How to maintain 12volts even the source drops to 9volts

[gokulak]

Dear All,
Need your expertise advice.
Problem:
I have fixed a new music player in car which needs 12volts but the existing or old music player runs in 9volts. When the car is running or ideal the voltage is comings as 12volts
But
Whenever the car engine is restarted, the voltage drops to 9volts and comes back to 12volts.
The new music system power off because the voltage drops to 9volts and it need 12volts.

 

[Strima]

How old is your battery?

Are all connections from the battery to the radio in good condition and secure? What's the rest of the vehicle wiring system like?

A car battery should be at around 12.5 volts, with the engine running should be around 14.5 volts. Batteries tend to have a life expectancy of around 5 years depending on usage and driving conditions/style.

 

[gokulak]

Thank you so much for your response.

The car is 2 years old and so the battery life could be around 2 years.
Yes the connections are in good condition. The inbuilt music player runs in 9 volts so even when the voltage drops when the engine starts, the inbuilt music player works fine at 9 volts.

The technical person who fitted the new music player found that the voltage drops to 9 volt at the source itself, means the battery output itself drops to 9 volts.
But the new music player needs 12 volts so it switches off when the engine is restarted.

please suggest any solution.

Thank you in advance.

 

[marconi]

Try connecting the + and - of the music player directly across the battery via a fuse. The fuse must be in the lead to the battery terminal which is not connected to the car body - see my diagram.

If the + was connected to the car body put the fuse in the - lead.

The problem with this way of connecting the music player to the battery is that if you forget to turn off the music player you risk discharging the battery.

You could do this way of wiring to the battery to check what Strima's #2 said that the problem could actually be a poor connection of the battery to the car's electrical system.

 

[gokulak]

Thank you so much for your message.
Please note that the power from battery output is dropping from 12.5volts to 9volts.
The problem is that the battery output is dropping to 9volts when the engine starts.

Please suggest, how to handle this drop in voltage at the source itself.
Will a capacitor help ?

Thank you in advance.

 

[static zap]

If it gets to 9V with a warm engine.
A ) Is it the correct battery ..especially if its a diesel
B) Is battery old.
C) If slow cranking (and A+B-check-out ) Poor connections -should get Hot.
D) Same conditions as C .... -starter motor may be on its way out.
E) Plenty of vehicles -Disable AUX supply while cranking ..(Fused-Direct power-path-Needed)

( I know my battery is on the way our -as it dislikes this cold weather. )
( My Battery may be failing it self in 3 years ..Modern tech .... has de-skilled )

 

[marconi]

Let us assume it takes 5 seconds to start the engine. During these 5 seconds the battery voltage drops to 9V. Let us assume the minimum voltage to operate the music player is 10V and at this voltage it draws 2A; thus its effective resistance R is 10/2 = 5 Ohms.

We can now imagine a capacitor charged to 12V before the engine starts and then being discharged over a 5 second period supplying the current for the music player during the cranking of the engine. During this period the voltage of the capacitor drops by 10/12 = 0.83 (I have used 10V to have a 1V margin over 9V).

In the attachment you will see the capacitor connected across a resistance R (the music player).

In such a simple CR series circuit the voltage across the capacitor Vc at time t is:

Vc = 12 x e (exp -t/CR) .... minus because the voltage is decaying.

Vc min/12 = 0.83.

Thus 0.83 = e (exp -t/CR) = e (exp -5/CR)

Therefore, ln(0.83) = -5/CR = -0.19

But R is 5Ohms.

5/5C = 0.19 (call it 0.2 for now).

1/C = 0.2 so C = 5F

Such a capacitor is a supercapacitor.

The affordable supercapacitors I found from RS only had a rated operating voltage of 3V dc.

The car's electrical system is 12V so I doubled this to suggest the required operating voltage of the supercapacitor ie:24V

When two identical capacitors are wired in series the effective end to end capacitance is halved.

Since the maximum voltage capacitor is 3V and 24/3 = 8 = 2 exp3, one would need to wire 8 supercapacitors in series. This means each SC would have to be 40F since 40 halved, halved again and then halved again (ie 1/2 exp3) is 5F.

MAL223091001E3 | Vishay 40F Supercapacitor EDLC -20 %, +50 % Tolerance, 230 EDLC-HV 3V dc, Through Hole | RS Components - https://uk.rs-online.com/web/p/electric-double-layer-capacitors/1802317/

The circuit then is as attached. The power diode is to isolate the circuit from the battery during the cranking of the engine. I have not yet done the sums on the required peak current handling of this diode and don't have time right now. Perhaps someone else could have a go?

I have not checked my sums but think I am right.

https://www.electronics-tutorials.ws/rc/rc_1.html

 

[marconi]

Assume the positive lead to music centre is 1.5mm2 with a resistance of 0.012mOhms/m and 2m long..

