Just baffled.  - Page 1
Forum section: electronics / beginners

author: sureshot
author_rank: Regular Contributor
timestamp: August 10, 2018, 11:38:08 pm
content: With electronics I've mostly put circuits together straight on copper strip board. Really I should have used bread boards more often. It's only due to the power supply circuits drawing a fare bit of current, I thought bread boarding would just melt jumper wires.So this evening I tried something I thought would be straight forward, but it was far from that.It was just a simple L7812 with a pnp pass transistor, I was trying to limit the current through the L7812. But no mater what value resistor from 10 ohms to 150 ohms the measured current done by the regulator was still 720mA there where probably losses as my load was a 20 watt halogen lamp. I stripped it back multiple times, checked and double checked all the connections. I tried different power rated resistors from 0.5 watt to 10 watts, but nothing limited the current between the base emitter chain. I was starting to think the breadboard was dodgy. Another strange thing, I opened the circuit to see the current the regulator was passing which showed the 720mA current it was passing. But when I removed the series meter connection the lamp was still alight but dimmly. Almost like there was some kind of low current path with the base and input to the regulator completely removed. To be honest I feel a bit silly asking, but I've no idea why I couldn't get this regulator to drop it's work load.Any help appreciated. What I was trying to do was get the regulator regulating with out a heatsink, and the pass transistor doing most of the current driving the lamp. If I disconnected the regulator from the heatsink,, it heated up just short of thermal throttling itself. As I said I'm completely at a loss as to why it wouldn't current limit the LM7812. I've got some L7812L 100mA regulators, and some LM317L 100mA regulators I was hoping to use. But no chance of that if I can't keep a TO220 device cool.Thanks for reading.It was the circuit below with out R2.

author: piguy101
author_rank: Regular Contributor
timestamp: August 10, 2018, 11:54:32 pm
content: Quote from: sureshot on August 10, 2018, 11:38:08 pmWhich resistor do you mean by R2? Which resistor were you changing? The 10 Ω resistor limits the pass transistor base current and the 1 Ω resistor is what biases the transistor. A larger resistor than 1 Ω will mean that the pass transistor will be taking a larger share of the current instead of the regulator.The pass transistor turns when the base-emitter voltage is greater than about 0.6 V, so 600 mA going through the regulator will develop 0.6 V on the pass transistor base, meaning it will start to conduct then.

author: Andy Watson
author_rank: Super Contributor
timestamp: August 10, 2018, 11:58:27 pm
content: Quote from: sureshot on August 10, 2018, 11:38:08 pmWhich resistor were you changing ? Can you explain why you think changing this resistor will affect the regulator?QuoteHowever you configure this circuit, a linear controller must dissipate the excess energy - as heat. Can you explain how this circuit diverts heat energy from the regulator to the pass-transistor ? More specifically, how is the current "shared" between the regulator and the transistor ?

author: StillTrying
author_rank: Super Contributor
timestamp: August 11, 2018, 12:07:29 am
content: 750mA is the L78012's short circuit current, which is not enough to allow the transistor to switch on, you'll have to increase the 1R to 2.7R or 3.3R.The cold resistance of the 20W halogen will be very low ~0.5R.

author: sureshot
author_rank: Regular Contributor
timestamp: August 11, 2018, 12:56:39 am
content: I've built this circuit exactly as it is in the schematic, and it works fine. What I was trying to achieve is the regulator doing around 50mA of the work and the transistor doing the rest. I was only using a single resistor on the input to the base and regulator input. So not using the 10 ohm resistor, but trying different values of the R1 resistor. But nothing I tried limited the current through the regulator, I tried 10 ohm 0.5 watt 10 ohm 3 watt, 22 ohm 3 watt, 50 ohm 3 watt, 100 ohm 3 watt, and 150 ohm 3 watt, nothing changed with the 20 watt load. I'm not sure if I'm missing something. I usually just build these circuits straight on copper strip board and they work fine if I don't alter anything. This was an attempt to get there regulator to function with out a heatsink, and the transistor do most of the work. The transistor was conducting as it showed 1 volt across it under load, that was a 100 ohm 10 watt resistor I tried. But removing the regulator from the heatsink was not really achievable with the amount of current it was drawing. The voltage across the resistors I tried ranged from 0.5 volts to 1 volt across the resistors. I know it's unlikely to start conducting around or under 0.6 volts. Is it not possible to run the regulator with no heatsink, and limit the current it passes ?Thank you for your replys and help.

author: sureshot
author_rank: Regular Contributor
timestamp: August 11, 2018, 01:03:51 am
content: In a previous thread I was told I didn't need the ballast resistor, and a single resistor would limit the current the regulator is doing. Maybe I do need the 10 ohm base resistor, but need to increase the value of that 1 ohm ballast resistor as suggested above. But I don't no what value that ballast resistor needs to be if I want the regulator to do just say 50mA of the work load. Figure not set in stone, just a current limit by where the regulator doesn't need a heatsink.

author: StillTrying
author_rank: Super Contributor
timestamp: August 11, 2018, 01:30:51 am
content: Yes, it can't be done with that amount of load current.Assuming the 20W bulb lights up, it needs about 1.7A @ 12VAssuming the TR has a gain of 20, it needs 85mA of base current.As the base current has to be flowing though the Reg.+any current left flowing through the (1R to 100R) there's no way to get it down to just 50ma through the Reg. with that load.

author: sureshot
author_rank: Regular Contributor
timestamp: August 11, 2018, 02:34:07 am
content: So I see a picture clearer now, the varying load alters the parameters of the components. And there is no set value resistor to achieve a low content stable drive through the regulator. And a higher gain transistor looks more favourable for getting the regulators work load down. I was hoping to not use a heatsink on the L7812 over varying loads, but from the above I don't think it's possible. The closest it gets is with a high gain Darlington I suspect. I was using a TIP2955 for the above breadboard put together. It really was border line on the gain side of things.Thank you again for the help. If my maths was better maybe I could get it done on paper. I usually just measure voltages and current and work it out from there.

author: Jwillis
author_rank: Super Contributor
timestamp: August 12, 2018, 10:52:07 pm
content: Not sure where you got that schematic from.Just looking at it your limiting more current to the transistor than the regulator.Since a transistor is current controlled at the base it needs to be able to pass current before the regulator so it turns on.The 10 ohm resistor prevents that.Get rid of the base resistor.Here's a schematic of a higher load regulator at 12V but it shows what needs to be done.The 100 ohm resistor is limiting both the regulator and transistor base to 120 mA this means that the transistor/s will turn on but the regulator is limited to only 120mA .If you want only 50mA then you need a 240 ohm resistor instead. The 0.1 ohm resistors there are to control thermal runaway when using parallel transistors.If you also need to limit the base resistor but also want to limit the regulator.The base resistor must be before the regulator resistor.

author: David Hess
author_rank: Super Contributor
timestamp: August 13, 2018, 01:27:53 am
content: A more advanced and stable form of the PNP boosted linear regulator uses a power diode (1N4001, 1N4933, whatever) to compensate for the Vbe voltage drop of the transistor and then a pair of ballast resistors on the input which set the ratio between the transistor current and regulator current as shown below.

author: sureshot
author_rank: Regular Contributor
timestamp: August 13, 2018, 06:55:46 pm
content: Thank you for your replys, I've put together the three transistor circuit with both two and four transistors, that works well. Thank you for the explanations, it's starting to sink in a bit more now. I always think of a series resistor and a lamp or motor scenario. Find it a bit more complicated in a chain like the emitter follower configuration, especially when multiple components parameters have to be factored into the equation as well.Thank you again.