Please humor us and try a voltage divider with 1K and 4K and show us the result. I can see no reason why that would not work.

Bob

 

Please humor us and try a voltage divider with 1K and 4K and show us the result. I can see no reason why that would not work.

Bob

If he did as normal procesure then the problem could be in the output part that we don't know.

 

What about a transistor acting like an open collector? Supply the desired voltage to the collector via a resistor. When the square wave goes high the transistor conducts and pulls the collector to ground (an inverter circuit).

 

Like this ? ? ? [edit] oops - ignore that little plus sign. [end edit] [edit #2] the base of the transistor will also need a resistor. Forgot that when I drew this. [end edit #2]

 

Quick note - the pulses will be inverted. If the 5V pulse is high for 80% of the period then the 1V pulse will be LOW for 80% of the period. If you're working with a square wave then you shouldn't have any problems. But if the high side versus the low side is not symmetrical then my drawing won't work like you want.

 

I would like to know why a voltage divider doesn’t work before proposing any other solution.

Bob

 

25 posts and we're going nowhere fast. Your question needs some work, and some information.

I tried using a simple voltage divider (with two resistors) put that didn't work. If the values of resistors are too high, in the Kilo ohm range , then there is nothing in the output.

What does that mean?

If I use low value resistors, then the signal output is really bad.

What does that mean?

The AD835 has an input impedance of 100 K. The 6903 is characterized for high output currents of 1 mA and 4 mA. A 2-resistor attenuator with a Thevenin equivalent impedance of less than 1 K will introduce an amplitude error of less than 1% at the 835 input, without overloading the 6903 output stage. Is this 10 times better than what you need, 1000 times worse than what you need, or something else?

ak

 

Like this ? ? ? [edit] oops - ignore that little plus sign. [end edit] [edit #2] the base of the transistor will also need a resistor. Forgot that when I drew this. [end edit #2]
View attachment 256922

Where did you get the 1V, add another LM317?

 

I’m not very experienced, but why not an inverting op amp configured with a gain of 0.2? If the pulse width is critical, use a transistor to invert the output?

Am I crazy?

 

If he did as normal procesure then the problem could be in the output part that we don't know.
View attachment 256908

A part of the Internal structure of the Programmable divider is linked from here.

 

Am I crazy?

That is beyond the scope of this forum.

The oscillator output is a 50% duty cycle square wave, so inverting it should not cause any downstream issues.

The 835 is specified for bipolar supplies only, so you need to either pee a little offset current into the summing node, or couple the output through a capacitor.

Still waiting for details about why a 2-resistor attenuator doesn't work.

ak

 

Where did you get the 1V, add another LM317?

I left certain things out because I don't know what the TS has available. However, it's quite possible a voltage divider could provide sufficient amperage for whatever he is running. He DID say that too low a resistance results in a dirty square wave and too high results in nothing happening. Somewhere in between is where the answer lay. I wouldn't recommend another regulator. I think the 5V could be divided down sufficiently. The easiest solution consists of very few parts. If those options don't produce the desired results then something more is needed. Since I don't have part-in-hand I can only offer minor guidance.

With my drawing above, the square wave would be inverted. A second transistor to invert the signal back would also potentially solve the problem.

 

Man, 32 replies and still no resolution!

 

That's how it gies when TS is uncooperative.

Bob

 

To design a simple voltage divider that will reduce the 5.00 volt square wave to 1.00 volts I need to know the resistance of both the source and the load. To get a voltage divider that will reduce the 5 volt square wave to a 1 volt square wave we can guess that the source resistance is zero and that the load resistance is very high. Then, because precision resistors are not handy, I can build a series string of five 1000 ohm 1% tolerance resistors, and apply the 5 volts to the top of the string and find the one volt across the bottom 1000 ohm resistor.
Realize now that two guesses were made here, and both of them matter quite a bit. If the driving circuit has feedback to keep the voltage at 5 volts independent of load, the the guess of zero source resistance is reasonable, and if the load has a reasonably high input resistance, such as a stereo amplifier, then the high resistance input guess is reasonable. But if it is the 50 ohm impedance input of an oscilloscope, then the input will be much less.

In my first post I mentionned the output of the LTC6903 goes directly into the AD835 input. From the datasheet of the AD835 ports are : High input impedances.

You can try the circuit below.

Or you can measure the Vout resistor from Ve(Pin e of bjt) and Ground (You should turn off the power, and separate the SVR from the transistor), and then to find two fixed resistors to replace the SVR, if you can find the appropriate values of resistors.

View attachment 256885

this is the best solution for me, The resistor divider works but with low ohmic values not in the K ohm values. I was down to 2ohm with 10 ohm resistors, But I needed a Pot and as soon as I introduce that in the circuit, then the output was unstable. The 2N3904 is a low cost transistor anyway. thank you

 

The resistor divider works but with low ohmic values not in the K ohm values. I was down to 2ohm with 10 ohm resistors.

Then you were doing something wrong. Show us how you connected your voltage divider.

Bob

 

Output resistance of LTC6903 is 45Ω.
Input resistance of AD835 is 100kΩ.

You have a wide range of resistance to choose in order to design a working 5:1 attenuator.



Hence the design criteria are:

R1 + R2 > 450Ω
R2 < 10kΩ
R1 = 4 * R2

For example,
R1 = 4kΩ
R2 = 1kΩ

Another example,
R1 = 1000Ω
R2 = 250Ω

If you want to be able to adjust it, use a 1000Ω trim-pot.

 

Maybe the resistor divider worked with extremely low resistor values but not with high resistor values because it was built on a breadboard and the high resistances and their wires picked up interference.

 

We have no idea what he built since he won’t give us a schematic..

Bob

 

He said he tried 2 ohms and 10 ohms resistors without seeing that it caused a severe overload.