single supply op amp non inverting level shifter for NTSC signal processing from 0-0.7Vpp to 2-2.5Vpp

Question

I am performing an RGB modding of an NTSC CRT TV. By performing tests I found out that the chroma chip will accept 2-2.5V Vpp RGB signal to get the proper colors. So I need to amplify and shift my input signal (0-0.7Vpp RGB signal).

The PCB has only a +5V supply available, so I chose to follow the SINGLE-SUPPLY OP AMP FORMS NONINVERTING LEVEL SHIFTER configuration as indicated on Application note 4836 from Maxim integrated, with some slight modifications

https://www.maximintegrated.com/en/design/technical-documents/app-notes/4/4836.html

To simulate 0.7Vpp I used a 1.5V battery and applied a voltage divider to which I added a 75-ohm line resistor and replaced the 75-ohm terminal resistor with a 43 ohm one so to lower the voltage to 0.5V

To simplify the problem I decided to shift the signal by the full 3.3V of the voltage regulator (gain near unity)

Rf is 100 ohm and Rg is 4700 ohm to obtain a gain near 1

Here is the wiring diagram

I tested the circuit on a breadboard the circuit. I used a MAx4383 (4 channel version of 4380)

I only get about 2V on the output instead of 3.8 V (3.3V + 0.5V). I can't figure out why.

I measure the voltage on some points: I only got 2V on the IN+ pin of the op-amp. I believe the problem is because the current flows back from the 3.3V to the 43-ohm ground resistor, but on the other hand I don't understand how different my summing arrangement is from what is indicated on the application note? I would appreciate some help on that.

As a second step, I tried to isolate the 0.5V signal from the 3.3V supply by inserting a voltage follower with another MAx4383 op-amp. I get a clean 0.5V output signal from the voltage follower. Yet again when I sum up this clean 0.5V signal with the 3.3V (no ground resistor this time on my input line), instead of getting 3.8V I still get 2V.

If I disconnect the 0.5V input, I get on the output of the second op-amp about 3.3V, which is what I expect.

here below is the diagram

What did I miss and what is the best solution to this problem?

Answer 1

Let's go through the schematic in the Maxim app note to see how it works:

^{simulate this circuit – Schematic created using CircuitLab}

The equation for V+ is \$\frac{1}{2}(V_{\text{ref}}+V_{\text{in}})\$. The op amp is set for a gain of 4, so Vop will be 4V+ and Vout will be \$\frac{1}{2}V_{\text{op}}\$. Combining all of these results we have:

$$ V_{\text{out}} = \frac{1}{2}V_{\text{op}} = \frac{1}{2}\cdot 4\cdot V_{\text{+}} = \frac{1}{2}\cdot 4 \cdot \frac{1}{2}(V_{\text{ref}} + V_{\text{in}}) = V_\text{ref} + V_\text{in} $$

So you can see why you need amplification in this case -- it offsets the effect of the voltage dividers.

Now, if your input section looks like this:

^{simulate this circuit}

Then the equation for V+ becomes (approximately):

$$ V_{\text{+}} = \frac{1}{4}V_{\text{in}} + \frac{1}{2}V_{\text{ref}} $$

To go from 0.7 Vpp to around 2.0 Vpp you want to amplify \$V_{\text{in}}\$ by around a factor of 3. This means the opamp gain needs to be 24. The equation for \$V_{\text{out}}\$ becomes:

$$ \begin{align} V_{\text{out}} &= \frac{1}{2}V_{\text{op}} \\ &= \frac{1}{2}\cdot 24 \cdot V_{\text{+}} \\ &= \frac{1}{2}\cdot 24 \cdot (\frac{1}{4}V_{\text{in}} + \frac{1}{2}V_{\text{ref}}) \\ &= 3 V_{\text{in}} + 6 V_{\text{ref}} \\ \end{align} $$

Now you choose Vref to achieve your desired DC bias.

Note that a lot depends on how you are measuring Vin and where the target output voltage is measured.

Answer 2

Based on James and ErikR's feedback here is the arrangement. Sorry for my comments in previous replies, I got caught with the 5-minute limit so they were not clear.

I tested it and it does give me 0.5V on the VGA line and 1V on the reference line and eventually 2-2.5Vpp on the output, which is what I was looking for.

Although a bit dull, the picture is clean with a better color grade than before, I need to adjust the 2-2.5Vpp range and eventually do a few other tweaks on the TV board.

I have some noise on the picture which I did not have previously, even though I soldered all the components on a generic PCB (although not the best workmanship here). Thx for everyone help and hopefully this post can help ppl who try to do similar mods.

Answer 3

Gain of circuit = (2.5 - 2)/(0.7 - 0) = R1/(R2+R3) = R6/(R4+R5) = 1/1.4

Cermet multi-turn pots, VR1 & VR2, must be set to the indicated reference voltages to achieve required dc shift.

2.25V is the mid-range of the output voltage range and 0.35V is the mid-range of the input voltage range.