Hi, I have created simple oscilloscope circuit taken from magazine, when I use 6V battery (4 pcs AA type), the input voltage is 6V, but when I am using regulator circuit 7806, the input voltage will drop. Why this happen?, my electrics basic is not so good, sorry. TIA Yanto
Assuming you are powering the 7806 from the batteries, the output will be Vbatt - V dropout (typically 2V for the 78xx series). So if you power it with 6V, you'll get about 4V out, and not regulated properly either.
The regulator requires a "head room" of 4 volts to stay in regulation
- the input voltage needs to be 4 volts higher than the output or 10 volts if you use the 7806 and want 6 volts out. It will be in the datasheet on the part you have and you can look it up on-line easily.
Search: 7806 datasheet
Low dropout regulators require less differential to stay in regulation (different part number) LDO's work well but are more prone to oscillate if the output or input isn't bypassed with a cap - read the application notes to find out how to use the parts - any parts.
And don't forget the ripple if you are using a transformer for mains power - A rule of thumb is 1,000 microfarads per 1/2 amp of current, for about 10% ripple at 60 HZ. If the "valley of the ripple" drops lower than the differential the regulator requires - ripple shows up on the output.
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Simply put, the input voltage has to be a couple of volts higher than the output voltage. The output of the 7806 is 6 volts when regulating properly. This means the input must be 8 volts or higher for proper operation. For this regulator, use a 9 volt battery. Or, don't use any regulator at all, just hook your circuit to the 6 volt battery stack.