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7805 regulator capacitors

HellasTechn

Apr 14, 2013
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Not completely. In this circuit the resistor will help to dissipate power which then doesn't have to be dissipated by the regulator, thus keeping the regulator cooler. The output will still be regulated.
Exactly.

So why not simply use a lower voltage supply to start with?
9v is fairly common and cheap.
Resistor defeats the whole purpose of using a regulator .

Yes but in my case 9V is not an option. I use 12V as input in my project which is used for driving orhter circuits and then it is stepped down to 5V to drive the microcontroller.

Hello,

The 20 Ohms resistor will be to large, as it will drop 20 X 0.3 = 6 Volts, leaving 5.3 volts at the input of the regulator when the input voltage is 12 Volts and the diode drop is 0.7 Volts.
The regulator needs at least 2 volts overhead to be able to regulate.

View attachment 46390

Bertus

I am going to double check with my multimeter and get back to you on that. 300ma is peak current consumpion. The average current consumption is about 210ma i think.
I didnt mention that in post 1 because my actual question is about Capacitors value and type selection :)
 
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bertus

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Hello,

When you lower the resistor to 15 Ohms, you will not reach the critical values for regulation.

Bertus
 

hevans1944

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@HellasTechn, you seem to be ignoring the datasheet recommendations concerning the capacitor values you should use. The input capacitor is a high-frequency bypass capacitor, needed ONLY if there is substantial distance between the 12 V "smoothing" capacitor and the three-terminal "7805" linear regulator input. It assumes there is NO resistor between the linear regulator input and the raw, filtered, DC output power supply. The output capacitor is a deliberately small value selected (only if needed) to avoid changing the linear regulator characteristics when the circuit is operating normally under load.

The raw DC should be derived from a full-wave bridge rectifier and "smoothed" or filtered with an electrolytic capacitor whose value depends on the current being supplied and the allowable ripple on the input. At the peaks of the input voltage, this capacitor will charge up and then linearly discharge into the voltage regulator input until the input voltage begins to increase again. The maximum and minimum values of this pseudo-sawtooth wave form must be within the allowable input voltage ranges of the the linear regulator.

So, knowing the discharge current of the filter capacitor is linear, and knowing the constant current of this discharge (why constant? because the output voltage is constant across a presumably constant resistance load, therefore the load current and capacitor discharge current is constant), and knowing something about the period of the pseudo-sawtooth wave form (about 1/120 seconds for 60 Hz line supply), you can calculate a conservative value for the smoothing capacitor. It is only if this capacitor must be physically located some distance from the regulator that the 0.33 μF input bypass capacitor is required.

The output bypass capacitor is only recommended for noise reduction. Because the output and its load is an integral part of the closed-loop negative feedback linear voltage control circuit, you should not add large values of "smoothing" capacitors to this output, lest you disturb the feedback characteristics so carefully designed into this integrated circuit. In particular, DO NOT add 100 μF aluminum or tantalum electrolytics between the output and ground.

Gather together and READ as many datasheets as you can find, from different semiconductor manufacturers, on the 78xxx series of three-terminal voltage regulators. IIRC, at least one of these datasheets explains why you do not want to add a large capacitance to the output terminal because it can actually cause instability in regulator operation.
 

HellasTechn

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Hello,

When you lower the resistor to 15 Ohms, you will not reach the critical values for regulation.

Bertus
do you mean i will reach the critical values for regulation ?

For now i can say for sure that the system is working. Like i said i will check again with the dmm.
 

HellasTechn

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@HellasTechn, you seem to be ignoring the datasheet recommendations concerning the capacitor values you should use. The input capacitor is a high-frequency bypass capacitor, needed ONLY if there is substantial distance between the 12 V "smoothing" capacitor and the three-terminal "7805" linear regulator input. It assumes there is NO resistor between the linear regulator input and the raw, filtered, DC output power supply. The output capacitor is a deliberately small value selected (only if needed) to avoid changing the linear regulator characteristics when the circuit is operating normally under load.

The raw DC should be derived from a full-wave bridge rectifier and "smoothed" or filtered with an electrolytic capacitor whose value depends on the current being supplied and the allowable ripple on the input. At the peaks of the input voltage, this capacitor will charge up and then linearly discharge into the voltage regulator input until the input voltage begins to increase again. The maximum and minimum values of this pseudo-sawtooth wave form must be within the allowable input voltage ranges of the the linear regulator.

