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I am currently reverse engineering a circuit which requires controlling of a magnetic field. For that, the circuit has a pair of D882 and B772 each. The PCB traces suggest that the transistors are arranged as shown in the picture below: Transistor arrangement This arrangement does not make any sense at all for me. Wouldn't applying a voltage to any of the control signals result in current through both transistors rather than through the coils?

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    \$\begingroup\$ reverse clamp zener or diodes are needed to protect Vbe reverse voltage on each \$\endgroup\$ – Sunnyskyguy EE75 Sep 30 '18 at 15:27
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    \$\begingroup\$ One is NPN, one is PNP. Both are emitter followers. This is a class B amplifier (times 2) with no correction for the crossover point. \$\endgroup\$ – immibis Oct 1 '18 at 0:41
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That is called an "H-Bridge."

It is often used to drive motors forwards as well as backwards.

In your case, it allows you to generate a magnetic field whose polarity and intensity you can vary using "control signal 1" and "control signal 2."

When both are high (or both are low,) no current flows through the coil.

If one is high and the other is low, then current will flow in a particular direction.

If you swap the high and lows, it will flow in the opposite direction.

Now, if you hold one steady and pulse the other you will get a pulsed current through the coil. It will be smoothed (somewhat) by the coil to a steady magnetic field whose strength is propotional to the duty cycle of the pulses.

Switching the polarity of the current also changes the polarity of the magnetic field.


That is very much a simplified description, but I think it contains enough key words that you should be able to locate more details on your own.

It is a common circuit with many uses - and plenty of tricks and traps that go into making, using, and controlling it.


A bit more on how it operates:

The key to the whole thing is how pnp and npn transistors function.

When the voltage on the base of an npn transistor is more than 0.7 volts above the voltage on the emitter, then current will flow through the collector to the emitter.

When the voltage on the base of a pnp transistor is more than 0.7 volts below the voltage on the collector, then current will flow through the collector to the emitter.

So, looking at the H-bridge, putting a high signal on one of the control signals will turn off the pnp and turn on the npn - that side of the bridge is connected to the positive supply voltage.

Now, if you put a low signal on the other control line the npn transistor will turn off and the pnp will turn on. That side of the bridge is connected to the ground.

Current can now flow from V+ on one side of the bridge, through the coils, to ground on the other side of the bridge.

So, which control signal is high and which is low dictates the direction of the current flow through the load in the middle of the bridge.


You also asked it is possible for both transistors on one side to turn on and cause a short circuit.

It can happen, and is called shoot through. Part of the design and operation of an H-bridge goes into making sure that it doesn't happen.

In the design you've posted, I don't think it can happen.

It looks to me like the transistors on each side can never be on at the same time. But, I'm not an engineer and may well have overseen something (though Tony is an engineer and doesn't think it can happen with this circuit.)

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    \$\begingroup\$ It might be helpful to mention that NPN transistors are on when base is higher than emitter, while PNP transistors are on when base is lower than emitter. Thus high control voltage switches on the NPN and low control voltage switches on the PNP. \$\endgroup\$ – jpa Sep 30 '18 at 14:16
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NO

The Vbe has a dead-zone for drive levels of < |+/-0.7V| however, back EMF during the time of load L/R=T(63%V) will occur where R is the DC resistance of coils.(DCR)

beware of the need to clamp inductive spikes to opposite rail with zener+diode pairs across motor or reverse Vce diodes across each transistor. In more advanced designs they use active clamps. Beware of reactive energy and current loop area in the layout. Keep it tight pairs from driver, power, ground to L to minimize CM noise.

However when commutating left right for forward and reverse. You must stop by having both top or bottom drivers high ( or low) to shunt the L/R =T time constant with another brake dead-time before direction reversal. This is done by your smart controller using Sig1=Sig2 = either 0 or 1. If this is not a motor, disregard.

When regulating current if the left side is high, right side is used for PWM average voltage to control surge current or velocity at steady state. THen when reversing load polarity, the opposite is done. Right side high and left side with ramped PWM towards full Vavg in the opposite polarity. If this is a motor, then the same is true for deaccelerating. Often a current shunt is used for current sensing, where the load inertia affects current during g time duration.

Also keep in mind these simple transistor switches have an hFE of about 10~5% of max hFE during saturation so input current and heat dissipation ought to be computed. while control signal ought to be above +12V or additional drop occurs due to Vbe. This is why MOSFETs are preferred but have shoot thru issues just like if these were open-collectors instead of emitter followers. THen the 2 inputs must be separated to 4 inputs with controlled dead-time.

