Dual polarity power supply circuit question

[rolerbe]

I'm having trouble with the power supply portion of an opamp circuit I'm working on. The supply is +/- 5 Volt using 7805/7905's. Half wave rectifiers provide the unregulated input with 1000uF input caps and 100uF output caps. AC is supplied by a 13VAC 'wall wart' with no center ground tap.

Any load I draw from the +5 supply (even as little as 50mAmp) drops the +5V side down to 3.9V or so.

The issue seems to be that the 7805 is being starved for input voltage. When I compare input voltage after the half waves relative to my DC ground, I get just barely 5V on the plus side (no load) and just over -11V on the negative side.

Do I need a voltage divider or something (say two 1MOhm resistors) between the Vreg inputs, connected to the output side ground to center the input? Or some other solution?

Thanks.

 

[Redbelly98]

Yes, a voltage divider will probably fix this. What values to use will require experimentation. It may be that unequal resistors are what you need, given the tendency for ground to be closer to the positive input voltage.

My best guess is that, when your regulators are behaving properly, they have a different current at the ground pin (Iq in datasheets). But connecting the grounds together, with no other current path at that connection, forces the two currents to be equal. One regulator must reduce it's Iq somehow in order to match the currents, so one of them (in your case the 7805) reduces the voltage between Input and Ground in order to reduce it's Iq.

Iq can be 5-8 mA, according to
http://www.fairchildsemi.com/ds/LM/LM7805.pdf

If a few mA mismatch were sent through a 1M resistor, that would be a kV, which won't happen. So 1M is too big. You want

R = (say 3 V) / (3 mA) = 1k.

A 2 x 1k bridge will draw, of course, 16V/2k = 8 mA additional from your input supply. Experiment to find the largest resistors you can get away with (to minimize the current drawn by the bridge).

Hope this helps, good luck!

Mark

 

[rolerbe]

Thanks for the quick feedback. It's just breaboard now, so I'll go give it a try.

 

[Redbelly98]

Cool. I'm curious to find out what you end up with to get it working.

 

[rolerbe]

Thanks for all the input. Problem solved. Simple dumb @#$ wiring error. :(

7905 pin out is not the same as 7805. Surprised more stuff didn't smoke.

 

[Redbelly98]

7905 pin out is not the same as 7805.

Wow. Go figure.

 

[smuscat]

I would like to know how the resistors are connected . i didn't understand clearly because i have the same problem as this guy.Can you make the calculation more clearly for me?please.

thank you in advanced.

Yes, a voltage divider will probably fix this. What values to use will require experimentation. It may be that unequal resistors are what you need, given the tendency for ground to be closer to the positive input voltage.

My best guess is that, when your regulators are behaving properly, they have a different current at the ground pin (Iq in datasheets). But connecting the grounds together, with no other current path at that connection, forces the two currents to be equal. One regulator must reduce it's Iq somehow in order to match the currents, so one of them (in your case the 7805) reduces the voltage between Input and Ground in order to reduce it's Iq.

Iq can be 5-8 mA, according to
http://www.fairchildsemi.com/ds/LM/LM7805.pdf

If a few mA mismatch were sent through a 1M resistor, that would be a kV, which won't happen. So 1M is too big. You want

R = (say 3 V) / (3 mA) = 1k.

A 2 x 1k bridge will draw, of course, 16V/2k = 8 mA additional from your input supply. Experiment to find the largest resistors you can get away with (to minimize the current drawn by the bridge).

Hope this helps, good luck!

Mark

 

[Redbelly98]

Welcome to PF.

First, double-check the pin configurations for the 7805 and 7905. The configurations are different for the two, which ultimately was rolerbe's problem.

If you go the resistor route, each resistor goes from the Vout pin to the common pin. I had suggested starting with a 1k resistor on each regulator. If that balances the outputs, try something larger and see if they are still balanced. If you can balance them with 10k's, then that's only 0.5 mA load on a 5V regulator, and you probably don't need to go any higher than that.

 

[Redbelly98]

Check that, the resistors would go between regulator Vin and common ... hey, it's been 6 months since I thought about this, and we didn't make any schematics since the OP found it was a simple wiring error.

But a better solution would be either to use separate + and - supplies to the regulators, or a transformer with a center tap that can connect to the common pins of the regulators.

 

[rolerbe]

A center tap transformer is by far the simplest and best solution. But, you don't really need the CT as the transformer gives you isolation from the AC source, so the fact that the DC side ground is 'floating' relative to AC ground is OK. You can safely tie the DC ground to the power GROUND (green) terminal to connect to 'local' Earth ground.

In practice, the two regulators should match quite well with very little bias (offset) on your floating DC ground, without balance resistors.

If this is not true in your as-built circuit, I've learned (the hard way) that A) either you have a wiring error (no matter how many times you think you've looked at it), or B) one of the regulators is already bad.

NOTE: not only are the pin-outs different for the 78XX and 79XX, in the TO-220 case, the case heat sink is NOT ground for the 79XX. Short the case heat sink to ground, however momentarily, and the IC is DEAD.

 

[smuscat]

Many thanks friends, the circuit is working perfectly now.The problem was about where the resistors are connected ,that was from Vreg to ground.Thankyou for your cooperation.

Gooday