The circuit in post 5 is OK then. It can pass up to about +-50mA.
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The circuit can be improved a bit though. Do you know something about electronics, so you can finish the design?
What do you mean by "voltage is dropped"?
What is output voltage with no load?
What is output voltage connected to load?
Do you know how much current is passed? Maybe the load is drawing more current than you expected.
How is the amplifier connected to the cuvette? Can you show a picture?
Cuvette is made of glass or plastic, isn't it? How do you connect the amplifier to that, or to the liquid inside? Are there metal plates inside on two sides or at the top and bottom, or do you stick two electrodes in at the top, or what?cuvette id connected to circuit's output directly
Do you mean 18V RMS? That is only 50V pk-pk. I thought you want 100V pk-pk.output voltage with no load is 18v
The load must be drawing a high current. Did you measure it?and with load drop to 8v.
Cuvette is made of glass or plastic, isn't it? How do you connect the amplifier to that, or to the liquid inside? Are there metal plates inside on two sides or at the top and bottom, or do you stick two electrodes in at the top, or what?
Do you mean 18V RMS? That is only 50V pk-pk. I thought you want 100V pk-pk.
The load must be drawing a high current. Did you measure it?
If you want higher voltage, why don't you increase the output from the signal generator?
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What kind of power supply are you using? If the voltage is too low, the amplifier may not work properly.
i read something about MCF-7:it's impedance is decreased form 3000 t0 1450 ohm by 11-101 kHz
cuvette is plastic with metal plates inside and outside on two sides and electrodes stick two sides,
power supply can support to 60v and i can use higher voltage from signal generator but i want to solve dropping voltage problem.
i don't know more about impedance of this load and want to match impedance to over problem.
Most likely: Q3 or Q4 got too hot. They should be mounted on heatsinks.i used +-40v power supply and had good output but when used +-50v, R2,Q3 and Q4 are damaged.
output voltage with no load is 18v and with load drop to 8v.
That's not easy to fix. The amplifier has an output impedance of about 100 Ohms. If the output voltage drops to about half when the load is connected, that means the load is also about 100 Ohms.i want to solve dropping voltage problem.
Most likely: Q3 or Q4 got too hot. They should be mounted on heatsinks.
Also possible: Q1 failed first and caused the others to fail. Did you use the right transistors, or others with a high voltage rating?
The amp is designed to work best with a supply of +-60V. +-50V is OK, but +-40V is too low. It will still work with +-40V, but the output will be distorted. If you want to use lower power supply voltages, then the design should be changed a bit.
That's not easy to fix. The amplifier has an output impedance of about 100 Ohms. If the output voltage drops to about half when the load is connected, that means the load is also about 100 Ohms.
If you don't want the voltage to drop, then the amplifier must have a much lower output impedance. That means you need a different circuit - this one can not be changed to have very low output impedance.
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I have a concern here. You said that you want to pass a current of 2mA to 7mA, but now it seems you are passing a current 10 times higher than that.
I estimate 8V / 100 Ohms = 80mA. Is that what you want? If you put 100V pk-pk into that cuvette, you are likely to cook the cell culture, instead of stimulating the cancer. Perhaps it would be better to work at lower voltage?
Which is more important to you - the voltage applied or the current that is passed. Maybe it would be more useful to control the current instead of the voltage?
btw, did you try to measure the current that was passed through the load? This is quite easy - just connect a resistor (say 10 Ohms) in series with the load, and measure the voltage across the resistor.
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