This is the simplest bridge driver, but compromises Vdrop on each switch but ok for small bridges at 12V. Even though it may work at 5V, not recommended for poor efficiency.

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On each side you have an NPN and a PNP transistor. If the control voltage levels are selected correctly, the NPN and PNP transistor will not be switched on at the same time.

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    \$\begingroup\$ if dV/dt is sufficiently high and there is trace inductance, it is possible during transitions in poor layouts for both to be on. But abnormal. Worst case is Vbe reverse is exceeded \$\endgroup\$ – Sunnyskyguy EE75 Sep 30 '18 at 15:25
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Is a PWM signal on control or an analog design from OPAmp? This circuit resemble an analog Bridge class B Booster. An equivalent complementary H PWM generally need each transistor to be driven separately and to saturation, this one is forever in linear zone, VCE never can reach saturation. On PWM H bridges Common Emitter is preferred to common collector; it is more simple to saturate each bridge transistor without additional supply voltages. Common Collector has disadvantage of propagating BEMF to BASE driving, this can destroy driver.

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    \$\begingroup\$ I'm having some trouble parsing your sentences. Do you think you could organize your answer a little more, and perhaps add some detail? \$\endgroup\$ – Elliot Alderson Sep 30 '18 at 20:21
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Some of the previous answers make correct statements, but no single answer satisfactorily answers the question.

@JRE is correct that we call this circuit topology an H-bridge, that it is commonly used to control motors, and how you would set the control lines to operate a motor.

@TonyEErocketscientist is correct that you need something to dissipate the current when the inductive load is turned off. His suggestion of back-to-back zener diodes, in parallel with the load, is the best solution. If the current is small, you could also get away with a non-polarized capacitor.

In a comment, @immibis correctly states that each individual transistor is connected in an emitter follower. In other words, the output is connected to the emitter of the transistor, rather than the collector. The output follows the voltage of the input, within a diode voltage drop.

The transistors in emitter followers DO stay on, except when the input voltage is close to the supply rails. Because of this, emitter followers are notorious for wasting power and needing heat sinks. The heart of a linear voltage regulator is an emitter follower, and these regulators are notorious for being inefficient and requiring heat sinks. Emitter-coupled logic (such as was used in Cray supercomputers) uses emitter followers to switch digital signals. Heat production in the Cray was so bad that the refrigeration unit was larger than the electronics! And the third example of emitter followers is a...

Class B amplifier, which @RRomano010 points out. They are made by two emitter followers, with an NPN transistor pulling to the high rail and a PNP transistor pulling to the low rail. That's what we have here. They are commonly used as the output stage of audio amplifiers to drive loudspeakers, are inefficient, and require ample heat sinking.

If you absolutely must drive your inductive load with an analog signal (i.e. PWM is not acceptable), then the circuit presented in the question is a okay design will barely work (although I would add @TonyEErocketscientist's protection diodes). You will get some crossover distortion because of the diode voltage offsets; these can be compensated the same way it's done in a class AB amplifier.

If you are driving your load on/off or with PWM, then it is an inefficient design. The usual way to make an H bridge is with PNP transistors pulling to the high rail and NPN transistors pulling to the low rail. In other words, swap the NPN transistors in this circuit with PNP, and vice versa. However, you will then need resistors on each transistor base. Perhaps the designer of this circuit was trying to avoid the extra components -- that would also explain the lack of protection diodes. Make sure you put those protection diodes in, too.

Or you could just use an H bridge chip, where someone else has taken care of these problems for you.

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    \$\begingroup\$ That particular arrangement of PNP and NPN makes it pretty much impossible for shoot through to happen. I expect that is the reason for the "upside down" setup. When "off" (both control signals at the same level,) the coils will be shorted through the transistors - this takes care of the inductive spike that would otherwise occur when switching off the current through a coil. This circuit is probably not worth spit for driving the coils with an analog signal. \$\endgroup\$ – JRE Oct 1 '18 at 17:23
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    \$\begingroup\$ @JRE: You're being too kind to the designer of the circuit. Either (1) didn't know what they were doing, or (2) wanted to use as few components as possible. As far as killing the inductive current during an off condition, that will fail if the circuit suddenly loses power. I'd rather include back-to-back zeners. \$\endgroup\$ – DrSheldon Oct 2 '18 at 4:09
  • \$\begingroup\$ @DrSheldon Please remove the comments from the answer, answers should not be a reply or for conversation but answer only the question. \$\endgroup\$ – laptop2d Oct 2 '18 at 4:58
  • \$\begingroup\$ @laptop2d: I don't understand what you're saying. Can you give an example of what you would like removed? \$\endgroup\$ – DrSheldon Oct 2 '18 at 5:07

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