sufficient bandwidth to handle a 50 kHz signal.

This is definitely NOT a trivial design task.
 
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