designing high frequency amplifier with low output impedance

[mahin91]

hi
i want to stimulate capacitive load ( about 2uF) by a signal generator with 600 ohms output impedance and frequency range of 100-500khz. i should design amplifier for amplify output voltage of signal generator and a buffer(i think) for matching impedance.
could anyone help me?
thank you,

 

[FvM]

Are you sure that you need to drive the capacitive load with constant voltage respectively low impedance? It seems like a bad idea. The natural way to drive a capacitive load is a current source.

 

[chuckey]

Xc= 1/ 2 *pi *f * C, XC = 1/6 ohm ~.159 ohms, this is a very low value,its series losses/inductance/lead resistance will swamp this? Even if you build an amplifier, it won't have such a low output impedance, so you will need to have a transformer to match it to the cap. So the thing to do is to build the transformer first, designed to match .2 ohms to 10 ohms (1:7 turns). Make the .2 ohm winding of a half turn of copper tape. Then do some experiments to find out whether your capacitor is still a capacitor. How much current or voltage do you want to put into this capacitor?, If transformer#1 worked, build a 600 ohm to 10 ohms (8:1 turns), use them in series with the output of your signal generator.
Frank

 

[mahin91]

Are you sure that you need to drive the capacitive load with constant voltage respectively low impedance? It seems like a bad idea. The natural way to drive a capacitive load is a current source.

load is a glass covett (2.5*7.5)(height=2.5) and two Alu plate such a capacitor, with salt water in it(4.5ml), i measured it's capacitance(=97nF). i want to pass current (only about uA) and voltage of 10-50vp-p.with frequency range of 100-500kHz. see picture below plz

 

[FvM]

100 nF sounds much more comfortable than 2 µF. 10 - 50 Vpp nevertheless involves AC currents of A rather than µA at 500 kHz. Something seems to be wrong with your calculation.

In any case, a high voltage respectively power OP (according to the actual required output current) seems to be the best solution.

 

[mahin91]

100 nF sounds much more comfortable than 2 µF. 10 - 50 Vpp nevertheless involves AC currents of A rather than µA at 500 kHz. Something seems to be wrong with your calculation.

In any case, a high voltage respectively power OP (according to the actual required output current) seems to be the best solution.

i know the impedance is very low but my references is articles that has 20v with uA current. then how can i limite current? can i put a resistor in series with load?

 

[FvM]

As said, your calculations must be wrong somehow. 100 nF and 500 kHz makes Xc = 3.2 ohm. 10 Vpp (3.5 Vrms) results in I = 1.1 Arms.

To get µA current, the voltage has to be reduced to mV or even µV levels. Your 100 nF capacitance measurement may be too high by several orders of magnitude.

 

[godfreyl]

@FvM: Hi,

Mahin's previous thread on exactly the same topic is here: voltage amplifier designing with 300kHz.

I finally gave up on that in frustration after waiting (in vain) several weeks for any sensible response to my comments and questions in post 16. Nevertheless there may be some useful information in that thread, if you're keen to help.

Regards - Godfrey

 

[chuckey]

mahin91, looking at the description of your load, I would describe it as being predominantly resistive, yet you do not mention the resistive component. I can't find the dielectric constant for water, but presuming its relatively low (or else why no water filled capacitors?), the capacitance of of your two plates will be in the order of 100s of PFs, but the shunt resistance will be in the order of 10s of ohms (was this what you were measuring?)
Frank

 

[mahin91]

@FvM: Hi,

Mahin's previous thread on exactly the same topic is here: voltage amplifier designing with 300kHz.

I finally gave up on that in frustration after waiting (in vain) several weeks for any sensible response to my comments and questions in post 16. Nevertheless there may be some useful information in that thread, if you're keen to help.

Regards - Godfrey

hi, and thank you for your help in past topics,
at that topics i had covette as load but now i have much bigger covette as you can see at pic. and now i want to have low current. i tried the circuite at topic#5 but i have mA current now i need another designing.

 

[mahin91]

hi, in fact i want to stimulate serum physiology(like salt water) with electric field, so i put it in a dish () such capacitor, then i named it capacitive load.
now i need amplifier to amplify signal generator's output with 600ohms output impedance at 100-500kHz.

 

[FvM]

I can't find the dielectric constant for water, but presuming its relatively low (or else why no water filled capacitors?)

It's quite high, about 78 up to 1 GHz, but with a considerable loss factor, thus no water filled capacitors.

I don't know what's the actual DUT capacitance and loss resistance. I think it's a essential part of the project to characterize the device and verify the results applying elementary electrical laws. Definitely not our job.

 

[chuckey]

I think K =78 is low compared to a high K dielectric of 5000. looking at the device in the absence of the water the capacitance would be in the order of pFs, so adding in the dielectric constant would bring the capacitance up to the region of 100s.
mahin91 feed the thing with a 10K resistor from a low voltage AC source, say 6V at 50 HZ, measure the volt drop across it with your DVM. Change the value of the resistor until the volt drop is 1/2 of the supply voltage. This resistor value will give you the resistance of the cell. Once you have established this, change to your signal generator and repeat but make sure that the voltmeter you are using is OK at that frequency. Between the two sets of readings you can calculate what the capacitance appears to be.
Frank