How Can I Maximize Output with LM331 Multiplier for Power Meter Design?

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In summary, the conversation discusses designing a power meter using the LM331 multiplier to multiply voltage and current signals. However, due to the transfer function of the LM331, the desired output of 6W cannot be achieved. The need for conditioning and scaling the inputs is mentioned in order to obtain the desired output. The conversation ends with a question about the maximum power that the circuit is expected to measure.
  • #1
syee10
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Hi,

I am designing a power meter which using the multiplier LM331 to multiply the Voltage and Current signal. The voltage and current signal will be go through the sample and hold circuit and the output of the sample and hold will be go into the LM331 multiplier. Power, P=VIcos(angle) and the LM331 having a transfer function of output W = [(X1-X2)(Y1-Y2)/10V]+Z. The problem is for example V=3V and I=2A, Output should be 6W(ignoring the cos(angle)). Due to the transfer function of LM331 i will never get the output 6W that i want. So anyone there can help me out? =)
 
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  • #2
syee10 said:
Hi,

I am designing a power meter which using the multiplier LM331 to multiply the Voltage and Current signal. The voltage and current signal will be go through the sample and hold circuit and the output of the sample and hold will be go into the LM331 multiplier. Power, P=VIcos(angle) and the LM331 having a transfer function of output W = [(X1-X2)(Y1-Y2)/10V]+Z. The problem is for example V=3V and I=2A, Output should be 6W(ignoring the cos(angle)). Due to the transfer function of LM331 i will never get the output 6W that i want. So anyone there can help me out? =)

What do you get for the output in your example? It looks like full scale is 10V? So what are the limits of your input signals?
 
  • #3
The multiplier scale the output to 10V..means the output i suppose to get is 10 but it scale to 1. What i need is a multiplier that multiply a signal with another signal and i get the exact output of both inputs. My task is actually get the power of input Voltage and input Current by using a multiplier.
 
  • #4
syee10 said:
The multiplier scale the output to 10V..means the output i suppose to get is 10 but it scale to 1. What i need is a multiplier that multiply a signal with another signal and i get the exact output of both inputs. My task is actually get the power of input Voltage and input Current by using a multiplier.

You need to condition and scale your inputs, so that you get the max of 10V at the output for the max power that you expect to be measuring. Scaling is all part of making a circuit like you are describing. What is the max power you want to measure (which will give you the 10V full scale output)?
 

1. What is the AD633 Multiplier Problem?

The AD633 Multiplier Problem is a common math problem encountered in engineering and science fields. It involves using an Analog Devices AD633 multiplier chip to perform multiplication of two input signals.

2. How does the AD633 Multiplier work?

The AD633 Multiplier chip uses an internal circuit composed of four analog multiplier blocks to perform the multiplication of two input signals. It takes two inputs, X and Y, and produces an output signal that is the product of the two inputs, Z = X * Y.

3. What are the applications of the AD633 Multiplier?

The AD633 Multiplier is commonly used in electronic circuits for signal processing, control systems, and instrumentation. It is also used in applications where analog multiplication is required, such as in analog computers and audio processing equipment.

4. What are the advantages of using the AD633 Multiplier?

The AD633 Multiplier offers high accuracy and precision in multiplication operations, with a typical accuracy of 1%. It also has a wide operating range and can handle input signals from DC to a few MHz. Additionally, it is a low-cost and compact solution for analog multiplication.

5. Are there any limitations of the AD633 Multiplier?

One limitation of the AD633 Multiplier is that it requires a dual power supply. It also has a limited output range, typically between -10V to +10V. Additionally, it is not suitable for high-frequency applications as it has a limited bandwidth of a few MHz.

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