impedance matching

[mahin]

hi every body
i have problem...i want to pass current (<10mA) from a load (Capacitive impedance) with very low impedance(about 12ohms at 300KHz) and want voltage across load be almost 10 volt or a bit higher , my signal generator have 600 ohms output impedance with 20vpp.
how can i do this?

 

[MikeMl]

((20/2.8)^2)/600 = 85mW

((10/2.8)^2)/12 = 106mW, so you need some power gain (cant do it with just a passive network).

Class C tuned amplifier with a pi-network output.

 

[ccurtis]

Do you mean 10 volts rms or peak-to-peak? Because if you want 10v rms you can't do it without active gain. If you mean 10vpp, then you can place a parallel inductor across the capacitor such that the parallel combination is 600 ohms impedance, splitting the 20vpp of your generator between the 600 ohm source resistance of your generator and the 600 ohm parallel combination.

 

[mahin]

load is capacitive...is this formula is correct?
and....would you help me design the circuit?
and what about current?
or if current be more important than voltage (about 3-5 mA) what can i do?

 

[ccurtis]

Try the online calculator here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rlcparc.html#c1
The formulas are also given there in a separate link.

You didn't answer the question about the 10 volts, so I'll assume 10 vpp is what you are looking for across the capacitor load. From the information you gave, your capacitor is .044 μF (C=1/(2∏FXc). Therefore, a parallel inductor of 6.33 μH across the capacitor will give an impedance of about 600 ohms at 300 KHz (I used the calculator for this, and a little trial and error, but you can use the formulas too). The current will be 20/(600+600)= 17 mA(p-p). Place this parallel inductor and capacitor directly across your generator output.

I hope that helps.

 

[mahin]

thank you,
If it is necessary to restrict the current (about 2.5mA-4mA) what can I do?
i have a question......
is it possible to control the voltage and current in a capacitor Simultaneously? for example have 10vpp across the load and pass 3mA current.
let's explain my problem: i have a load ,when connect to signal generator (output impedance=600ohms) with 10vpp, voltage across the load drop to 0.2vpp, so i calculate load impedance=12ohms (if correct).
now i want to have 10v with 3mA across the load.

 

[ccurtis]

thank you,
If it is necessary to restrict the current (about 2.5mA-4mA) what can I do?

Your welcome. For 3 mA(p-p) current you need a total impedance of 20/0.003=6667 ohms. So the total load impedance needs to be 6667-600≈6000 ohms. To get that put a 6.4 uH coil in parallel with a 6000 ohm resistor in parallel with your capacitor. The capacitor and inductor will resonate at 300 KHZ leaving a pure 6000 ohm load impedance, and the 3 mA(p-p) current will flow through the 6000 ohms resistor. The voltage across the capacitor//resistor//inductor will be 6000*.003=18vpp. You could just as well eliminate the capacitor and inductor and accomplish the same thing, though. So, I'm not sure if that's what you want to do.

You could also modify what I described in my earlier post and put a series 5466 ohm resistor between the generator and the parallel LC and the current will be 3 mA(p-p) also. But, again, you will no longer have 10vpp across the capacitor.

 

[mahin]

Your welcome. For 3 mA(p-p) current you need a total impedance of 20/0.003=6667 ohms. So the total load impedance needs to be 6667-600≈6000 ohms. To get that put a 6.4 uH coil in parallel with a 6000 ohm resistor in parallel with your capacitor. The capacitor and inductor will resonate at 300 KHZ leaving a pure 6000 ohm load impedance, and the 3 mA(p-p) current will flow through the 6000 ohms resistor.


.[if this current will flow through my load too? because 6000 ohm resistor is parallel with load ./QUOTE]

 

[ccurtis]

.[if this current will flow through my load too? because 6000 ohm resistor is parallel with load ./QUOTE]

18/12=1.5A(p-p) will flow through the capacitor. 3mA (p-p) will flow from the signal generator.

 

[mahin]

so we cannot control voltage and current Simultaneously across a load. isn't it?
i measured capacitance of load=27uF then calculate load impedance that will be Z=R+j/wc=11.9+j/(2*pi*f*c), and |Z|~=12 ,

 

[ccurtis]

so we cannot control voltage and current Simultaneously across a load. isn't it?

You mean to say control them independently? I think not. Take the simplest example of a load - a resistor. Ohms Law indicates that the current is a function of both the voltage and the resistance values. If you change the voltage, the current changes. A capacitor load is no different, except that you might be able to obtain a specific voltage and current at one instant in time only.

 

[ccurtis]

Here is one thing you could do. The current through the capacitor averages 3 mA and the voltage across the capacitor is almost 10 volts. The values can be adjusted by changing the resistor values.

 

[mahin]

thanks a lot, what about negative values i need whole signal,

 

[mahin]

thanks a lot, what about negative values i need whole signal,
or...can i use op amp as buffer in this frequency?

 

[ccurtis]

thanks a lot, what about negative values i need whole signal,
or...can i use op amp as buffer in this frequency?

Like this?

 

[mahin]

Like this?

i tried it when 100 ohm series with load but voltage drop to 14mvpp (vi=20vpp)

 

[ccurtis]

Hi. I'm not sure where you are putting that 100 ohm resistor, or why you are using it. The 600 ohm resistor in the schematic I provided is internal to your signal generator and should not be added outside the generator. The node on top of the 600 ohm resistor is one of the two output terminals of your signal generator. The other terminal is the shown connected to the bottom of C1.

 

[mahin]

no no , i didn't use 600 ohm i know about it, but because of voltage dropping and changing waveform of voltage across the load i used 100 ohm resistor series with load.

 

[ccurtis]

I'm sorry. I think I am just not understanding very well. Is everything fine with the circuit I posted except for the very small amount of voltage ripple across C1? If so, it cannot be removed entirely. You cannot have a varying current through a capacitor without some finite amount of voltage ripple across it (that I know of). What is the maximum amount of ripple voltage you can tolerate? I think I can do something to reduce it, but not eliminate it. An external power source, such as a battery, might be needed. Is that the only problem remaining and is a battery or batteries okay to use?

 

[mahin]

let me explain more....i have a plate with two small electrode inside it such a capacitor(d=1cm) and fill with salt water. (this is my load) Its capacity was measured using LC meter show 27uF and less...now i want to pass about 3-5 mA current and have about 5-10vpp across this load....my signal generator have 600 ohm impedance....voltage will drop across the load when connecting to generator
help me....