H bridge Troubleshooting - Page 1
Forum section: electronics / beginners

author: SilverHawk
author_rank: Contributor
timestamp: April 19, 2017, 04:43:42 am
content: Dear Experts,I need some help in building H-bridge circuit.In A = B = 5V and Power in 12V.Motor A and Motor B are two opposite terminals of the same motor.When we tested the circuit with only IN A (with IN B ignored as we only want the motor to turn in one direction), we could not get it to run. Hence, we did some preliminary troubleshooting.The set up is the same as in the schematic.We are expecting Q7 to be ON as in the simulation [Capture.PNG]. However, it is OFF in our hardware and subsequently the rest of the circuit does not work as intended.A surprising thing to note is that if Power IN is switched off momentarily with IN A maintained at 5V, the transistor Q7 works as expected, we get a short between collector and emitter leg.Any suggestions will be much appreciated!Thanks!

author: Zero999
author_rank: Super Contributor
timestamp: April 19, 2017, 05:19:38 pm
content: Have you tried using a resistor as a dummy load?How much current does it have to drive?What about using an H-bridge IC such as the L293D, SN754410, DRV8829 etc.?

author: SilverHawk
author_rank: Contributor
timestamp: April 20, 2017, 04:12:17 am
content: Haven't tried the resistor,it's part of our project to build the H bridge

author: nuno
author_rank: Frequent Contributor
timestamp: April 20, 2017, 03:16:51 pm
content: Check that you have GND from the 5V driving circuit and GND from the 12V circuit connected together.You can also pick a DMM and measure the several circuit nodes and compare to the sim, see where they disagree will help you nail it.

author: Ian.M
author_rank: Super Contributor
timestamp: April 20, 2017, 03:25:04 pm
content: No clamping diodes for the motor back-EMF means that if you ever do get it working, one of the output Darlingtons will soon fail and then it will go *BANG* as soon as the other Darlington in that leg turns on.

author: Zero999
author_rank: Super Contributor
timestamp: April 20, 2017, 04:58:27 pm
content: Quote from: Ian.M on April 20, 2017, 03:25:04 pmThere are clamping diodes but they aren't shown on the schematic. The PDF attached to the first post, shows the part numbers for the transistors: TIP120 & TIP125 which have built-in clamping diodes, along with some pull-down resistors to speed up the turn-off time.https://www.onsemi.com/pub/Collateral/TIP120-D.PDFAnother thing: it isn't the turn on which causes the back-EMF pulse but the turn off, when the current attempts to carry on flowing and generates a high voltage.Anyway, I'd probably use MOSFETs for such a low voltage application and consider a high side N-channel drive but it depends on what level the original poster is studying and it's best to get the basics right first.

author: Ian.M
author_rank: Super Contributor
timestamp: April 20, 2017, 05:23:42 pm
content: Ok. so they've got internal diodes and wont die immediately. However the datasheet line:Unclamped Inductive Load Energy (Note 1) E 50 mJ (max)is worrying - it doesn't sound like they are very robust clamping diodes, and in a motor driving application, with the possibility of overrrun, it would probably be a good idea to add beefy external Schottkys.Yes, its the turn-off that produces the back-EMF, and kills transistors but its when the other transistor in the same leg to the just failed one turns on that it goes *BANG* as it attempts to crowbar the DC bus.

author: Zero999
author_rank: Super Contributor
timestamp: April 20, 2017, 05:47:14 pm
content: Quote from: Ian.M on April 20, 2017, 05:23:42 pmIt makes sense to me. The unclamped inductive load energy, applies to a single transistor driving a solenoid or relay, when the internal diode would be useless anyway. The fact that the internal diode's current rating is not specified is more of a concern.The diodes will not conduct when the motor is over-running, unless the back EMF generated by the motor is greater than the power supply voltage (this can't happen due to inertia alone, unless the external power is removed) or you're doing dynamic breaking, which is a possibility.

