# Using a DC battery to power 20 V AC, 1 A circuit

and thanks in advance for not blasting me for this question.

I am trying to find a way to power a 20V 1A Ac circuit. using 12vdc Li-on or LI-po cells. The device has a 120vac to 20v 1 amp transformer, So I was thinking I could use a dc-ac converter and just tie into the primary circuit of the transformer, but I am assuming that using a dc-ac inverter will have a lot of wasted energy in the conversion, so I think that the dc-low voltage ac would be an easier and longer running circuit.

Any thoughts? any Ideas?

thanks all

So i ordered a replacement 20v AC 1 A transformer, didnt realize but it has 2 primary leads and 3 secondary, so i powerd the primarys and check the leads to see which two had 20V AC. Soldered them into the board and fliped the switch and pop went the fuse. Tried the 11V circuit of the other leads and no fuse pop, but the amp does not produce any sound and it appears one of the transistors is getting very very hot. Any ideas?

Additionally i took the old transformer apart and found the parimary lead wires to the coil burnt but rhe secondary seems intack can i check the se ondary windjngs in anyway to make sure we specked the transformer properly?

Thanks for all the help everyone!

• If the transformer feeds a rectifier there is a good chance that a DC/DC converter will do the job. Please clarify and if you could read the DC voltage inside the device it would be a big help. – Transistor Jul 31 '16 at 21:50
• So basically your device is powered by 120 VAC, not 20 VAC, and you happened to see a transformer? – pipe Jul 31 '16 at 21:51
• I have a schematic, the device is 120v input to the primary side of the transformer, and 20v 1A at the secondary side of the transformer, It is actually 25.5 V AC. The 20V an is run through a series of 4 diodes. Thanks for the questions. – Xander K Jul 31 '16 at 22:43
• Does the 25.5 Vac output of the xformer connect to anything in this equipment besides the "series of 4 diodes"? Is this 60 or 50 Hz AC? Does the 120 VAC input connect to anything other than the 25.5 V transformer? Are there any other secondary windings on this xformer? Does the secondary winding have 2 wires or 3? – FiddyOhm Jul 31 '16 at 22:58
• @XanderK If you have a schematic then why not post the schematic? – user253751 Jul 31 '16 at 23:03

Figure 1. AC input circuit showing DC connection points.

simulate this circuit – Schematic created using CircuitLab

Figure 2. Battery power circuit.

How it works:

• Two series 12 V batteries will give you 24 V DC.
• The fuse will protect the batteries and the amplifier should any faults occur.
• D1 will prevent backfeed from the AC supply if the mains is connected.
• IOD-IX4 (the bridge rectifier) will prevent the battery backfeeding into the transformer.

Run time

How long do you suppose 1 Ah would run this circuit?

Let's look for some clues on the mains side.

• In Figure 1 we can see a 0.5 A fuse on a 120 V supply. From this we can calculate that the maximum power from mains would be $P = VI = 120 \cdot 0.5 = 60~W$.
• Redoing the calculation for 24 V and rearranging to solve for the current we get $I = \frac {P}{V} = \frac {60}{24} = 2.5~A$.
• Run time is $\frac {Ah}{A} = \frac {1}{2.5} = 0.4~h$ at full power.

And let's look for clues on the audio side.

• The amplifier needs to output an AC signal (positive and negative) to the 4 Ω speaker while running from a single-rail power supply. To do this we would expect the TR8 and TR9 junction to sit at mid-supply voltage while the amplifier is quiet. This is confirmed by the 12.3 V reference reading on the schematic. The 1000 µF, 50 V capacitor on the output blocks the DC reaching the speaker.
• When the signal goes positive TR8 will conduct and pull the output high. When the signal goes negative TR9 will pull output low. Allowing for some losses in the transistors we might see about 20 V peak to peak or 10 V peak.
• From Ohm's law we can calculate the current into the speaker as $I = \frac {V}{R} = \frac {10}{4} = 2.5~A ~peak$.

The two calculations are crude but agree. 0.4 h = 24 minutes. If you're playing heavy metal with the volume at '11' then that's about it. If you're playing occasional jazz fills at civilised volume levels your power demand will be much reduced.

Output power

You may be interested in the output power of the amplifier. We saw that the peak voltage was 10 V. The RMS (root-mean-square) voltage is given by

$$V_{RMS} = \frac {V_{PEAK}}{\sqrt {2}} = \frac {10}{\sqrt {2}} = 7~V_{RMS}$$

The maximum speaker power for an undistorted sinewave is given by

$$P = \frac {V^2}{R} = \frac {7^2}{4} = 12~W$$

• how long do you suppose 1 Ah would run this circuit? thanks for the circuit. – Xander K Aug 1 '16 at 23:58
• See the update. – Transistor Aug 2 '16 at 8:17

According to the schematic diagram, the internal power bus is nominally 25.5VDC.

The easiest way to power this from 12V batteries (of whatever chemistry) would likely to be using TWO of the "12V" batteries in series to directly produce a DC power bus of 24V directly without loss from any conversion process. Of course, there are dozens of details missing from the question which would help us know whether this is a practical suggestion or not.

• Richard, thanks for the reply, What exactly would you need to know in order to make an informed decision? Runtime, weight, packaging? I will get you any info you need, or if you can point me to the equations I will do all the leg work myself. I am not trying to be lazy here. additionally am I to assume the power supply before the diodes is AC and after is DC? I bought a 120 V AC power transformer to 20V 1 A AC is this the correct transformer? I am assuming that after the diodes it would be converted to DC. Again thanks – Xander K Aug 1 '16 at 21:10
• Is this permanent or temporary? Can you modify the amp, or must you keep it "stock"? Do you need "hot-standby" where the battery takes over from mains failure? Or can the user select either-or? Is using TWO 12V batteries an option in your scenario? How do you plan to re-charge them? Yes, the diodes turn the AC into DC. Not clear why you bought another transformer when there is already one in the amp? Using any kind of DC-AC or DC-DC inverter/converter is a major inefficiency and waste of power. It is so much easier to apply your 24V DC directly to the power rail of the amp (internally). – Richard Crowley Aug 1 '16 at 22:10
• Richard, I would like to keep this as stock as possible, I could always make a pack to use the amp. basically id like it to be mobile, portable. the power transformer I ordered was to replace the stock unit that was faulty – Xander K Aug 1 '16 at 22:20
• It would be easy enough to make an external battery pack that could plug into the amp (which automatically disconnects the internal AC supply transformer/diodes). But how are you going to recharge two 12V batteries connected in series? You may need another "charging plug" which disconnects them from series and puts one 12V battery on each charger, etc. – Richard Crowley Aug 1 '16 at 22:25
• yes. building a pack for 24v should be easy enough, and they make 24v chargers, my problem would be, with this circuit how long would 1AH produce as far as runtime? – Xander K Aug 1 '16 at 22:42