Problems with resistance based amplitude shifting networks ?
2:37 PM
Consider Fig. 1 where a voltage source / current source is applied to a load (A box is shown and it is assumed to be a resistance R). If we do not use power electronics for amplitude shift then efficiency apart, we can only step down the amplitude.
Both of these two methods exhibits low efficiency when the ratio between the input and output is low.
Efficiency = (Po/Pin)= (VoIo)/(VinIo). For the current source the (Vo/Vin) is constant therefore, the efficiency= (Io/Iin). Same is the case with the voltage source.
Now, since the output is always less than 1 and that if the network has to step down the input current of 10A to 1A, the efficiency is only 10% as shown in Fig.2. The red part accounts for the power loss.
 |
| Figure 1 Resistance based dc-dc voltage converter (click to zoom) |
For voltage source the output voltage can be expressed as
Vo= (Vs R)/(Rs+R)
Therefore, the value of series resistance Rs can be calculated as
Rs= R((Vin/Vo)-1)
Therefore, in order to have a real value of Rs the output voltage Vo can never be greater than the input voltage.
Similarly, for the current source network shown , the output current can be expressed as
Io=Iin((Rsh/Rsh )+R)
Therefore, the value of shunt resistance Rshcan be written as
Rsh=R/((Iin/Io)-1)
This shows that in order to have the value of Rsh greater than zero the output current amplitude cannot be greater than the input current.
Both of these two methods exhibits low efficiency when the ratio between the input and output is low.
Efficiency = (Po/Pin)= (VoIo)/(VinIo). For the current source the (Vo/Vin) is constant therefore, the efficiency= (Io/Iin). Same is the case with the voltage source.
Now, since the output is always less than 1 and that if the network has to step down the input current of 10A to 1A, the efficiency is only 10% as shown in Fig.2. The red part accounts for the power loss.
 |
| Figure 2 Ratio of input and output (Click to zoom) |
0 comments