7805 -- 5 volt output at 4.8vdc?

[tom86951]

I am using a 7805 5 volt regulator in a very simply ciruit that draws very little current. My problem is that the output of the regulator is not staying at 5 volts when the circuit is active. (Tests at a perfect 5vdc on the bench.) Instead, it drops anywhere from 4.9 to 4.75 depending on the load. It is not a transitory dip, it will stay at that voltage as long as the circuit is active. The supply voltage is a 12+ volt car battery. The circuit is a voltage divider -- essentially two 1K resistors in series from the output of the 7805 going to ground (with the voltage in between the resistors tapped for a low impedance data logger). One of the resistors is actually a thermistor, but it stays around 1000 ohms most of the time.

So, how do I keep the voltage at 5 without dropping like that?

 

[MrDEB]

post a schematic

they will help in solving problems

 

[blueroomelectronics]

What are you powering the 7085 with?

 

[MikeMl]

I am using a 7805 5 volt regulator in a very simply ciruit that draws very little current. ...

Are you aware that there is minimum load current spec on a 7805 of 5mA?
(The output voltage is not speced at all for loads less than 5mA)
Read the **broken link removed**

Also, are you feeding the regulator with 12V?

Do you have the recommended bypass caps on it?

 

[Hero999]

There's no minimum load requirement for the LM7805, it has a nominal quiescent current of 5mA.

How much current does the circuit use?

What's supplying the LM7805?

Does it have a heatsink?

 

[MikeMl]

There's no minimum load requirement for the LM7805, it has a nominal quiescent current of 5mA.

Hero

go and read the data sheet! All specs related to Vo and Δ Vo are valid only when Io is between 5mA ≤ Io ≤ 1A (or 1.5A)! This has nothing to do with the quiesent current, which 6mA. Heat is not the OP's problem. His problem could be that his load is too small.

 

[Hero999]

Hero

go and read the data sheet! All specs related to Vo and Δ Vo are valid only when Io is between 5mA ≤ Io ≤ 1A (or 1.5A)! This has nothing to do with the quiesent current, which 6mA.

You're correct about what it says on the datasheet.

Although all it means is that the load regulation isn't guaranteed if the current is below 5mA. It doesn't specify any minimum load current.

I've just tested a couple of LM7805 with no load and they both worked perfectly, one was a genuine branded part than the other was a no name part. One regulator gave 5V bang on and the other gave 5.09V which is well within tolerance. I've used the LM7805 in many applications with a minimum load below 5mA and have had no problems. I have seen many circuits for benchtop 5V PSUs based on the LM7805 and none have minimum load resistors.

I can understand why you thought what you thought but datasheets aren't always to be trusted.

Heat is not the OP's problem. His problem could be that his load is too small.

Even if the LM7805 did have a minimum load requirement the output voltage would be too high, not too low, which is what the Op is experiencing.

 

[MikeMl]

You're correct about what it says on the datasheet.

Although all it means is that the load regulation isn't guaranteed if the current is below 5mA. It doesn't specify any minimum load current.

I've just tested a couple of LM7805 with no load and they both worked perfectly, one was a genuine branded part than the other was a no name part. One regulator gave 5V bang on and the other gave 5.09V which is well within tolerance. ...

My former employer has used tens of thousands of LM340/78XXs, mostly made by National. Most do output the same voltage unloaded as loaded. However, the ones with date codes prior to about 2001 did not, so it depends on the date code when they were manufactured.

The OP is only drawing either 0 or 2.5mA, and is noticing that the voltage drops from 5.0V to 4.75V. That is toward the most extreme change I ever saw when comparing unload Vo to loaded Vo, but I routinely saw > 100mV change.

A much better chip to use for this application is a REF02, or just add a 1K parallel min resistor to the LM7805.

 

[Hero999]

Does the brand make any differance?

Looking at the Fairchild datasheet. . .
https://www.electro-tech-online.com/custompdfs/2009/07/LM7805.pdf

It specifies 4.8V to 5.2V at a junction temperature of 25°C, with presumably a 500mA load.

With a load current between 5mA to 1A it specifies 4.75V to 5.25V.

If the regulator is outputting 4.8V@2.5mA and 5V at no load, then I would be suspicious: I think it's probably a cheap no brand Chinese copy.

 

[crutschow]

If the 7805 is not properly decoupled with input and output bypass capacitors, it's possible it could be oscillating. That can result in odd shifts in the DC output voltage.

 

[MrDEB]

Where is a schematic of your circuit??

Several items can be the problem.
My first thought was capacitors but without a schematic ??
a picture is worth a thousand words

 

[kchriste]

A 78L05, or other low power regulator, would have been the better choice here. I agree with the comments above that the problem is most likely the lack of bypass caps due to the fact that the voltage is too low.

 

[Hero999]

I'm not saying it isn't possible but I've not seen a 7805 oscillate because the output capacitor is missing. I tested the regulators without a bypass capacitor but I didn't 'scope them so I can't be sure.

 

[MikeMl]

I'm not saying it isn't possible but I've not seen a 7805 oscillate because the output capacitor is missing. I tested the regulators without a bypass capacitor but I didn't 'scope them so I can't be sure.

The data sheet application section says that the output capacitor is optional. It improves transient response by supplying a sudden load current spike until the regulator can ramp-up it's output. The bandwidth of the control loop inside the regulator is in the tens of kHz range, so if a step load is applied, it takes 10s to 100s of usec for the regulator to catch up.

A low ESR capacitor on the input is required to prevent oscillation if there is more than a few inches of wire between the regulator and the power source. It is especially needed if the power source has some internal series resistance, like a battery...