[SOLVED] Understanding Input Impedance (probably impedance in general)

Hello,

I have what is a really basic question.

I am working with an Voltage Controlled Amplifier which states the following about its control ports (ports 2 and 3):


Hopefully this quote is sufficient context. If not, the datasheet us here (reference page page 8 under "Control Port Drive Impedance"):


My plan was to feed one of the control ports with the output of a voltage divider that brings the voltage down to a maximum that the control port of the VCA will accept.

So my questions are:

If I make a connection from the output of my voltage divider to the control port, is my impedance effectively zero? Of course, ignoring impedance introduced by the connection.

Assuming the answer to the above is true, would a way of introducing impedance (which I do not want to do in this case) be to add a resistor in series between the output of the voltage divider and the control port?

My concern is that in order to calculate the impedance feeding the control port I need to do calculations on everything that leads up to the voltage out of the voltage divider.

Thanks,

John

johna said:

If I make a connection from the output of my voltage divider to the control port, is my impedance effectively zero?

Click to expand...

No. Assuming you're using a simple resistive divider, the impedance depends on the resistors used, and the choice of resistors depends on where the control signal is coming from and how much it needs to be attenuated.

Thanks for the response...

godfreyl said:

No. Assuming you're using a simple resistive divider, the impedance depends on the resistors used, and the choice of resistors depends on where the control signal is coming from and how much it needs to be attenuated.

Click to expand...

Here is the portion where the control signal (5v DC) enters the circuit (pin 6 of the DIN), passes through the voltage divider and connects to the control pin (3).



It may be hard to read, but the voltage divider is comprised of a 6.8K and a 1K resistor. This drops the voltage from 5v to around 0.56v.

Also, the control signal comes from a device which is a black box to me. There at least is circuitry to bring 12v to the 5v that enters my circuit. For that matter there is also a battery used which also contributes to the impedance. Reading into your response it would seem this is all relevant to calculating impedance.

John

The main contributor to the impedance is the 1K resistor. That starts by setting the impedance to 1K. Then the 6.8K resistor reduces the impedance a bit. The total impedance seen at the control input = (1K * 6.7K) / (1K + 6.7K) = 0.87K.

If you can change the 1K resistor to 47 Ohms, the impedance will be OK. If you do that, you will also have to make the other resistor much smaller.

The datasheet is already considering a higher control port drive impedance than 50 ohm by suggesting a RC filter with 100 ohm series resistor when connecting an OP outout.

In fact the low driver impedance requirement is related to achieving an exact exponential control characteristic. If you don't need it, a higher source impedance won't hurt. That's probably the explanation for the about 0.9k source impedance of the quoted schematic.

godfreyl said:

The total impedance seen at the control input = (1K * 6.7K) / (1K + 6.7K) = 0.87K.

Click to expand...

Thanks. I should be able to use lower value resistors (need to double check the power draw).

You used the formula for parellel resistors. However, a voltage divider looks like resistors in series to me. I will look into this more.

John

The resistor calculation assumes zero source impedance driving the divider, then the output impedance is in fact identical to parallel resistors. Zero source impedance may be incorrect of course.

FvM said:

The datasheet is already considering a higher control port drive impedance than 50 ohm by suggesting a RC filter with 100 ohm series resistor when connecting an OP outout.

Click to expand...

Thanks for this information. I am currently prototyping and will keep it in mind