Why does having a high input impedance and low output impeda

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Connecting a circuit with low output impedance to one with high input impedance minimizes voltage drop due to the nature of voltage dividers, where the voltage seen by the second stage is less affected by the first stage's output. When high output impedance is connected to low input impedance, significant voltage drop occurs because the output voltage is divided across the impedances, leading to reduced input voltage. The discussion highlights the importance of impedance matching for maximum power transfer versus minimizing signal loss. Understanding voltage dividers is crucial for grasping these concepts. Overall, the configuration of circuits significantly impacts voltage transfer efficiency.
alexdr5398
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Homework Statement


Say you connect Circuit A with low output impedance to circuit B with high input impedance. Why does this cause minimal voltage drop compared to connecting them the other way around (high output to low input).

Homework Equations


P = V^2/R
P = IV

The Attempt at a Solution


I am guessing that the power must stay the same when going from circuit A to circuit B, so when the resistance increases, the voltage drops to keep the power equal. But why does this cause a "minimal" voltage drop. Wouldn't matching the impedance be the best way reduce the voltage drop?
 
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alexdr5398 said:

Homework Statement


Say you connect Circuit A with low output impedance to circuit B with high input impedance. Why does this cause minimal voltage drop compared to connecting them the other way around (high output to low input).

Homework Equations


P = V^2/R
P = IV

The Attempt at a Solution


I am guessing that the power must stay the same when going from circuit A to circuit B, so when the resistance increases, the voltage drops to keep the power equal. But why does this cause a "minimal" voltage drop. Wouldn't matching the impedance be the best way reduce the voltage drop?
Think about Vin and Vout. When the output resistance of a stage is high into a low-input impedance stage, what happens to the Vin seen by the 2nd stage? Think voltage dividers... :smile:
 
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berkeman said:
Think about Vin and Vout. When the output resistance of a stage is high into a low-input impedance stage, what happens to the Vin seen by the 2nd stage? Think voltage dividers... :smile:

I've never learned about voltage dividers. Would the Vin in the 2nd stage be larger than the Vout in the 1st stage?
 
alexdr5398 said:

Homework Equations


P = V^2/R
P = IV

The Attempt at a Solution


I am guessing that the power must stay the same when going from circuit A to circuit B, so when the resistance increases, the voltage drops to keep the power equal. But why does this cause a "minimal" voltage drop. Wouldn't matching the impedance be the best way reduce the voltage drop?

I think you are confusing maximum power transfer and minimum signal (voltage) loss.
 
alexdr5398 said:
I've never learned about voltage dividers. Would the Vin in the 2nd stage be larger than the Vout in the 1st stage?
No, Google voltage divder circuits and do a little readng. You'll find that the Voltage Divider equaton is like this:

V_{out} = V_{in} \frac{R_d}{R_i + R_d}

Try that search, and try answering again. :smile:
 
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