# Anti-Crank Power Supply for Car

I have a circuit that draws a maximum on 1 amp @ 11v. Its an ATmega328P running PwM through some Mosefts to run some LED DRLs (Daytime Running Lights).

I want to prevent the situation where, during the engine starting, the LED rings flicker as the voltage drops below 11v on starter motor crank. I was thinking that a large capacitor on the input of the PSU would be sufficient but no idea how big! I estimate that the circuit would only need a 4 second buffer tops.

Any thoughts?

PS, I don't want to redesign the LED driver side of the circuit to make use of dedicated DRL controller chips as it massively over-complicates a simple circuit! :)

• Are you able to verify the minimum voltage down to which the LED power supply can continue to power the LEDs? If the LEDs are singly connected parallel strings, often they would not show appreciable flicker down to 6-7 Volts. If the LEDs are a series string on a constant current supply, some constant current LED driver ICs incorporate a voltage boost capability internally, to avoid drop-out during voltage dips. Jun 19 '13 at 8:43
• The LEDs will dim with a lower voltage but I want to stabilise the input voltage and iron out the cranks. As the MCU is also on the same power circuit I do not want to risk it corrupting the running routine when the car starts. Jun 19 '13 at 8:57
• Just so I understand: You want to avoid the complication of circuits designed to support the specific situation, therefore implying that you believe that those massively over-complicated devices don't really have any reason for existence? Hmm, ok. Jun 19 '13 at 9:01
• You misunderstand the statement, and have missed the original question. I need to protect both the LED's from flickering and an under voltage situation on the IC that may result in its program failing. The rings are driven at 11V from a DC DC Buck with a mosfet on each bank handling PwM switching from the IC. So I am as near as needed to a dedicated DRL driver such as ZXLD1371. What I need is to buffer the input to the DC DC buck circuit for 4 seconds to solve the problem. Jun 19 '13 at 9:05
• @JabawokJayUK Do you have an option to turn off the LEDs in the crank situation? That could be acceptable, since all the other car electronics stop at this time, too. Jun 19 '13 at 11:17

Device load is 1 Ampere at 11 Volts. Capacitor bank should sustain this power for 4 seconds, before dropping voltage below acceptable (unspecified) threshold voltage. Let's say a drop down to 10 Volts is acceptable.

• 1 Farad = 1 Ampere-Second / Volt

For 1 Volt of drop, over 4 seconds, at 1 Ampere,

• F = 1 x 4 / 1 = 4 Farads

Thus, required capacitors must total 4 Farads, with a recommended voltage rating of ~ 24 Volts or higher: Car electrical systems can see voltage spikes of over twice the nominal voltage in certain situations.

For charging the capacitor at start-up, the battery must support the inrush current the capacitor bank will draw. As car wiring and batteries are typically very low resistance, given that 100 Amperes or more may be drawn for starting the car. A discharged capacitor behaves pretty much like a short circuit initially when connected to a charging source, so 100 or even up to 1000 Amperes could well be drawn by it.

Solution:

Use a bank of car audio capacitors that incorporate soft-charge circuitry into the capacitor units. These are in the $100 to$200 range. Keep in mind, though, that car audio capacitors are typically rated far higher than their actual capacitance:

For instance, this 10 Farad rated capacitor was found to actually be a mere 0.227 Farads.

The typical naïve car mod customer never figures it out, so the manufacturers continue to sell them.

Like I have implied in comments, this capacitor bank approach is non-ideal, could be dangerous, and is actually strongly not recommended. I've put down the numbers simply because that's what the question asks for.

Both the answers posted at this time other than mine, actually point out the more practical approaches to the problem:

• Either turn off the lights at crank time, or
• use a different power supply, i.e. a SEPIC or buck-boost.

Both those much more practical solutions have been rejected by the OP.

• Thank you for this answer. It actually answered the question asked with sufficient information for me to learn and understand the approach flaws. The reason the prior answers were rejected is they 1) didn't answer the question asked, and 2) suggested methods (one of which was not suitable, i.e switch off), and one that was focussed on driving the LEDs only not the entire circuit. A buck Boost that could drive the whole circuit at 11v 1 amp would solve the problem. Jun 19 '13 at 11:47
• @JabawokJayUK Excellent. Glad it worked for you. Jun 19 '13 at 11:59

Do you have any analogue inputs on the MCU? If you do then use one to monitor the high side of the power supply. You can use a resistor potential divider to ratio down the voltage to about one-fifth so a 3V3 referenced ADC can adequately read 11V as 2.2V. Don't start the LED circuits until the voltage has risen back above 12V.

• Its an interesting idea, but it would not account for you turning on the ignition, and then waiting a few seconds before starting the car as the signal would see 12v HIGH and allow the DRL's to start, then when you start the car and the cranks occur, it would drop the voltage and flicker the rings. It may also effect the running program on the ATmega. Jun 19 '13 at 8:34
• "Don't start the LED circuits until the voltage has risen back above 12V". OK you probably missed what I meant but I see the code doing exactly that - power on and 12V is good so don't start the LEDs. Volts drop down to 11V (or maybe a bit below that) then wait till 12V returns to power the LEDs. Alternatively, if at any time the voltage is not 11.8V or above, disable the LEDs. Are you sure that during starting the engine the ATmega will be affected, because if you are then a different method should be adopted or use a diode in series with its supply and a power hold-up capacitor as well. Jun 19 '13 at 9:07