Constant current source

[gjoo]

Hi,
In this constant current circuit, I am not 100% sure on how it operates. More , specifically r5, r7 r9 and c4, what are there purpose in feeding the MOSFET. Any help appreciated. It's a constant current source for an electromagnet.

Thanks,
GJ

 

[crutschow]

specifically r5, r7 r9 and c4, what are there purpose in feeding the MOSFET

The are just to roll-off the feedback high-frequency response and improve the stability of the op amp feedback loop.

The op amp negative feedback loop always tries to keep the voltage between its (+) and (-) inputs very close to 0V (typically a few mV or less), thus it adjusts the MOSFET gate voltage so that the voltage across R14 is is equal to the op amp (+) input voltage (R9 carries no significant DC current so the op amp (-) input voltage equals the voltage at R14).

Since R14 is 1 ohm, the current through R14 (and thus the MOSFET drain load) is 1mV input generates 1mA MOSFET current, as noted on the schematic.

 

[danadak]

Hi,
In this constant current circuit, I am not 100% sure on how it operates. More , specifically r5, r7 r9 and c4, what are there purpose in feeding the MOSFET. Any help appreciated. It's a constant current source for an electromagnet.

Thanks,
GJ

R5 is to decouple the Cload presented by the MOSFET and its Miller C. This
helps to increase the phase margin due to Zo of the OpAmp and the MOSFET
C.



R7 provides a DC bias return path for the OpAmp.


Regards, Dana.

 

[gjoo]

The are just to roll-off the feedback high-frequency response and improve the stability of the op amp feedback loop.

The op amp negative feedback loop always tries to keep the voltage between its (+) and (-) inputs very close to 0V (typically a few mV or less), thus it adjusts the MOSFET gate voltage so that the voltage across R14 is is equal to the op amp (+) input voltage (R9 carries no significant DC current so the op amp (-) input voltage equals the voltage at R14).

Since R14 is 1 ohm, the current through R14 (and thus the MOSFET drain load) is 1mV input generates 1mA MOSFET current, as noted on the schematic.

 

[gjoo]

Thanks for the responses. But, what purpose does R9 serve, is it part of a voltage divide with R 14. Also, how is the size of r5 chosen?

 

[danadak]

R9 allows the tiny bias current of U2A inverting input to flow, a DC
path, so OpAmp output stays in linear region, eg acts as a follower at DC
so the OpAmp output does not saturate into a rail depending on
sign of bias current. And as crutschow points out is part of the
compensation loop along with C4.

R5 can be analytically determined, or in practice most folks pick a
value in the ~ 50 - 200 ohm region. Then look at transient response
as it decouples the MOSFET Cgate and miller from severe C loading
of U2A output which produces inadequate phase margin, ringing, even
oscillation. So it too is part of the compensation loop in a sense. Interesting
is datasheet has this chart which helps you (along with phase response
charts) -



A useful translation for determining R5 is to translate its series R & C to
an equivalent parallel R & C which allows you to see how much C
is reflected to the OpAmp output. Then refer to C loading graphs in
datasheet to see if you have adequate phase margin.

Series to Parallel Impedance Transformation

Summary of the series-parallel transformation for impedance matching in radio circuits

aaronscher.com

Regards, Dana.

 

[gjoo]

Thanks for all the feedback, I've been slow to respond because I'm battling the flu. But, one last question about r7, it's the pull down resistor to turn on the MOSFET, but how is its resistance value determined here?

 

[danadak]

Its actually to turn OFF MOSFET. And I am not sure why its needed as
the OpAmp will drive to the Vss rail, so curious myself why its there.


Regards, Dana.

 

[Harald Kapp]

R7 prevents any charge from building up when the circuit is not powered. It thus prevents unintentional turn-on of the MOSFET in this case.

 

[gjoo]

How is the size of that resistor chosen?

 

[Harald Kapp]

More or less arbitrarily and from experience. Large enough to not disturb the normal operation, small enough to keep the gate discharged when power is off.
Any value from 100 k to 1M may work equally well for this application.

 

[gjoo]

How much does the choice of r7 effect the size of r9 in this circuit?

 

[Harald Kapp]

Negligible.

 

[gjoo]

Is R9 even necessary

 

[Harald Kapp]

Yes. Together with C4 it forms a low pass filter for stability.

 

[gjoo]

I think I'm starting to understand it now. Thanks for all the help

 

[gjoo]

One thing that's been bugging me, I thought c4 and r5 form a high pass filter and c6 with or without r13 also is a high pass filter?

 

[danadak]

Do a sim and test out your thoughts.

RC high pass looks like :




Regards, Dana.

 

[gjoo]

I will thanks, your example is why on the surface it appears that way.

 

[danadak]

I will thanks, your example is why on the surface it appears that way.

The OpAmp is the source of signal so it drives R5 and C4.
But C4, its "effective" ground end, is R14 since its approaching
a V source due to its very low Z. So effectively C4 is from the OpAmp
output to ground, hence with internal Zout_opamp, forms a LPF.

But we have a path from V on R14 feeding back to OpAmp inverting
input, and with C4 strapped across OpAmp becomes an integrating
response, eg. another LPF. Note R14 acts effectively as a Vsource
because of its very low Z.

R5 along with parasitic and miller C of MOSFET form yet a 3'ed LPF.

So not sure where you visualize a HPF.

Note this form is not a HPF as the V source makes C4 irrelevant.




Regards, Dana.