So, knowing the discharge current of the filter capacitor is linear, and knowing the constant current of this discharge (why constant? because the output voltage is constant across a presumably constant resistance load, therefore the load current and capacitor discharge current is constant), and knowing something about the period of the pseudo-sawtooth wave form (about 1/120 seconds for 60 Hz line supply), you can calculate a conservative value for the smoothing capacitor. It is only if this capacitor must be physically located some distance from the regulator that the 0.33 μF input bypass capacitor is required.

The output bypass capacitor is only recommended for noise reduction. Because the output and its load is an integral part of the closed-loop negative feedback linear voltage control circuit, you should not add large values of "smoothing" capacitors to this output, lest you disturb the feedback characteristics so carefully designed into this integrated circuit. In particular, DO NOT add 100 μF aluminum or tantalum electrolytics between the output and ground.

Gather together and READ as many datasheets as you can find, from different semiconductor manufacturers, on the 78xxx series of three-terminal voltage regulators. IIRC, at least one of these datasheets explains why you do not want to add a large capacitance to the output terminal because it can actually cause instability in regulator operation.

Okay that makes sense.

In my case the 12V comes from an SMPS far from the regulators location. There are at least 5 meters of cable between.

So i have chosen wrong capacitor values. I should go with what the datasheet recommends.

I have to say though that i have seen 7805 regulators oscillating the output when no capacitors are used. I am not sure why that happens.
 

Harald Kapp

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I use 12V as input in my project which is used for driving orhter circuits and then it is stepped down to 5V to drive the microcontroller.
In that case a switch mode regulator is the less power hungry alternative. You can get them as 7805 plug-in replacements.

7805 regulators oscillating the output when no capacitors are used. I am not sure why that happens.
This is for LDOs but the theory is comparable to 7805's behavior.
 

HellasTechn

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Hello,
The 20 Ohms resistor will be to large, as it will drop 20 X 0.3 = 6 Volts, leaving 5.3 volts at the input of the regulator when the input voltage is 12 Volts and the diode drop is 0.7 Volts.
The regulator needs at least 2 volts overhead to be able to regulate.
View attachment 46390
Bertus

Here are the actual readings.
The actual resistor value is 26 Ohms and With 12V input under normal operation the circuit consumes 180ma, the voltge across the resistor is 2.4V and the input to the 7805 is 9.6V
The measurements where taken without the diode so subtracting 0.5V (actual diodes voltage drop).
Thathe actual readings indicate that im working within limits.

I cant mathematically explain it but readings are readings yes ?
 

bertus

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Hello,

The 26 Ohms and 180 mA do not match with the mentioned 2.4 Volts.
When I do the calculations the voltage accross the resistor would be 26 Ohms X 180 mA = 4.68 Volts.
I do not have an explanation why the voltage accross the resistor is 2.4 Volts.

Bertus
 

HellasTechn

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I know. I really do not know what to say. It probably has something to do with the 7805.
 

Hunter64

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You mentioned a maximum current of 300mA. With the 26 Ohm resistor you'll lose 7.8V.

Good luck with reliabillity.
 

HellasTechn

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The 26 Ohms and 180 mA do not match with the mentioned 2.4 Volts.
Bertus

You mentioned a maximum current of 300mA. With the 26 Ohm resistor you'll lose 7.8V.
Good luck with reliabillity.
Ive been thininking about the math... They didnt "add up" because i forgot that i took the current measurement at the input of the device, not at the 7805 output. So i was way off.
I have measured the current going out the 7805 (actually consumed by the 5V circuits) and goes between 70 and 90ma depending the state of the i/o pins. That explains the 9.6Volts on the 7805's input
 
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HellasTechn

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Thank you all four your advices. Special thanks to sir hevans1944 for post #23

Though i have seen many circuits using 7805 with aluminum electrolytic capacitor between the output and gnd, i guess that the people who wrote the datasheets know better thatn i do so i used the capacitors recommended by them. 300nf ceramic at the input and 100nf ceramic at the output.
 
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