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

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?

• The two 1.2k resistors form a voltage divider giving 1.9V which is half way between 0.5V and 3.3V or, looking at it another way, the two 1.2k resistors mix the 0.5V and 3.3V signals giving the average which is 1.9V. (Assume the non-inverting input draws no current). You say that there is only +5V supply available on the PCB but do you have 3.3V regulator output also available for a solution.
– user173271
Jun 2, 2021 at 4:34
• Those two 75R resistors at the output will halve the voltage swing and halve the dc level. Do you require 2V - 2.5V after that output voltage divider or before it?
– user173271
Jun 2, 2021 at 4:54
• Which chroma chip? Why are you trying to drive a 75 Ohm impedance-matched load? Jun 2, 2021 at 5:43
• Comment 1 : I have a 3.3Voltage regulator available on my bread board. I can also work straight from the 5V available on the PCB, but I was wondering that the power supply would be more steady. Maybe I am not correct Jun 2, 2021 at 13:58
• Comment 2 : Those are from the diagram provided by Max. I don't think I need them. I would feed straight to the Chroma Jun 2, 2021 at 15:09

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.

• Thanks for the explanation. You connected the dot as I could not understand the purpose of the x4 multiplier until now. Studying the second calculation now. My DC bias being 2V, I believe Vref would be 0.33V. Then for the 3xVin since I inject 0-0.7, I would hit anywhere between 0 to 2.1V so in total anywhere between 2V and 4.1V for Vout. Jun 3, 2021 at 2:06

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.

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.

• Thx very much for the diagram. That is 2 voltage followers (is the purpose to isolate the load? should I also apply a voltage follower for the 0.7V signal) and a summing /differentiator. I have all the components, except the VR, I guess I can substitute with resistors, am I correct? Also what is the purpose of C1 and C2? I will give test it. Jun 3, 2021 at 2:15
• Yes, the voltage followers (buffers) isolate the following resistors from the pots (trimmers) so that the reference voltages remain constant and are not affected (not varied) by the action of the following circuitry. If the 0.7V signal has a low impedance source then there is no need to buffer it but if it has any significant impedance then a buffer is necessary for the same reason as for the pots. You can use resistors instead of pots but it will sacrifice accuracy of the reference voltages as it will be difficult to achieve the exact required resistor values with standard value components.
– user173271
Jun 3, 2021 at 4:05
• ..... The purpose of the two capacitors is to filter the reference voltages (pot outputs) so that any noise/ripple on the +5V supply doesn't get passed through to the reference voltages and therefore through to the circuit's output. - Standard practice. You should include the capacitors even if you use discrete resistors instead of pots.
– user173271
Jun 3, 2021 at 4:11
• @BenjaminSAUTHIER ....Forgot to ping my comments.
– user173271
Jun 3, 2021 at 4:37
• Hi have tested a slightly different arrangement. 1) Input signal with voltage divider to lower to 0-0.5Vpp and voltage follower op amp. 2) 3.3Vref with capacitor (although 0.1microF only as I don't have 10microF capacitor) and voltage divider to get 2V through voltage follower op amp. 3) I sum up 1) and 2) with 13 Jun 6, 2021 at 2:55