# Define voltage level correctly in the output channels

**URL:** <https://forum.liquidinstruments.com/t/define-voltage-level-correctly-in-the-output-channels/880>\
**Category:** Moku Cloud Compile\
**Created:** [January 5, 2024, 6:56pm UTC](https://forum.liquidinstruments.com/t/define-voltage-level-correctly-in-the-output-channels/880 "2024-01-05T18:56:55Z")\
**Posts on this page:** 3\
**Page:** 1

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**Author:** ![roflo](https://avatars.discourse-cdn.com/v4/letter/r/90ced4/32.png) [@roflo](https://forum.liquidinstruments.com/u/roflo)\
**Post date:** [January 5, 2024, 6:56pm UTC](https://forum.liquidinstruments.com/t/define-voltage-level-correctly-in-the-output-channels/880/1 "2024-01-05T18:56:55Z")

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Hi guys, I’m in the process of building different Tigger Vairanten with the Muko: Go. And I have moved the DC\_Sequencer.vhd example. My code works from the triggers point of view, but the voltage on the output channels is only 800 mV and should be 5V in HI\_LVL.

Could you take a look at my code and help me further?

library IEEE;  
use IEEE.Std\_Logic\_1164.all;  
use IEEE.Numeric\_Std.all;

architecture Behavioural of CustomWrapper is

```
-- Constants for 5V and 0V in mV
constant HI_LVL : signed(15 downto 0) := to_signed(5000, 16); -- 5V in mV
constant LO_LVL : signed(15 downto 0) := to_signed(0, 16); -- 0V in mV
constant Vref : signed(15 downto 0):= to_signed(2500, 16); -- 2.5V in mV

signal Step : std_logic;
signal Trigger, TriggerDly : std_logic;
signal DCLevelAddr : unsigned(6 downto 0);

```

begin  
– Input Schmitt trigger functionality  
– to help reduce trigger on noise  
SCHMITT: process(Clk) is  
begin  
if falling\_edge(Clk) then  
if Reset = ‘1’ then  
Trigger \<= ‘0’;  
elsif InputA \>= Vref then – Compare with the reference voltage  
Trigger \<= ‘1’;  
elsif InputA \< Vref then  
Trigger \<= ‘0’;  
end if;

```
        TriggerDly <= Trigger;
    end if;
end process;

-- Step the DC index on each rising edge
Step <= Trigger and not TriggerDly;

ADDR_COUNTER: process(Clk) is
begin
    if falling_edge(Clk) then
        if Reset = '1' then
            DCLevelAddr <= (others => '0');
        elsif Step = '1' then
            -- Will wrap after 127
            DCLevelAddr <= DCLevelAddr + 1;
        end if;
    end if;
end process;

process(Clk) is
begin
    if falling_edge(Clk) then
        if Trigger = '1' and DCLevelAddr mod 2 = 0 then
            OutputA <= HI_LVL;
            OutputB <= LO_LVL;
        elsif Trigger = '1' and not DCLevelAddr mod 2 = 0 then 
            OutputB <= HI_LVL;
            OutputA <= LO_LVL;
        elsif Reset = '1' then
            OutputA <= LO_LVL;
            OutputB <= LO_LVL;
        end if;
    end if;
end process;

```

end architecture;

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**Author:** ![PaulC](https://yyz1.discourse-cdn.com/flex035/user_avatar/forum.liquidinstruments.com/paulc/32/23_2.png) [@PaulC](https://forum.liquidinstruments.com/u/PaulC)\
**Post date:** [January 6, 2024, 12:38am UTC](https://forum.liquidinstruments.com/t/define-voltage-level-correctly-in-the-output-channels/880/2 "2024-01-06T00:38:06Z")

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Hi roflo,

welcome to the Liquid Instruments forum.

This is a good question and relates to the scaling of the VHDL signed, 16 bit numbers and the corresponding output voltage.

A decimal value of 5000 will give approximately 770 mV on the output. If you refer to :

[https://compile.liquidinstruments.com/#/docs/intro](https://compile.liquidinstruments.com/#/docs/intro)

you will see the Moku:Go DAC (output) is scaled at 6550 ‘bits per volt’, thus a decimal of 5000/6550 = 763 mV.

If you want to drive 5V on the DAC output, you will need 5 \* 6550 = 32750 e.g.

> `constant HI_LVL : signed(15 downto 0) := to_signed(32750, 16); -- 5V scaled to signed 16 bits`

Regards,

Paul.

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<div class="post-metadata">

**Author:** ![roflo](https://avatars.discourse-cdn.com/v4/letter/r/90ced4/32.png) [@roflo](https://forum.liquidinstruments.com/u/roflo)\
**Post date:** [January 9, 2024, 1:23pm UTC](https://forum.liquidinstruments.com/t/define-voltage-level-correctly-in-the-output-channels/880/3 "2024-01-09T13:23:29Z")

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Hi Paul, Thanks for your answer. I try the way you showed but there is no change in the output voltage. I solved it know with the PID afterwards to rise the signal, but there must be a more eloquent way.