another regulator question

[daz]

Sorry about starting a new thread, but because i wrote the last post in the other one where i suggested i was done i figured no one would know i posted a new question. So to those like mick and enzo, G1, Juan etc who were helping me, i have a question when implementing enzo's suggestion to put a diode in series with the regulator's ground leg to change the 8v out to 8.5. Theres a 22uf cap that goes from the regulator's ground leg to the + out and what i am wondering is should the cap go to the regulator's negative leg or to the cathode of the diode same as the battery negative ?

[The Dude]

The cap should go to ground or battery negative.

[daz]

Just to be clear, is this correct?

[daz]

oops

[The Dude]

No. The cap's negative lead should go to ground.

[daz]

Yes, i realized that and was about to repost a corrected one only to find you beat me to it. This one

[daz]

In any case, thank you.

[daz]

Is there any way this could cause the battery to drain faster? I ask because today i measure the battery at 9.3v and within about 30 minutes of it being on it died. Measured it at 4v at that point. |They never did that w/o the regulator.

[Greg Robinson]

Quiescent current of the LM2940T you're using is up to 60mA max, twice the 30mA draw of your pedals. Even the typical quiescent current (estimated from the graphs in figure 4 at an output current of 30mA) is going to be in the range of 15mA, a full 3rd of your current budget just to power the regulator. This was not a very good choice for battery operation.
Here's the datasheet

[daz]

Quiescent current of the LM2940T your using is up to 60mA max, twice the 30mA draw of your pedals. Even the typical quiescent current (estimated from the graphs in figure 4 at an output current of 30mA) is going to be in the range of 15mA, a full 3rd of your current budget just to power the regulator. This was not a very good choice for battery operation.
Here's the datasheet

great, NOW i hear this ! I still don't see how it drained from 9.3v to nothing so fast but i guess i'll pull the thing. Oh well. Thanks for the heads up.

[Helmholtz]

Measure input and output current of the regulator with full battery and load connected. The difference is the current through the diode.
Or directly measure the diode current by inserting your Ammeter between the middle pin and the diode.

What regulator are you using?

[daz]

Measure input and output current of the regulator with full battery and load connected. The difference is the current through the diode.
Or directly measure the diode current by inserting your Ammeter between the middle pin and the diode.

What regulator are you using?

LM2940 8v. Measured 11.6 ma with the 2 pedals i am powering on. one is a 4ma pedal, the other is a tube screamer that i think is around 7 maybe. Should i have measured it with pedals disconnected? The battery i just put in is at 10.4v and the regulator output is 8.6v with the diode

[Helmholtz]

LM2940 8v. Measured 11.6 ma with the 2 pedals i am powering on. one is a 4ma pedal, the other is a tube screamer that i think is around 7 maybe. Should i have measured it with pedals disconnected? The battery i just put in is at 10.4v and the regulator output is 8.6v with the diode

Is this regulator input or output current? We need both.

(As always, unloaded battery voltage doesn't tell much about charge condition.)

[daz]

Is this regulator input or output current? We need both.

I measured current by putting the meter set to mA in series with the diode and the rest was battery and output voltage. Battery measured by itself is 10 v (was wrong about 10.4) and at the power jacks of the pedals 8.6v.

edit: between in and out reg legs i get 8 mA with pedals connected and 1.3mA without them connected. Don't know if thats how u wanted me to measure but there ya go whatever that matters.

[Helmholtz]

I measured current by putting the meter set to mA in series with the diode and the rest was battery and output voltage.

I see, so that would be the additional quiescent current. That's considerable but hardly explains the fast drain with a fully charged battery.
Please also measure the battery current (= regulator input current) with load.

Again, always measure loaded battery voltage. Unloaded voltage may fool you.