My design to a large degree follows that of Fig. 27 on p.16 of the data sheets for AD737 (rev. K) that I'm including a link to.

1. Can you spot any errors in or improvements that should be made to my circuit?

2. What is the purpose of the two diodes in the circuit?

3. Can I construct this circuit on strip board? I have the AD737 in the DIP package.

Thanks in advance for your suggestions and comments.

-Pete

https://www.analog.com/media/en/technical-documentation/data-sheets/ad737.pdf

 

Resistor R2 in my circuit I believe doesn't need to have a power rating anywhere near the 2 Watts that I show. In Fig. 27 of the data sheets, the resistor in the same place as my R2 has a wattage rating of 1W which I also don't understand.

 

R2 and the two 1N4148s are to protect over-voltage input. In case the switch is mistakenly put in the top position (right after C4) and the input is 100V, you do need R2 and D1/D2 to clamp the voltage to AD737's pin 2 to about V-/V+. In such case, 2W rating is reasonable.

 

For the voltage clamp function, I was recently reminded that the power coming into the positive bus has to go somewhere. And the 7805 will not sink it very well. I suggest a string of 1 amp diodes to be forward biased into full conduction just above 5 volts. And maybe a fast one amp fuse in series with the input.

 

For the voltage clamp function, I was recently reminded that the power coming into the positive bus has to go somewhere. And the 7805 will not sink it very well. I suggest a string of 1 amp diodes to be forward biased into full conduction just above 5 volts. And maybe a fast one amp fuse in series with the input.

With an unattenuated voltage of 100 VRMS at input, this would result in about a 100V drop across resistor R2 in my circuit. So then this would produce somewhat better than 20 mA of current through R2, isn't it true? Are you saying that the L7805 regulator is not capable of sinking 20 mA of current?

 

I do not believe that the LM7805 series regulator is capable of sinking ANY current from the output to the common. That will destroy it!

 

A three-terminal voltage regulator’s circuitry is designed to only source current, never ever to sink current.
The easiest solution would be to include a constant load. The simplest load would be a LED. 20 mA is a lot for a single indicator LED, so you could add a second one.

 

A three-terminal voltage regulator’s circuitry is designed to only source current, never ever to sink current.
The easiest solution would be to include a constant load. The simplest load would be a LED. 20 mA is a lot for a single indicator LED, so you could add a second one.

And a series stack of seven one amp diodes to start conducting at about 4.9 volts. A more serious clamp scheme.

 

A three-terminal voltage regulator’s circuitry is designed to only source current, never ever to sink current.
The easiest solution would be to include a constant load. The simplest load would be a LED. 20 mA is a lot for a single indicator LED, so you could add a second one.

What would say then about the over-voltage protection of Dick Cappel's adapter with the AD737? It was his circuit design that gave me the idea to obtain a reference voltage via a linear voltage regulator.

http://www.cappels.org/dproj/RMS-TO-DC_Adapter_For_DVM/TRUE_RMS-TO-DC_Adapter_For_DVM.html

 

I had missed that V+ goes to a battery, which itself can absorb significant amounts of current.
As long as this condition is maintained exactly, you should be fine.
EDIT; your schematic is missing the two zener diodes.

 

Here is my version 2. I did a rudimentary test of this version with hardware and it functioned correctly. For this version I drew on Rod Elliott's way of configuring AD737 which is somewhat different from how it is shown in the data sheets.

https://www.sound-au.com/project140.htm

In obtaining the resistance values of 90k, 9k, and 900 Ohms for the voltage divider of my adapter, I found that without too much difficulty I could use resistors of 5% tolerance in parallel and get the required resistance +/- 1%.

-Pete