Measuring maximum current. - Page 1
Forum section: electronics / beginners

author: Kalpo
author_rank: Newbie
timestamp: July 09, 2017, 07:23:59 pm
content: Hello Everyone!I have been googling around for a while and I just can't find a way to do this. I'm really sorry if I suck at searching, if it's the case, please do refer me to a link that might help.Let me explain the situation. I have an old stereo set that has three parts, Radio/Amp, Record player and Tape/CD. The record player is connects to the amp/radio part in two different ways: throw an double RCA and with a priority Sony power connector. Since the record player doesn't work well at all, I decided to take it apart and see if I can harness the power that would normally feed to the record player unit in order to power a small Bluetooth receiver.The voltage seems to be 12V, which can easily be transformed into 5V, but I have no idea how much current it can provide. Could you help me find out how to do this?I have converted the 12V to 5V using a 470 and 330 ohms resistors and connected it to a micro usb cable and to the bluetooth receiver. It doesn't seem to receive enough power, when meassuring the current it seems to use around 50mamps. And when I connect it to a real usb on my laptop it uses 200.Would a dc2dc step down converter allow for more current to be delivered? How can I tell what's the max current the stereo will give me??Thank you very much for your help

author: Zero999
author_rank: Super Contributor
timestamp: July 09, 2017, 09:23:24 pm
content: If you know the output power of the amplifier, then that can help you guess. Unfortunately, manufacturers often lie about how much power the amplifier can output.If it uses an old transformer, then the maximum power output can be estimated. If it has a switched mode power supply, then you can gradually increase the output current by connecting different resistors to the output, until the voltage drops by around 5% of the unloaded value.A potential divider is a poor voltage converter. The output impedance is equal to the two resistor values in parallel, so you were effectively connecting it to 5V, via a 194R resistor.You could use a linear regulator, such as the LM7805 but the extra voltage will be dissipated as heat, which might be acceptable for low power levels. A buck converter can achieve efficiencies of around 90%.

author: rstofer
author_rank: Super Contributor
timestamp: July 09, 2017, 09:54:47 pm
content: Study up a bit on Thevenin's Theorem...Put a load on the output such that the voltage drops from the normal open circuit voltage and calculate the current. Now you have a voltage divider and you can calculate the internal resistance from knowing the open circuit voltage, the load resistor voltage and the current through the loop.The current I is simply VLoad/ RLoad- the output voltage and the load resistanceNow that you have the current, you can calculate the internal resistance asVOpenCircuit- VLoad/ I - the internal voltage drop (across the internal resistance) divided by the current.As you reduce the resistor (subject to not burning out the supply), you will get better approximations. Study up a bit of the Maximum Power Transfer Theorem which will tell you that when the voltage drops in half under load, the internal and external resistance are the same. You might not want to load the power supply this much. You certainly can't do it if the source is a voltage regulator of almost any kind. They will sense the overload and roll back the output voltage (or equivalent).Your resistor voltage divider will never work for a number of reasons including a variation in the load causing a variation in voltage. I would look at a 7805 with a heatsink. Assuming the internal supply can deliver 200 mA at 7V or more.

author: bson
author_rank: Unknown
timestamp: July 09, 2017, 10:20:41 pm
content: A divider only gives a specific voltage when open circuit. Under load it will always droop. You can view it as a series current limiter and a parallel minimum load - anything you add increases the load and the divider droops. It's impossible to tell what the maximum output current of a socket is without knowing what's inside. Thevenin equivalent resistance isn't a useful analytical tool for active regulators and transformers. The absolutely easiest is to wire up a trivial regulator with a 7805 and a couple of capacitors and simply seeing if it works - I'm pretty sure if it ran the motor and vintage electronics (which weren't power efficient) of a record player it'll have no difficulty with a highly integrated BT transceiver.

author: Kalpo
author_rank: Newbie
timestamp: July 10, 2017, 02:40:16 am
content: Thank you very much!After some testing it seems the receiver gets enough power from the voltage divider. I will buy myself some 7805 though, for what you are saying it seems to be a better option. I guess I could try the buck converter as well.@rstofer, Thank you for that detailed explanation. I am afraid I am rather new to all this and it's very hard for me to understand the content of what you explained. I will try to do some research and see if I manged to understand it. I really appreciate your input.