# Limiting/Driving AC Current

Discussion in 'The Projects Forum' started by SCSU_Student, Jun 6, 2012.

1. ### SCSU_Student Thread Starter New Member

Jun 6, 2012
3
0
Hi, I have a school project that will be used for medical applications. I need to limit an AC current to a max of about 500uA. I would like the current's p-p to be nearly constant no matter the load but that is not a necessity. I will be working with a large frequency range during testing, 1Mhz - 100Mhz, in order to find the best frequency. I use a DDS to generate the wave in that frequency range. It outputs a wave at about 100mV. The design I have tried will work pretty well in simulation, but when I build it, the circuit does not work at all like the simulation. I will post a picture of the current driver circuit I have soon. Any suggestions would be appreciated.

2. ### SCSU_Student Thread Starter New Member

Jun 6, 2012
3
0
I attached a picture of the circuit I am using. The op-amp is an AD8099.

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3. ### wayneh Expert

Sep 9, 2010
12,368
3,224
Show your build. Those are very high, radio frequencies and very hard to design and build for a DIY project. You need short leads, circuit traces, shielding, all kinds of things to consider. But there are experts here that can look at your build and find the sore spots.

4. ### crutschow Expert

Mar 14, 2008
13,472
3,359
You have a resistor-capacitor network (C1, C2, R7) with the capacitor values being impractically small, going from the minus supply to a nameless input. What's that supposed to do?

That circuit won't work at the high frequencies you want. Does the circuit work at low frequencies?

The easiest way to get a quasi constant current at high frequencies is to use a relatively high voltage fed through a high value resistor.

5. ### SCSU_Student Thread Starter New Member

Jun 6, 2012
3
0
Here is what those three components are for. This is from the data sheet for the op-amp. The input is the external compensation pin.

C2The compensation capacitor decreases the open-loop gain at higher frequencies where the phase is degrading. By decreasing the open-loop gain here, the phase margin is increased and the amplifier is stabilized. Typical range is 0 pF to 5 pF.

R7The series lead inductance of the package and the compensation capacitance (CC) forms a series resonant circuit. RC dampens this resonance and prevents oscillations. The recommended value of RC is 50Ω for a closed-loop gain of 2. This resistor introduces a zero in the open-loop response and must be kept low so that this zero occurs at a higher frequency. The purpose of the compensation network is to decrease the open-loop gain. If the resistance becomes too large, the gain will be reduced to the resistor value, and not necessarily to 0 Ω, which is what a single capacitor would do over frequency. Typical value range is 0 Ω to 50 Ω.
C1To lower the impedance of RC , C1 is placed in parallel with R. C1 is not required, but greatly reduces peaking at low closed-loop gains. The typical value range is 0 pF to 2 pF.

I cant get it to work correctly at any frequencies. It is designed to drive the current to be about 250-300uA p-p. In simulation it sticks around that range with loads from 100ohms to 1000ohms. But when I actually test the circuit when the load is increased the p-p current goes down to much.
I did not design this circuit, I don't fully understand how it works. It was given to me by my professor. Really what I need to do is limit the current to be below 500uA p-p. I do not need to keep the current at a constant p-p current, but I don't want it dropping below 100uA.