7805s measuring 4.85V and 4.9V no load

Thread Starter

spinnaker

Joined Oct 29, 2009
7,830
I have a couple of 7805s. One is in circuit but just with a couple of filter caps on the input (still building the rest of the circuit).

The other I bread boarded with nothing but the regulator.

I connected walwart. The input on both chips measured 16.6 VDC.

On the 7805 on my circuit board I am only measuring 4.85 VDC. The bread boarded one does just a bit better at 4.9 VDC.

Both of these regulators are from the same source.


I have a third 7805 hat I got from the Shack (a 2nd source) that I have soldered up on a perf board. I am getting 5.01 VDC unloaded. That is more like I would expect.


Do I have a bad batch of 7805s from the first source?

I see from the datasheet on I'm one the low end of it's normal operating range?
 

SgtWookie

Joined Jul 17, 2007
22,230
Did you see on the datasheet that you have to have a load of at least 5mA up to 1A in order to obtain guaranteed regulation?

Did you see on the datasheet that the acceptable range for output voltage is 4.65v to 5.35v?

Do you know that you can increase the output voltage by adding resistance from the GND terminal to actual ground?

If the current from the GND terminal to GND is nominally 5mA, how much resistance do you think you will need to add between the GND terminal and GND in order to adjust the two low regulators to get a 5v output?
 

Thread Starter

spinnaker

Joined Oct 29, 2009
7,830
Did you see on the datasheet that you have to have a load of at least 5mA up to 1A in order to obtain guaranteed regulation?

Did you see on the datasheet that the acceptable range for output voltage is 4.65v to 5.35v?

Do you know that you can increase the output voltage by adding resistance from the GND terminal to actual ground?

If the current from the GND terminal to GND is nominally 5mA, how much resistance do you think you will need to add between the GND terminal and GND in order to adjust the two low regulators to get a 5v output?

Yeas I know it requires a load but shouldn't the voltage actually be lower with a load? Don't answer that I guess I will need to experiment. Anyway that is where I was going.

As for bumping up the voltage with a resistor. Guess I will need to live with it because my PCB is already designed but would I calculate for the difference I want to increase at 5ma? What can be done if the voltage is too high?
 

SgtWookie

Joined Jul 17, 2007
22,230
Yeas I know it requires a load but shouldn't the voltage actually be lower with a load? Don't answer that I guess I will need to experiment. Anyway that is where I was going.
Don't expect a regulator to output within it's advertised range without it's rated load.

What you are reading is well within published specifications.

As for bumping up the voltage with a resistor. Guess I will need to live with it because my PCB is already designed
You can always cut a trace and solder in components.

but would I calculate for the difference I want to increase at 5ma?
Ask yourself that question, and then go look up Ohm's Law. See what Ohm's Law has to say about current across a resistor. This isn't rocket science; it's basic math.

You tell me what resistances you will need. I already know.
What can be done if the voltage is too high?
If the voltage is too high, you will need a negative voltage reference for the GND pin.

Don't worry too much about it; you're building a general purpose power supply. If you wanted better accuracy, you would have used different parts. It's a learning experience.
 

Thread Starter

spinnaker

Joined Oct 29, 2009
7,830
Ask yourself that question, and then go look up Ohm's Law. See what Ohm's Law has to say about current across a resistor. This isn't rocket science; it's basic math.
OK now that I think about it, ground is at zero volts (obviously) and my output is at 4.9. I'll take a guess and say if I raise the reference on the "ground" pin I also raise the output by the same amount. If it is 4.9V output. I need to raise the output .1 volts. . To do that, R= V/I

.1/.005 = 20 ohms.
 

Mike33

Joined Feb 4, 2005
349
You can also stick 1 or more forward-biased diodes in there, say a 1N914, between the GND pin and actual ground. It will raise your output voltage by .7V for each diode. But you definitely need to have the rated min. load on the output, and the regulator is, as stated, well within its tolerances. The diode trick is more for turning a 7805 into a 7806.4, LOL...
Don't sweat .1V unless this is for a nuclear reactor or something. Good learning, tho!
 

SgtWookie

Joined Jul 17, 2007
22,230
Yes, for the 4.9v regulator, 20 Ohms is the number I was looking to see. 30 Ohms for the regulator that measures 4.85v.

Note that the 5mA current from the ground terminal to GND is nominal. Your mileage may vary.

Try measuring your GND terminal current. Put a load on the output of the regulator, somewhere between 200 and 300 Ohms to ground. Connect a meter set to the 20mA range from the GND terminal to ground. Apply 7v-12v to the input terminal, and see what your actual current reading is.

The 78xx series regulators aren't terribly accurate, but they're good enough for most TTL circuits.

Using a diode to boost the output voltage can be done, but the diode Vf will change (decrease) over temperature; whereas the voltage drop across a resistor will be constant for the same current.
A 1N400x series diode will have a Vf of around 650mV at 5mA current
A 1N4148/1N914 diode will have a Vf of around 700mV at 5mA current.
Schottky diodes like a 1N5817 will have a much lower Vf (around 200mV @ room temp for 5mA)
 

AlexR

Joined Jan 16, 2008
732
One further point to consider, how do you know that its the regulator that is out and not your meter.
A 0.1V error in 5V is a 2% error and 2% accuracy is a about as good as you will get in a reasonably priced meter.
I have 4 or 5 digital multimeters scattered around the workshop and I don't think that any 2 of them read the same.
Digital meters are nice and easy to read but they do give you a false sense of precision that may very well be totally unwarranted.
 
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