Negative return is through chassis and of lower resistance than positive - assume half value ie: 1m long length of 1.5mm2.

Thus resistance of path to bank of supercapacitors is (2 + 1 ) x 0.012 = 0.036mOhms - call it 0.04

Peak current through diode is when supercapacitor is completely discharged and then connected to 12 V battery. Ignore internal resistance of battery for now. I peak is 12/0.04 = 300A

This 20A diode would suit because it has working reverse voltage of 100V and short term surge current rating of 350A:

Diotec FT2000AB Superfast Rectifier Diode Two Polarities 100V 20A TO-220AC | Rapid Online - https://www.rapidonline.com/diotec-ft2000ab-superfast-rectifier-diode-two-polarities-100v-20a-to-220ac-64-9095

 

[gokulak]

If it gets to 9V with a warm engine.
A ) Is it the correct battery ..especially if its a diesel
B) Is battery old.
C) If slow cranking (and A+B-check-out ) Poor connections -should get Hot.
D) Same conditions as C .... -starter motor may be on its way out.
E) Plenty of vehicles -Disable AUX supply while cranking ..(Fused-Direct power-path-Needed)

( I know my battery is on the way our -as it dislikes this cold weather. )
( My Battery may be failing it self in 3 years ..Modern tech .... has de-skilled )

Thank you so much for your message.
It is 2 year old car. The battery is also 2 years old.
As per you message I will check in the service center.
Thank you again !
Thank you.

 

[gokulak]

Let us assume it takes 5 seconds to start the engine. During these 5 seconds the battery voltage drops to 9V. Let us assume the minimum voltage to operate the music player is 10V and at this voltage it draws 2A; thus its effective resistance R is 10/2 = 5 Ohms.

We can now imagine a capacitor charged to 12V before the engine starts and then being discharged over a 5 second period supplying the current for the music player during the cranking of the engine. During this period the voltage of the capacitor drops by 10/12 = 0.83 (I have used 10V to have a 1V margin over 9V).

In the attachment you will see the capacitor connected across a resistance R (the music player).

In such a simple CR series circuit the voltage across the capacitor Vc at time t is:

Vc = 12 x e (exp -t/CR) .... minus because the voltage is decaying.

Vc min/12 = 0.83.

Thus 0.83 = e (exp -t/CR) = e (exp -5/CR)

Therefore, ln(0.83) = -5/CR = -0.19

But R is 5Ohms.

5/5C = 0.19 (call it 0.2 for now).

1/C = 0.2 so C = 5F

Such a capacitor is a supercapacitor.

The affordable supercapacitors I found from RS only had a rated operating voltage of 3V dc.

The car's electrical system is 12V so I doubled this to suggest the required operating voltage of the supercapacitor ie:24V

When two identical capacitors are wired in series the effective end to end capacitance is halved.

Since the maximum voltage capacitor is 3V and 24/3 = 8 = 2 exp3, one would need to wire 8 supercapacitors in series. This means each SC would have to be 40F since 40 halved, halved again and then halved again (ie 1/2 exp3) is 5F.

MAL223091001E3 | Vishay 40F Supercapacitor EDLC -20 %, +50 % Tolerance, 230 EDLC-HV 3V dc, Through Hole | RS Components - https://uk.rs-online.com/web/p/electric-double-layer-capacitors/1802317/

The circuit then is as attached. The power diode is to isolate the circuit from the battery during the cranking of the engine. I have not yet done the sums on the required peak current handling of this diode and don't have time right now. Perhaps someone else could have a go?

I have not checked my sums but think I am right.

https://www.electronics-tutorials.ws/rc/rc_1.html

Thank you so much for your solution.

I shall try this solution and see if it works.
I will sure update on the solution.
Thank you again !
Thank you.

 

[gokulak]

Assume the positive lead to music centre is 1.5mm2 with a resistance of 0.012mOhms/m and 2m long..

Negative return is through chassis and of lower resistance than positive - assume half value ie: 1m long length of 1.5mm2.

Thus resistance of path to bank of supercapacitors is (2 + 1 ) x 0.012 = 0.036mOhms - call it 0.04

Peak current through diode is when supercapacitor is completely discharged and then connected to 12 V battery. Ignore internal resistance of battery for now. I peak is 12/0.04 = 300A

This 20A diode would suit because it has working reverse voltage of 100V and short term surge current rating of 350A:

Diotec FT2000AB Superfast Rectifier Diode Two Polarities 100V 20A TO-220AC | Rapid Online - https://www.rapidonline.com/diotec-ft2000ab-superfast-rectifier-diode-two-polarities-100v-20a-to-220ac-64-9095

Thank you so much.

I will share this details also to the technician and try it out.
Thank you again !
Thank you.