author: mfratus2001
author_rank: Contributor
timestamp: April 20, 2017, 07:01:17 pm
content: It is the voltage from emitter to base that determines whether it is going to be on or off. Actually, it is the current flowing into or out of the base-emitter junction, but a simple measurement of voltage will tell you if it should be on.Also, what is this supposed to do? Where did you get the design? It really would not do anything (if it worked) except run the motor in one direction. If that is all you want to do, you sure don't need an H-bridge for that.The way it is designed, until you have it debugged it will keep threatening to blow transistors from being shorted. I would start off without the output transistors in place, or at least a larger resistor, say, 100 Ohms, in place of the motor. There are lots of reasons why it is difficult to go from simulation to actual circuit, one of which is errors in wiring, wrong parts values, and the fact that models and reality are different.You said, "if I turn on the 12V without the 5V, it works..." but does it? are you saying that because the motor runs? Because a two shorted transistors can make the motor run.Go a step at a time.

author: Zero999
author_rank: Super Contributor
timestamp: April 20, 2017, 07:28:34 pm
content: I think the idea is to make an h-bridge which will drive both forward and backwards. At the moment, it's only hooked up to run in one direction only for test purposes, not for the final design.

author: Audioguru
author_rank: Super Contributor
timestamp: April 20, 2017, 08:27:07 pm
content: The circuit design is wrong.The collector current of Q7 is only 55mA because its base current is only 4mA. For Q7 to saturate with the very low values for R7 and R8 its collector current must be at least 160mA and its base current must be at least 16mA.The output current of the darlingtons must be only 1.5A so their base current must be only 6mA so increase the values of R7 and R8 to about 1k ohms.

author: Zero999
author_rank: Super Contributor
timestamp: April 20, 2017, 10:44:30 pm
content: Quote from: Audioguru on April 20, 2017, 08:27:07 pmNo, it isn't Q7 which is being underdriven here but Q6 and Q8. The values of R2 & R5 need to be decreased to 330R to give the Darlington pairs about 13mA of base drive. R1 and R4 aren't needed as R7 & R8 limit the base current due to emitter degradation.

author: Audioguru
author_rank: Super Contributor
timestamp: April 21, 2017, 12:17:32 am
content: Q6 and Q8 do not turn on any darlingtons, instead they turn off the lower NPN darlingtons. Then Q6 and Q8 can simply be resistors to ground and 4 base resistors can be removed.When the left input turns on Q7 then its collector turns on darlington Q1 and its emitter turns on darlington Q4.When the right input turns on Q5 then its collector turns on darlington Q2 and its emitter turns on darlington Q3.Resistors are missing to turn off darlingtons Q1 and Q2.

author: nuno
author_rank: Frequent Contributor
timestamp: April 21, 2017, 09:39:41 am
content: The Darlingtons have internal base-emitter resistors (as Hero already pointed out).

author: Zero999
author_rank: Super Contributor
timestamp: April 21, 2017, 10:28:14 am
content: Quote from: Audioguru on April 21, 2017, 12:17:32 amI was incorrect earlier. Q6 and Q8 are there to prevent both Q1 & Q3 or Q2 & Q4 from turning on at the same time and short circuiting the power supply.Quote from: Audioguru on April 20, 2017, 08:27:07 pmNo the design is fine. It does have its downsides: mainly high saturation voltage inherent with Darlington pairs and slow switching speed, but it will work and is no worse tha and IC such as the L293D.QuoteNo, Q7 does not need to saturate, just pass enough current to switch on Q4 & Q1 and 55mA of collector current is more than enough to achieve this. The low value of R7 is necessary to provide sufficient base drive to the Darlington output stage.QuoteI don't recall a requirement for limiting the output current to 1.5A. Anyway, restricting the base current is a very unreliable of limiting the output current. The minimum current gain is specified to be 1000, but it's typically 2500 and could me much higher than that, especially when the transistors have heated up, due to the short circuit current.The simplest way to add current limiting is to add another transistor current source/sink to the circuit. Unfortunately it will increase the voltage loss but that can be minimised by selecting a current limit of double the nominal current.