I'm designing a power supply using a 7805 regulator and I need a very low output noise voltage at 1 Hz, hopefully around 1 mVpp. Will increasing the size of the output capacitor help me do this, or should I be adding additional circuitry?
Electronic – Voltage regulator output ripple
voltage-regulator
Related Solutions
If I understand you correctly, you are trying to adjust the output voltage of your 7805 regulator to output 5V - 10V by adjusting the voltage at it's ground pin between 0V and 5V.
If you are doing this, the pin needs to be able to sink current in order to maintain regulation, so whatever the power source used, it needs to be capable of this. Many supplies will only source current, and if this is the case what you are seeing will occur.
A simple resistor divider with potentiometer can be used, or an opamp to drive the 7805 ground pin is maybe the best solution, preferably with a output including its negative rail if you want to go all the way down to 5V. You will find many example adjustable supply circuits on the web, and in the datasheets of the regulators also.
Here are a couple of example circuits:
Pot adjust, from this page with more info:
Opamp adjust, from datasheet pg.24 (don't use a 741 if you can possibly avoid it, as they are obsolete):
You are correct in that the linear regulator appears to be using 17mA at 37V so that is a power of:
Vin * Iregulator
37V * 0.017A = 0.629W
But you need to also consider that the linear regulator that you are using is also dissipating power of:
(Vin - Vout) * Iload
(37V - 5V) * 0.031A = 0.992W
Thus you have a total power loss in the linear regulator of:
0.629W + 0.992W = 1.621W !!!
You really need to be using a switching buck regulator for this job. If you do not want to build such a thing you can find a plethora of ready made modules on web sites the like of Amazon or eBay. They do not cost a lot either.
Related Topic
- Electronic – Switching Voltage Regulator Problems
- Electronic – DC voltage ripple help
- Electronic – LM317 regulator with NPN pass transistor alternatives and ripple consideration
- Electronic – Why do voltage regulator ICs have big ripple rejection ratio
- Electrical – 7805 voltage regulator reverse polarity
Best Answer
The impedance of a capacitor is \$ 1/(2 \pi f C)\$ so a 1000µF cap will have 160 ohms impedance at 1Hz.
Your regulator has an output impedance too. This is due to the internal feedback loop which adjusts the output transistor drive to respond to changing load currents. This loop is not infinitely fast, and it does not have infinite gain, so it will not hold the output voltage perfectly at the set point, but at low frequencies output impedance is pretty low. Say, 0.01 ohms.
Thus, considering the two impedances in parallel, the regulator will control the output completely at 1Hz. The cap only comes into play at much higher frequencies.
The low-frequency output ripple will depend entirely on regulator noise, and the input ripple divided by regulator ripple rejection.
Why do you need 1mV ripple on 5V?