# amplify/increase current from 9v battery

I have a working circuit that generates a signal (a range of slow and fast tones) and sends them to a speaker. I tried to switch to a piezo driver and speaker and was successful using the MAX9788 as the driver, however I was unable to make it loud enough.

The MAX9788 requires 2.7V to 5.5V. If I supply it with 5V from a power supply and use a regular speaker then I am able to create the desired volume during which the power supply reads between 250mA to 850mA depending on the speed of the tone, i.e. steady and slow to very fast.

My problem comes from when I supply my circuit with a 9V battery and use a linear DC-DC regulator to send 5V to the MAX9788 (I am using a L7805). I am able to hear the slower tones but at a much lower volume and never the faster ones. I have been able to increase the volume using a voltage follower with a darlington pair, my current during this time has never exceeded 50mA.

I believe my mistake was using a linear DC-DC regulator and my next step is to use a buck convertor to hopefully get more current to the MAX9788.

TLDR - my systems works if I supply the MAX9788 with a power supply but not with a 9V.

My question is will this work? is there a better way to see the current I need on the driver using a 9V battery?

• 9V batteries aren't made to deliver high current.
– JRE
Oct 15, 2019 at 19:20
• Use a wall wart. Oct 15, 2019 at 19:24

If you use a buck switching regulator instead of a linear regulator like the 7805, you can get more current at 5 V than your battery puts in at 9 V.

But you can't get more power out than you put in. So to get 850 mA at 5 V, you will need something like 500-550 mA at 9 V, which is more than it's reasonable to expect from a 9 V battery.

You could either use a big honking lantern battery, or switch to using a mains-powered supply (wall wart). If you need to limit yourself to using a single 9 V battery, you'll have to re-design your circuit to use much lower power. That probably means much quieter output.

In your case I would use a simple two cell 28650 and use a buck converter to keep a steady voltage of 5V. Using those cells is better and can deliver more current to your circuit - way better than a 9V battery.

regards

8 Ohm speakers can be 4 Ohms DC so a big cap to block DC reduces wasted power as they show in MAX9788 spec.

TO generate 5.5V max use only good batteries. e.g. 3x C cells 4.5V or 4 x1.2V = 4.8V NiMH. or DIY discrete design below.

The driver should be a push-pull FET bridge going direct to the battery. A Piezo Speaker or something >32 OHms would work better.

The fresh Battery battery can supply >1A short circuit which at <9V means the source ESR is <9 OHms. So you can achieve max power transfer with an 8 OHm speaker but at 50% loss of energy (MPT Theorem).

A 2Ah LiPo or Li-Ion string of 3 cells can supply 10A with a 10% drop from 11.2V with an ESR of 0.1~0.2 Ohm from the battery pack, resulting in less drop.

A simple CMOS Schmitt Oscillator and Biased FET Half bridge can work.

Dual Nch FET bridge needs a boost Cap-diode for the low side switching and high side Vboost > Vbat.

Dual channel PCh+Nch needs steering bias so that both FETS are off when control voltage is at 50% to prevent shoot thru.(power short) This is also called dead-time FET commutation control.